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\(4x^2\left(x+y\right)-x-y\)
\(=4x^2\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(4x^2-1\right)\)
\(=\left(x+y\right)\left(2x-1\right)\left(2x+1\right)\)
\(16ty^2+6xt-9t-tx^2\)
\(=t.\left(16y^2+6x-9-x^2\right)\)
\(=t.\left[\left(4y\right)^2-\left(x^2-2.x.3+3^2\right)\right]\)
\(=t.\left[\left(4y\right)^2-\left(x-3\right)^2\right]\)
\(=t.\left(4y-x+3\right)\left(4y+x-3\right)\)
\(x^2-9xy+20y^2\)
\(=\left(x^2-4xy\right)-\left(5xy-20y^2\right)\)
\(=x.\left(x-4y\right)-5y\left(x-4y\right)\)
\(=\left(x-4y\right)\left(x-5y\right)\)
\(\left(1+x^2\right)^2-4x\left(1-x^2\right)=1+2x^{ }+x^4-4x+4x^3\)\(=\left(x^4+2x^3-x^2\right)+\left(2x^3+4x^2-2x\right)-x^2-2x+1=x^2\left(x^2+2x-1\right)+2x\left(x^2+x-1\right)-\left(x^2+2x-1\right)\)\(\left(x^2+2x-1\right)\left(x^2+2x-1\right)=\left(x^2+2x-1\right)^2\)
(x2 - 8)2 + 36 = x4 - 16x + 100 = x4 - 20x2 + 100 - 4x2
= (x2 - 10) - 4x2 = (x2 - 10 - 2x)(x2 - 10 + 2x)
(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 !
Sai đề nhé bạn
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(x^2+x+1=t\)
Đa thức trở thành \(t\left(t+1\right)-12\)
\(=t^2+t-12\)
\(=t^2+3t-4t-12\)
\(=t\left(t+3\right)-4\left(t+3\right)\)
\(=\left(t+3\right)\left(t-4\right)\)
Thay vào ta được
\(\left(x^2+x+4\right)\left(x^2+x-3\right)\)
Trả lời:
a) \(x^3+2x=x\left(x^2+2\right)\)
b) \(3x^3-12x^2=3x^2\left(x-4\right)\)
Ta có:
\(x^2+2xy+y^2-x-y-12=(x^2+2xy+y^2)-(x+y)-12\)
\(=(x+y)^2-(x+y)-12 \) \((*)\)
Đặt \(x+y=a\)
từ \((*)\Rightarrow a^2-a-12=(a^2+3a)-(4a+12)\)
\(=(a+3)(a-4)\)
Thay \(a=x+y\)
\(\Rightarrow (x+y+3)(x+y-4)\)
Ta có : \(4x^2-3x-1\)
\(=4x^2-4x+x-1\)
\(=4x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(4x+1\right)\)
Ta có : \(x^2-7x+12\)
\(=x^2-3x-4x+12\)
\(=x\left(x-3\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
\(x^2-x-12=x^2-x+\frac{1}{4}-\frac{1}{4}-12=\left(x-\frac{1}{2}\right)^2-\frac{49}{4}=\left(x+3\right)\left(x-4\right)\)
\(x^2-x-12=x^2-4x+3x-12=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)