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31 tháng 7 2023

p) \(\left(9-x\right)\left(x^2+2x-3\right)\)

\(=9\left(x^2+2x-3\right)-x\left(x^2+2x-3\right)\)

\(=9x^2+18x-27-x^3-2x^2+3x\)

\(=-x^3+7x^2+21x-27\)

n) \(\left(-x+3\right)\left(x^2+x+1\right)\)

\(=-x\left(x^2+x+1\right)+3\left(x^2+x+1\right)\)

\(=-x^3-x^2-x+3x^2+3x+3\)

\(=-x^2+2x^2+2x+3\)

o) \(\left(-6x+\dfrac{1}{2}\right)\left(x^2-4x+2\right)\)

\(=-6x\left(x^2-4x+2\right)+\dfrac{1}{2}\left(x^2-4x+2\right)\)

\(=-6x^3+24x^2-12x+\dfrac{1}{2}x^2-2x+1\)

\(=-6x^3+\dfrac{49}{2}x^2-14x+1\)

q) \(\left(6x+1\right)\left(x^2-2x-3\right)\)

\(=6x\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)

\(=6x^3-12x^2-18x+x^2-2x-3\)

\(=6x^3-11x^2-20x-3\)

r) \(\left(2x+1\right)\left(-x^2-3x+1\right)\)

\(=2x\left(-x^2-3x+1\right)+\left(-x^2-3x+1\right)\)

\(=-2x^3-6x^2+2x-x^2-3x+1\)

\(=-2x^3-7x^2-x+1\)

u) \(\left(2x-3\right)\left(-x^2+x+6\right)\)

\(=2x\left(-x^2+x+6\right)-3\left(-x^2+x+6\right)\)

\(=-2x^3+2x^2+12x+3x^2-3x-18\)

\(=-2x^3+5x^2+9x-18\)

s) \(\left(-4x+5\right)\left(x^2+3x-2\right)\)

\(=-4x\left(x^2+3x-2\right)+5\left(x^2+3x-2\right)\)

\(=-4x^3-12x^2+8x+5x^2+15x-10\)

\(=-4x^3-7x^2+23x-10\)

v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4x^3\right)\)

\(=-\dfrac{1}{2}x\left(2x+6-4x^3\right)+3\left(2x+6-4x^3\right)\)

\(=-x^2-3+2x^4+6x+18-12x^3\)

\(=2x^4-12x^3-x^2+6x+15\)

p: (-x+9)(x^2+2x-3)

=-x^3-2x^2+3x+9x^2+18x-27

=-x^3+7x^2+21x-27

n: (-x+3)(x^2+x+1)

=-x^3-x^2-x+3x^2+3x+3

=-x^3+2x^2+2x+3

o: (-6x+1/2)(x^2-4x+2)

=-6x^3+24x^2-12x+1/2x^2-2x+1

=-64x^3+49/2x^2-14x+1

q: (6x+1)(x^2-2x-3)

=6x^3-12x^2-18x+x^2-2x-3

=6x^3-11x^2-20x-3

r: (2x+1)(-x^2-3x+1)

=-2x^3-6x^2+2x-x^2-3x+1

=-2x^3-7x^2-x+1

u: =-2x^3+2x^2+12x+3x^2-3x-18

=-2x^3+5x^2+9x-18

s: =-4x^3-12x^2+8x+5x^2+15x-10

=-4x^3-7x^2+23x-10

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

31 tháng 7 2023

v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4c^3\right)\)

\(=-\dfrac{1}{2}\left(2x+6-4c^3\right)+3\left(2x+6-4c^3\right)\)

\(=-x^2-3x+2c^3x+6x+18-12c^3\)

\(=-x^2+3x+2c^3x+18-12c^3\)

f) \(\left(2x-5\right)\left(x^2-x+3\right)\)

\(=2x\left(x^2-x+3\right)-5\left(x^2-x+3\right)\)

\(=2x^3-2x^2+6x-5x^2+5x-15\)

\(=2x^3-7x^2+11x-15\)

w) \(\left(3x+1\right)\left(x^2-2x-5\right)\)

\(=3x\left(x^2-2x-5\right)+\left(x^2-2x-5\right)\)

\(=3x^3-6x^2-15x+x^2-2x-5\)

\(=3x^3-5x^2-17x-5\)

x) \(\left(6x-3\right)\left(x^2+x-1\right)\)

\(=6x\left(x^2+x-1\right)-3\left(x^2+x-1\right)\)

\(=6x^3+6x^2-6x-3x^2-3x+3\)

\(=6x^3+3x^2-9x+3\)

y) \(\left(5x-2\right)\left(3x+1-x^2\right)\)

\(=5x\left(3x+1-x^2\right)-2\left(3x+1-x^2\right)\)

\(=15x^2+5x-5x^3-6x-2+2x^2\)

\(=-5x^3+17x^2-x-2\)

z) \(\left(\dfrac{3}{4}x+1\right)\left(4x^2+4x+4\right)\)

\(=\dfrac{3}{4}x\left(4x^2+4x+4\right)+\left(4x^2+4x+4\right)\)

\(=3x^3+3x^2+3x+4x^2+4x+4\)

\(=3x^3+7x^2+7x+4\)

f: =2x^3-2x^2+6x-5x^2+5x-15

=2x^3-7x^2+11x-15

w: =3x^3-6x^2-15x+x^2-2x-5

=3x^3-5x^2-17x-5

x: =6x^3+6x^2-6x-3x^2-3x+3

=6x^3+3x^2-9x+3

y: =(5x-2)(-x^2+3x+1)

=-5x^3+15x^2+5x+2x^2-6x-2

=-5x^3+17x^2-x-2

z: =3x^3+3x^2+3x+4x^2+4x+4

=3x^3+7x^2+7x+4

30 tháng 6 2021

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

30 tháng 6 2021

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x2-8x+6x-4x2=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

21 tháng 7 2021

a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28

b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10

c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x

d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1

1 tháng 4 2020

e, 3x(2-x) =15(x-2)

\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy..

f, (x+5)(x+4)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)

Vậy..

g, x(x+4)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

,h, (2x -4)(x-2)=0

\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

i, (x+1/5)(2x-3)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)

k, x²-4x=0

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

m, 4x²-1=0

\(\Leftrightarrow\left(2x\right)^2-1^2=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)

n, x²-6x+9=0

\(\Leftrightarrow x^2-2.x.3+3^2=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)

<=> x=3

l, (3x-5)²-(x+4)²=0

\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)

\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy ..

o, 7x(x+2)-5(x+2)=0

\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)

Vậy....

p, 3x(2x-5)-4x+10=0

\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)

\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy...

q, (2-2x)-x²+1=0

\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)

\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

Vậy ....

r, x(1-3x)=5(1-3x)

\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)

s, 2x-3/4+x+1/6=3

\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)

1 tháng 4 2020

r, x(1-3x)=5(1-3x)

➜x(1-3x)-5(1-3x)=0

➜(x-5)(1-3x)=0

\(\left[{}\begin{matrix}x-5=0\\1-3x=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)

Mk lười lắm mai nha!!!~~~~~~~~~~~~

5 tháng 7 2018

1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)

\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)

\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)

\(=27x^3-4x^2+20x-1\)

b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)

\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)

\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)

\(=13x-28x^2-21-x^3\)

c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)

\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)

\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)

\(=16x^2-17+x^3\)

d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)

\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)

\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)

\(=-27x^2+63x-46\)

e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)

\(=12x^2-24x-6x^2-10x-4x^2\)

\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)

\(=2x^2-34x\)

f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)

\(=30x^2-25x-36x+30-3x^2-10x\)

\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)

\(=27x^2-71x+30\)

5 tháng 7 2018

2) a)\(x\left(x+3\right)-x^2=6\)

\(\Rightarrow x^2+3x-x^2=6\)

\(\Rightarrow\left(x^2-x^2\right)+3x=6\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

Vậy x=2

b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)

\(\Rightarrow2x^2-10x-2x^2-x=6\)

\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)

\(\Rightarrow-11x=6\)

\(\Rightarrow x=-\dfrac{6}{11}\)

\(\)Vậy \(x=-\dfrac{6}{11}\)

c) x(x+5)-(x+1)(x-2)=7

\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)

\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)

\(\Rightarrow6x=5\)

\(\Rightarrow x=\dfrac{5}{6}\)

Vậy x=\(\dfrac{5}{6}\)

d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)

\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)

\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)

\(\Rightarrow10x-10=10\)

\(\Rightarrow10x=20\)

\(\Rightarrow x=2\)

Vậy x=2

14 tháng 9 2021

1) \(\dfrac{3x}{4x-8}\)

\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)

2) \(\dfrac{2x}{x^2-9}\)

\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

3) \(\dfrac{6}{x^3+1}=\dfrac{6}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)

(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))

4) \(\dfrac{6x^2}{x^2-2x+1}=\dfrac{6x^2}{\left(x-1\right)^2}\)

\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)

5) \(\dfrac{x-2}{x^2+3}\)

Do \(x^2+3>0\forall x\in R\)

Vậy biểu thức trên xác định với mọi x

6) \(\dfrac{2x}{x^2+3x+2}=\dfrac{2x}{\left(x+1\right)\left(x+2\right)}\)

\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)

1: \(=x^4-8x^3+24x^2-32x+16+x^4-12x^3+54x^2-108x+81-1\)

\(=2x^4-20x^3+78x^2-140x+96\)

\(=2\left(x-3\right)\left(x-2\right)\left(x^2-5x+8\right)\)

2: \(=x^4-4x^3+6x^2-4x+1+x^4+12x^3+54x^2+108x+81-512\)

\(=2x^4+8x^3+60x^2+104x-430\)

\(=2\left(x^4+4x^3+30x^2+52x-215\right)\)

3 tháng 8 2020

Bài 1:

a) \(4x\left(3x-1\right)-2\left(3x+1\right)-\left(x+3\right)\)

\(=12x^2-4x-6x-2-x-3\)

\(=12x^2-11x-5\)

b) \(=\left(-2x^2-1xy+2y^2\right)\left(-1x^2y\right)\)

\(=\left[\left(-1x^2y\right)\left(-2x^2\right)\right]-\left[\left(-1x^2y\right).1xy\right]+\left[\left(-1x^2y\right).2y^2\right]\)

\(=\left(2x^4y\right)-\left(-1x^3y^2\right)+\left(-2x^2y^3\right)\)

\(=2x^4y+1x^3y^2-2x^2y^3\)

c) \(4x\left(3x^2-x\right)-\left(2x+3\right)^2\left(6x^2-3x+1\right)\)

\(=\left(4x.3x^2\right)-\left(4x.x\right)-\left[\left(2x\right)^2+2.2x.3+3^2\right]\left(6x^2-3x+1\right)\)

\(=12x^3-4x^2-\left(4x^2+12x+9\right)\left(6x^2-3x+1\right)\)

\(=12x^3-4x^2-\left[4x^2\left(6x^2-3x+1\right)+12x\left(6x^2-3x+1\right)+9\left(6x^2-3x+1\right)\right]\)

\(=12x^3-4x^2-\left[\left(24x^4-12x^3+4x^2\right)+\left(72x^3-36x^2+12x\right)+\left(36x^2-27x+9\right)\right]\)

\(=12x^3-4x^2-24x^4+12x^3-4x^2-72x^3+36x^2-12x-36x^2+27x-9\)

\(=-48x^3-8x^2-24x^4+15x-9\)

3 tháng 8 2020

Bài 2 ạ