K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

a) Ta có: \(x^4+64\)

\(=x^4+16x^2+64-16x^2\)

\(=\left(x^2+8\right)^2-\left(4x\right)^2\)

\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

b) Ta có: \(81x^4+4y^4\)

\(=81x^4+36x^2y^2+4y^4-36x^2y^2\)

\(=\left(9x^2+2y^2\right)^2-\left(6xy\right)^2\)

\(=\left(9x^2-6xy+2y^2\right)\left(9x^2+6xy+2y^2\right)\)

c) Ta có: \(x^5+x+1\)

\(=x^5+x^2-x^2+x-1\)

\(=x^2\left(x^3+1\right)-\left(x^2-x+1\right)\)

\(=x^2\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)

 

28 tháng 10 2021

a: \(=5a\left(x-2y\right)\)

b: \(=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)

c: =(x-1)(x-7)

28 tháng 10 2021

a)\(5ax-10ay=5a\left(x-2y\right)\)

b) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x+1\right)\left(x-y\right)\)

c) \(x^2-8x+7=\left(x-7\right)\left(x-1\right)\)

a: \(3x^4-4x^3+1\)

\(=3x^4-3x^3-x^3+1\)

\(=3x^3\left(x-1\right)-\left(x-1\right)\left(x^2+x+1\right)\)

\(=\left(x-1\right)\left(3x^3-x^2-x-1\right)\)

b: \(x^3-19x-30\)

\(=x^3-4x-15x-30\)

\(=x\left(x-2\right)\left(x+2\right)-15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-2x-15\right)\)

\(=\left(x+2\right)\cdot\left(x-5\right)\left(x+3\right)\)

a: \(A=x^3y-12xy-x^2y\)

\(=xy\cdot x^2-xy\cdot12-xy\cdot x\)

\(=xy\left(x^2-x-12\right)\)

\(=xy\left(x^2-4x+3x-12\right)\)

\(=xy\left[x\left(x-4\right)+3\left(x-4\right)\right]\)

\(=xy\left(x-4\right)\left(x+3\right)\)

c: \(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)

=(x+1)(x+4)(x+2)(x+3)-120

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-120\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)-96\)

\(=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)

\(=\left(x^2+5x+16\right)\left(x+6\right)\left(x-1\right)\)

d: \(D=x^5-x^4+x^2-1\)

\(=\left(x^5-x^4\right)+\left(x^2-1\right)\)

\(=x^4\left(x-1\right)+\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)\left(x^4+x+1\right)\)

s không có câu b ạ

 

21 tháng 12 2020

a) \(x^2 (x+1)-2x(x+1)+x+1 \\ =(x+1)(x^2-2x+1)\\=(x+1)(x-1)^2\)

b) \(4x^2 -8x+3 \\= (2x^2)-2.2x .2 + 2^2 -1 \\=(2x-2)^2-1^2\\=(2x-2+1)(2x-2-1)\\= (2x-1)(2x-3)\)

28 tháng 7 2021

a) (x3-x2)+(8x-8)=x(x-1)+8(x-1)=(x2+8)(x-1)

b) 8x3-8x2y+2xy2=2x(4x2-4xy+y2)

c) (x2+y2-z2)2 - 4x2y2=(x2+y2-z2)2 - (2xy)2=(x2+y2-z2-2xy)(x2+y2-z2+2xy)

24 tháng 8 2021

a) \(40x^4-10x^2=10x^2\left(4x^2-1\right)=10x^2\left(2x-1\right)\left(2x+1\right)\)

b) \(16x^4-20x^2-y^2-5y=\left(4x^2-\dfrac{5}{2}\right)^2-\left(y-\dfrac{5}{2}\right)^2=\left(4x^2-\dfrac{5}{2}-y+\dfrac{5}{2}\right)\left(4x^2-\dfrac{5}{2}+y-\dfrac{5}{2}\right)=\left(4x^2-y\right)\left(4x^2+y-5\right)\)c)\(64a^2-9b^2-16a+1=\left(8a-1\right)^2-9b^2=\left(8a-1-3b\right)\left(8a-1+3b\right)\)d) \(5x^2+23x-10=5\left(x-\dfrac{2}{5}\right)\left(x+5\right)\)

a: \(40x^4-10x^2\)

\(=10x^2\left(4x^2-1\right)\)

\(=10x^2\cdot\left(2x-1\right)\left(2x+1\right)\)

b: \(16x^4-20x^2-y^2-5y\)

\(=\left(4x^2-y\right)\left(4x^2+y\right)-5\left(4x^2+y\right)\)

\(=\left(4x^2+y\right)\left(4x^2-y-5\right)\)

c: Ta có: \(64a^2-9b^2-16a+1\)

\(=\left(8a-1\right)^2-9b^2\)

\(=\left(8a-1-3b\right)\left(8a-1+3b\right)\)

d: Ta có: \(5x^2+23x-10\)

\(=5x^2+25x-2x-10\)

\(=\left(x+5\right)\left(5x-2\right)\)

a: \(x^4-4x^3-8x^2+8x\)

\(=x\left(x^3-4x^2-8x+8\right)\)

\(=x\left[\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\right]\)

\(=x\left(x+2\right)\left(x^2-6x+4\right)\)

b: \(x^2-1-xy+y\)

\(=\left(x-1\right)\left(x+1\right)-y\left(x-1\right)\)

\(=\left(x-1\right)\left(x-y+1\right)\)

c: Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)^2\cdot\left(x-2\right)\)

\(=\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3-x-1\right)\)

\(=2\cdot\left(x-1\right)\cdot\left(x-2\right)^2\)

a: \(x^4+4=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

b: \(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

c: \(x^8+x^4+1\)

\(=\left(x^8+2x^4+1\right)-x^4\)

\(=\left(x^4-x^2+1\right)\cdot\left(x^4+x^2+1\right)\)

\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)

 

26 tháng 8 2021

a)\(x^4+4\\ =\left(x^2\right)^2+4x^2+4-4x^2\\ =\left[\left(x^2\right)^2+4x^2+4\right]-\left(2x\right)^2\\ =\left(x^2+2\right)^2-\left(2x\right)^2\\ =\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)