Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1
a) (6x4y2 - 3x3y3) : 3x3y2 = 6x4y2 : 3x3y2 - 3x3y3 : 3x3y2 = 2x - y
b) (2x - 1)(x2 - x + 3) = 2x3 - 2x2 + 6x - x2 + x - 3 = 2x3 - 3x2 + 7x - 3
Bài 2
1) (x - 2)2 - (x - 3)2 = (x - 2 - x + 3)(x - 2 + x - 3) = 2x - 5>
2) 4x2 - 4xy + 2y2 + 1 = (4x2 - 4xy + y2) + y2 + 1 = (2x - y)2 + y2 + 1 > 0
vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\y^2\ge0\end{cases}}\)
a) x2 - 5x - y2 -5y
= ( x2 - y2 ) + ( -5x - 5y)
= ( x - y ) ( x + y) - 5( x + y )
= ( x + y ) ( x - y -5)
b) x3 + 2x2 - 4x - 8
= x2 ( x + 2 ) - 4 ( x + 2 )
= ( x +2 ) ( x2 -4 )
= ( x+2)2 ( x-2)
Bai 2 :
a, \(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(=x^2+6x+9+x^2-4x+4-2\left(x^2-2x+3x-6\right)\)
\(=2x^2+2x+13-2x^2-2x+12=25\)
b, \(B=\left(x-2\right)^2-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)
\(=x^2-4x+4-x\left(x^2-3x-x+3\right)+3x^2-9x+8\)
\(=4x^2-13x+12-x^3+4x^2-3x=-16x+12-x^3\)
$A=x^2+2xy+y^2-4x-4y+1$
$=(x+y)^2-4(x+y)+1$
Thay $x+y=3$:
$A=3^2-4\cdot3+1$
$=9-12+1$
$=-2$
$A=x^2-3x+5$
$=\left(x-\dfrac32\right)^2-\dfrac94+5$
$=\left(x-\dfrac32\right)^2+\dfrac{11}{4}$
Vì $\left(x-\dfrac32\right)^2\ge0$ nên:
$A_{\min}=\dfrac{11}{4}$
Dấu "=" khi $x=\dfrac32$.
bài 1 : a, sai đề . đề đúng là x3 + x2 + 22 = x3 + 2x2- x2+ 4
= x2 ( x + 2 ) - ( x - 2 ) ( x + 2 ) = ( x + 2 ) ( x2 - x + 2 )
b, xy -y2 -x -y = ( xy -x ) + ( y2 - y ) = x ( y -1) + y ( y -1) = ( y -1 ) ( x +y )
Bài 1:
a: \(A=\frac{x^2}{\left(y+1\right)^2}:\frac{2x}{y+1}:\frac{2x}{y+1}\)
\(=\frac{x^2}{\left(y+1\right)^2}\cdot\frac{y+1}{2x}\cdot\frac{y+1}{2x}\)
\(=\frac{x^2}{4x^2}=\frac14\)
b: \(B=\frac{x^2}{\left(y+1\right)^2}:\left(\frac{2x}{y+1}:\frac{2x}{y+1}\right)\)
\(=\frac{x^2}{\left(y+1\right)^2}:\left(\frac{2x}{y+1}\cdot\frac{y+1}{2x}\right)\)
\(=\frac{x^2}{\left(y+1\right)^2}:1=\frac{x^2}{\left(y+1\right)^2}\)