Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2-6x+5\)
\(=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-1\right)\left(x-5\right)\)
\(x^2-x-12=x^2-x+\frac{1}{4}-\frac{1}{4}-12=\left(x-\frac{1}{2}\right)^2-\frac{49}{4}=\left(x+3\right)\left(x-4\right)\)
\(x^2-x-12=x^2-4x+3x-12=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)
(1+x2)2−4x(1−x2)
= \(-\left(1-x^2\right)^2-4x\left(1-x^2\right)\)
đặt \(\left(1-x^2\right)\)= a
ta có :
- a . a - 4x .a
= a ( - a - 4x )
thay a = \(\left(1+x^2\right)\) ta có
\(\left(1+x^2\right)\left(1-x^2-4x\right)\)
phân tích tiếp nhé !
\(\left(1+x^2\right)^2-4x\left(1-x^2\right)=1+2x^{ }+x^4-4x+4x^3\)\(=\left(x^4+2x^3-x^2\right)+\left(2x^3+4x^2-2x\right)-x^2-2x+1=x^2\left(x^2+2x-1\right)+2x\left(x^2+x-1\right)-\left(x^2+2x-1\right)\)\(\left(x^2+2x-1\right)\left(x^2+2x-1\right)=\left(x^2+2x-1\right)^2\)
1/ Đặt x2+x=t
=>(x2+x)2+4(x2+x)-12=t2+4t-12=t2+6t-2t-12=t(t+6)-2(t+6)=(t-2)(t+6)=(x2+x-2)(x2+x+6)=(x2-x+2x-2)(x2+x+6)
=[x(x-1)+2(x-1)](x2+x+6)=(x-1)(x+2)(x2+x+6)
2/ Đặt x2+x=t
=>(x2+x)2+9x2+9x+14=(x2+x)2+9(x2+x)+14=t2+9t+14=t2+2t+7t+14=t(t+2)+7(t+2)=(t+2)(t+7)=(x2+x+2)(x2+x+7)
3/ Đặt x2+5x=t
=>(x2+5x)2+10x2+50x+24=(x2+5x)2+10(x2+5x)+24=t2+10t+24=t2+4t+6t+24=t(t+4)+6(t+4)=(t+4)(t+6)=(x2+5x+4)(x2+5x+6)
=(x2+x+4x+4)(x2+2x+3x+6)=[x(x+1)+4(x+1][x(x+2)+3(x+2)]=(x+1)(x+4)(x+2)(x+3)=(x+1)(x+2)(x+3)(x+4)
(x2 - 8)2 + 36 = x4 - 16x + 100 = x4 - 20x2 + 100 - 4x2
= (x2 - 10) - 4x2 = (x2 - 10 - 2x)(x2 - 10 + 2x)
b) \(\dfrac{2}{3}x^2y-2xy^2+4xy=2xy\left(\dfrac{1}{3}x-y+2\right)\)
c) \(4x^2-2x-3y-9y^2\)
\(=\left(2x-3y\right)\left(2x+3y\right)-\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x-3y-1\right)\)