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a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
1, a,= (x+2)^2/3.(x+2) = x+2/3
b, = 3x.(x+4)/2x.(x+4) = 3/2
k mk nha
a) Để rút gọn biểu thức (x+2)(x^2+4x+4)-(x-2)(x^2-4x-4)-12x^2-x, ta thực hiện các bước sau:
(x+2)(x^2+4x+4) = x(x^2+4x+4) + 2(x^2+4x+4)
= x^3 + 4x^2 + 4x + 2x^2 + 8x + 8
= x^3 + 6x^2 + 12x + 8
(x-2)(x^2-4x-4) = x(x^2-4x-4) - 2(x^2-4x-4)
= x^3 - 4x^2 - 4x - 2x^2 + 8x + 8
= x^3 - 6x^2 + 4x + 8
Thay vào biểu thức ban đầu, ta có:
(x+2)(x^2+4x+4)-(x-2)(x^2-4x-4)-12x^2-x
= (x^3 + 6x^2 + 12x + 8 - (x^3 - 6x^2 + 4x - 12x^2 - x
= x^3 + 6x^2 + 12x + 8 - x^3 + 6x^2 - 4x - 8 - 12x^2 - x
= 8x + 8 - 4x - 8
= 4x
Vậy biểu thức đã được rút gọn thành 4x.
b) Để rút gọn biểu thức (x-2)(x+2)(x+3)-(x+1)(x^2-x+1), ta thực hiện các bước sau:
(x-2)(x+2) = x^2 - 2^2 = x^2 - 4
Thay vào biểu thức ban đầu, ta có:
(x-2)(x+2)(x+3)-(x+1)(x^2-x+1)
= (x^2 - 4)(x+3) - (x+1)(x^2-x+1)
= x^3 + 3x^2 - 4x - 12 - (x^3 + x^2 - x + x^2 - x + 1)
= x^3 + 3x^2 - 4x - 12 - x^3 - x^2 + x - x^2 + x - 1
= x^3 - x^3 + 3x^2 - x^2 - x^2 + 3x - 4x + x - 12 - 1
= 2x^2 - x - 13
Vậy biểu thức đã được rút gọn thành 2x^2 - x - 13.
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
=-10x-50
4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16\)
=-6x+25
5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)
\(=y^2-6y+9-y^2+6y-9\)
=0
6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
=12x+18
b) Ta có: \(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1\)
\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)
c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)
\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)
\(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1=\left(x+1\right)^2+\left(y-2\right)^2+1\)
\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2
\(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)
dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)
\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)
=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2-1-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2\left(2-x\right)-1\ge-1\)
dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Bài 1 yêu cầu gì em?
Bài 2:
\(a,x\left(x-1\right)+5\left(x-1\right)=\left(x+5\right)\left(x-1\right)\\ b,3x\left(x+1\right)+3\left(x+1\right)=\left(3x+3\right)\left(x+1\right)=3\left(x+1\right)\left(x+1\right)=3\left(x+1\right)^2\\ c,x\left(x-3\right)+xy\left(x-3\right)=\left(x+xy\right)\left(x-3\right)=x\left(y+1\right)\left(x-3\right)\\ d,2x\left(x-2\right)-6\left(x-2\right)=\left(2x-6\right)\left(x-2\right)=2\left(x-3\right)\left(x-2\right)\)
Bài 1:
a) \(3xy+6y\)
\(=3y\left(x+2\right)\)
b) \(3x^2+9x\)
\(=3x\left(x+3\right)\)
c) \(6x-9y^2\)
\(=3\left(2x-3y^2\right)\)
d) \(10xy^2-6x^2y\)
\(=2xy\left(5y-3x\right)\)
Bài 2:
a) \(x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x+5\right)\)
b) \(3x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(3x+3\right)\)
\(=3\left(x+1\right)\left(x+1\right)\)
\(=3\left(x+1\right)^2\)
c) \(x\left(x-3\right)+xy\left(x-3\right)\)
\(=\left(x+xy\right)\left(x-3\right)\)
\(=x\left(1+y\right)\left(x-3\right)\)
d) \(2x\left(x-2\right)-6\left(x-2\right)\)
\(=\left(2x-6\right)\left(x-2\right)\)
\(=2\left(x-3\right)\left(x-2\right)\)
c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)
\(=x^2+x-10\)
Bài 2:
a) \(=x^2-36y^2\)
b) \(=x^3-8\)
Bài 3:
a) \(=x^2+2x+1-x^2+2x-1-3x^2+3=-3x^2+4x+3\)
b) \(=6\left(x-1\right)\left(x+1\right)=6x^2-6\)