ai giúp mình với ạ, mình cảm ơn ạ <3
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) x² + xy
= x(x + y)
b) x³ - 4x
= x(x² - 4)
= x(x - 2)(x + 2)
c) x² - 9 + xy + 3y
= (x² - 9) + (xy + 3y)
= (x - 3)(x + 3) + y(x + 3)
= (x + 3)(x + y - 3)
d) x²y + x² + xy - 1
= (x²y + xy) + (x² - 1)
= xy(x + 1) + (x - 1)(x + 1)
= (x + 1)(xy + x - 1)
x2 - x - y2 - y
=x2 - y2 - x - y
=(x - y)(x + y) - (x + y)
=(x + y)(x - y - 1)
x^2 + 2y^2 - 2y - 2xy + 1 = (x^2 - 2xy + y^2) + (y^2 - 2y + 1) = (x - y)^2 + (y - 1)^2
\(x^2+2y^2-2y-2xy+1\)
\(=x^2-2xy+y^2+y^2-2y+1\)
\(=\left(x-y\right)^2+\left(y-1\right)^2\)
\(=\left(x-y\right)^2-\left(1-y\right)^2\)
\(=\left(x-y-1+y\right)\left(x-y+1-y\right)\)
\(=\left(x-1\right)\left(x-2y+1\right)\)
Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
\(=\left(x^2-6x+9\right)-4y^2\)
\(=\left(x-3\right)^2-\left(2y\right)^2\)
\(=\left(x-3-2y\right)\left(x-3+2y\right)\)
= ( x^2 - 4y^2 ) + ( 9 - 6x)
= [ x^2 - (2y)^2 ] + 3( 3 - 2x )
= (x - 2y)(x + 2y)+ 3(3 - 2x)
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
\(\left(x+3\right)^2-16\)
\(=\left(x+3-4\right)\left(x+3+4\right)\)
\(=\left(x-1\right)\left(x+7\right)\)