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a, =x4(x+2)-x3(x+2)+x2(x+2)-x(x+2)+(x+2)
=(x+2)(x4-x3+x2-x+1)
d)\(\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2+2xy\right)\left(x^2+y^2-z^2-2xy\right)\)
\(=\left[\left(x^2+2xy+y^2\right)-z^2\right]\left[\left(x^2-2xy+y^2\right)-z^2\right]\)
\(=\left[\left(x+y\right)^2-z^2\right]\left[\left(x-y\right)^2-z^2\right]\)
\(=\left(x+y-z\right)\left(x+y+z\right)\left(x-y-z\right)\left(x-y+z\right)\)
e)Đặt \(x^2+3x=a\)
Có: \(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)
\(=\left(a+1\right)\left(a-3\right)-5\)
\(=a^2-3a+a-3-5\)
\(=a^2-2a-8\)
\(=a^2+2x-4x-8\)
\(=a\left(a+2\right)-4\left(a+2\right)\)
\(=\left(a+2\right)\left(a-4\right)\)
\(=\left(x^2+3x+2\right)\left(x^2+3x-4\right)\)
\(=\left(x^2+x+2x+2\right)\left(x^2-x+4x-4\right)\)
\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x-1\right)+4\left(x-1\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x-1\right)\left(x+4\right)\)
\(d,\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+y^2-z^2-2xy\right)\left(x^2+y^2-z^2+2xy\right)\)
\(=\left[\left(x^2-2xy+y^2\right)-z^2\right]\left[\left(x^2+2xy+y^2\right)-z^z\right]\)
\(=\left[\left(x-y\right)^2-z^2\right]\left[\left(x+y\right)^2-z^2\right]\)
\(=\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)\left(x+y+z\right)\)
\(e,\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\left(1\right)\)
\(\text{Đặt }x^2+3x+\frac{1-3}{2}=t\)
\(\text{hay }x^2+3x-2=t\left(2\right)\)
\(\left(1\right)\Leftrightarrow\left(t+3\right)\left(t-1\right)-5\)
\(\Rightarrow t^2-t+3t-3-5\)
\(=t^2+2t-8\)
\(=t^2-2t+4t-8\)
\(=t\left(t-2\right)+4\left(t-2\right)\)
\(=\left(t-2\right)\left(t+4\right)\left(3\right)\)
\(\text{Thay (2) vào (3),ta được:}\)
\(\left(x^2+3x-2-2\right)\left(x^2+3x-2+4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x^2-x+4x-4\right)\left(x^2+x+2x+2\right)\)
\(=\left[x\left(x-1\right)+4\left(x-1\right)\right]\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)
chịu....................??????????????????????????????????????????????????
x^2(1-x^2)-4-4x^2
=x^2-x^4-4-4x^2
=x^2-(x^2+1)^2
=(x-x^2-1)(x+x^2+1)
4x(x+y)(x+y+z)(x+z)+y2z2=4(x2+xy+xz)(x2+xy+xz+yz)+y2z2=4(x2+xy+xz)2+4yz(x2+xy+xz)+y2z2=(2(x2+xy+xz)+yz)2=(2x2+2xy+2xz+yz)
đặt \(x^2+4x+8=a\)
=> \(A=a^2+3ax+2x^2=a^2+ax+2ax+2x^2=a\left(a+x\right)+2x\left(a+x\right)\)
\(=\left(a+x\right)\left(a+2x\right)\)
b) ta có
\(B=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
đặt \(x^2+8x+11=a\)
=> \(B=\left(a-4\right)\left(a+4\right)+15=a^2-16+15=a^2-1=\left(a-1\right)\left(a+1\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x^2+6x+2x+12\right)\)
\(=\left(x^2+8x+10\right)\left[x\left(x+6\right)+2\left(x+6\right)\right]=\left(x^2+8x+10\right)\left(x+6\right)\left(x+2\right)\)
a) Đưa về hằng đẳng thức số 3 , ta có :
\(\left(x^2+1\right)^2-4x^2\)
\(=\left(x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(x^2-1-2x\right)\left(x^2-1+2x\right)\)
b) \(x^2-y^2+2yz-z^2\)
\(=x^2-\left(y^2-2yz+z^2\right)\)
\(=x^2-\left(y-z\right)^2\)
Tương tự như câu a , áp dụng hằng số 3 , ta có :
\(=x^2-\left(y-z\right)^2=\left(x-y+z\right)\left(x+y-z\right)\)
1) \(\left(x^2+1\right)^2-4x^2\)
\(=\left(x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)^2\left(x-1\right)^2\)
\(=\left(x^2-1\right)\left(x^2-1\right)\)
\(=\left(x^2-1\right)^2\)