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\(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]=\left(x-1\right)\left(2\left(x^2-2x+1\right)-5x+5-1\right)=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)=\left(x-1\right)\left(2x^2-9x+6\right)\)
\(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\)
\(=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]\)
\(=\left(x-1\right)\left[2\left(x^2-2x+1\right)-5\left(x-1\right)-1\right]\)
\(=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)\)
\(=\left(x-1\right)\left(2x^2-9x+6\right)\)
1)
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:
\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)
2) Bạn xem lại đề!
1
(x2-8)2+36
=x4-16x2+64+36
=x4+20x2+100-36x2
=(x2+10)2-(6x)2
HĐT số 3
\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
\(x^2\left(x+1\right)-\left(x+1\right)\left(3x+1\right)+7x-x^2\)
\(=x^3+x^2-3x^2-4x-1+7x-x^2\)
\(=x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
\(x\cdot\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x+3\right)+1\)
\(=\left(x+1\right)\left(x+2\right)\left[x\left(x+3\right)\right]+1\)
\(=\left(x^2+x+2x+2\right)\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+2\right)\left(x^2+3x\right)+1\)
gọi \(\left(x^2+3x\right)=a\)
\(\Rightarrow\left(t+2\right)t+1\)
\(=t^2+2t+1=\left(t+1\right)^2\)
\(\Rightarrow=\left(x^2+3x+1\right)^2\)
\(\Rightarrow x\cdot\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x+3\right)+1\)\(=\left(x^2+3x+1\right)^2\)