Thực hiện phép tính:
a,(x^2-2x+3)(1/2x-5)
b,(x^3-2x^2+x-1)(5-x)
c,(x^3-x^2-7x+3):(x-3)
d,(x^4-x^3+x^2+3x):(x^2-2x+3)
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AH
Akai Haruma
Giáo viên
9 tháng 9 2021
Lời giải:
a.
$(2x-3)^2+(2x+3)(5-2x)=(4x^2-12x+9)-(-4x^2+4x+15)$
$=4x^2-12x+9+4x^2-4x-15$
$=24-8x$
b.
$3(2x-3)+5(x+2)=6x-9+5x+10=11x+1$
c.
$3x(2x-8)+(6x-2)(5-x)=(6x^2-24x)+(-6x^2+32x-10)$
$=6x^2-24x-6x^2-32x+10$
$=8x-10$
d.
$(x-3)(x+3)-(x-5)^2=(x^2-9)-(x^2-10x+25)$
$=x^2-9-x^2+10x-25=10x-34$
e.
$(x-y)^3-(x-y)(x^2+xy+y^2)=(x^3-3x^2y+3xy^2-y^3)-(x^3-y^3)$
$=-3x^2y+3xy^2=3xy(y-x)$
9 tháng 9 2021
a: ta có: \(\left(2x-3\right)^2+\left(2x+3\right)\left(5-2x\right)\)
\(=4x^2-12x+9+2x-4x^2+15-6x\)
\(=-16x+24\)
b: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)
\(=6x-9+5x+10\)
\(=11x+1\)
c: ta có: \(3x\left(2x-8\right)+\left(6x-2\right)\left(5-x\right)\)
\(=6x^2-24x+30x-6x^2-10+2x\)
\(=8x-10\)
18 tháng 12 2021
Bài 1:
\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)
Bài 2:
\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)
13 tháng 12 2021
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
24 tháng 2 2023
a: \(=\dfrac{2x^4+x^3-5x^2-3x-3}{x^2-3}\)
\(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)
\(=2x^2+x+1\)
b: \(=\dfrac{x^5+x^2+x^3+1}{x^3+1}=x^2+1\)
c: \(=\dfrac{2x^3-x^2-x+6x^2-3x-3+2x+6}{2x^2-x-1}\)
\(=x+3+\dfrac{2x+6}{2x^2-x-1}\)
d: \(=\dfrac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)
\(=x^2-2x-5\)
8 tháng 3 2023
a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10
=1/2x^5-3x^4-5x^3
b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x
=-15x^5+12x^4-9x^3+9x^2
c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)
d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
14 tháng 5 2022
a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
7 tháng 12 2021
\(a,\left(2x-5\right)\left(5-x\right)=5\left(2x-5\right)-x\left(2x-5\right)=10x-25-2x^2+5x=15x-2x^2-25\\ b,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}=\dfrac{3x+2-3x+2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{4}{\left(3x-2\right)\left(3x+2\right)}\)
\(c,\dfrac{3}{x-3}-\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9-6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
a: \(=x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15=x^3-6x^2+\dfrac{23}{2}x-15\)
b: \(=5x^3-x^4-10x^2+2x^3+5x-x^2-5+x\)
\(=-x^4+7x^3-11x^2+6x-5\)
c: \(=\dfrac{x^3-3x^2+2x^2-6x-x+3}{x-3}=x^2+2x-1\)