Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
a) Ta có: \(x^2y^2-x^2+6xy-9y^2\)
\(=x^2y^2-\left(x^2-6xy+y^2\right)\)
\(=\left(xy\right)^2-\left(x-3y\right)^2\)
\(=\left(xy-x+3y\right)\left(xy+x-3y\right)\)
b) Ta có: \(9-x^2+2xy-y^2\)
\(=9-\left(x^2-2xy+y^2\right)\)
\(=9-\left(x-y\right)^2\)
\(=\left(9-x+y\right)\left(9+x-y\right)\)
a)
(x-y+5)2-2.(x-y+5)+1
=(x-y+5-1)2
=(x-y+4)2
b)
(x2+4y2-5)2-16.(x2.y2+2xy+1)
=(x2+4y2-5)2-[4.(xy+1)]2
=(x2+4y2-5-4xy-4)(x2+4y2-5+4xy+4)
=(x2+4y2-4xy-9)(x2+4y2+4xy-1)
=[(x-2y)2-9][(x+2y)2-1]
=(x-2y-3)(x-2y+3)(x+2y-1)(x+2y+1)
=(x2+x-3x-3)((x-2y+3)(x+2y-1)(x+1)2
=[x(x+1)-3(x+1)](x-2y+3)(x+2y-1)(x+1)2
=(x+1)(x-3)(x-2y+3)(x+2y-1)(x+1)2
a) \(100x^2-\left(x^2+25\right)^2=\left(10x\right)^2-\left(x^2+25\right)^2=\left(10x-x^2-25\right)\left(10x+x^2+25\right)\)
\(=-\left(x-5\right)^2\left(x+5\right)^2\)
b) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1=\left(x-y+5-1\right)^2=\left(x-y+4\right)^2\)
c) \(\left(x^2+4y^2-5\right)^2-16\left(x^2+y^2+2xy+1\right)\)
Có lẽ bạn ghi sai đề rồi nha.
a. 2xy - x2 - y2 + 16
=(2xy-x2-y2)+16
=16-(x-y)2
=(4+x-y)(4-x+y)
b. x2 + x - 6
=x2+3x-2x-6
=x(x+3)-2(x+3)
=(x-2)(x+3)
c. x2 + 5x + 6
=x2+3x+2x+6
=x(x+3)+2(x+3)
=(x+2)(x+3)
a.
\(5x^2\left(x-2y\right)-15x\left(x-2y\right)\)
\(=\left(x-2y\right)\left(5x^2-15x\right)\)
\(=5x\left(x-2y\right)\left(x-3\right)\)
b.
\(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(3+5x\right)\)
x2 + y2 - 3x - 3y + 2xy
= ( x2 + 2xy + y2 ) - ( 3x + 3y )
= ( x + y )2 - 3( x + y )
= ( x + y )( x + y - 3 )
b) ( x2 - 4x )2 - 2( x - 2 )2 - 7
= ( x2 - 4x )2 - 2( x2 - 4x + 4 ) - 7 (*)
Đặt t = x2 - 4x
(*) <=> t2 - 2( t + 4 ) - 7
= t2 - 2t - 8 - 7
= t2 - 2t - 15
= t2 + 3t - 5t - 15
= t( t + 3 ) - 5( t + 3 )
= ( t + 3 )( t - 5 )
= ( x2 - 4x + 3 )( x2 - 4x - 5 )
= ( x2 - x - 3x + 3 )( x2 + x - 5x - 5 )
= [ x( x - 1 ) - 3( x - 1 ) ][ x( x + 1 ) - 5( x + 1 ) ]
= ( x - 1 )( x - 3 )( x + 1 )( x - 5 )
a) Ta có: \(x^2+y^2-3x-3y+2xy\)
\(=\left[\left(x^2+y^2+2xy\right)-2\left(x+y\right)+1\right]-\left(x+y+1\right)\)
\(=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]-\left(x+y+1\right)\)
\(=\left(x+y-1\right)^2-\left(x+y+1\right)\)
\(=\left(x+y-1\right)^2-\left(\sqrt{x+y+1}\right)^2\)
\(=\left(x+y-1+\sqrt{x+y+1}\right)\left(x+y-1-\sqrt{x+y+1}\right)\)
a: =(x-29)(5x+1)
b: =(x-y-4)(x-y+4)
a: =(x-29)(5x+1)
b: =(x-y-4)(x-y+4)