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b )=x4-2x3-2x3+4x2+4x2-8x-8x+16
=x3(x-2)-2x2(x-2)+4x(x-2)-8(x-2)
=(x-2)(x3-2x2+4x-8)
=(x-2)[x2(x-2)+4(x-2)]
=(x-2)2(x2+4)
a) đề thiếu ko bn?
b) \(x^4-4x^3+8x^2-16x+16=\left(x^4-4x^2\right)-\left(4x^3-12x^2+8x\right)-\left(8x-16\right)\)
\(=x^2\left(x-2\right)\left(x+2\right)-4x\left(x^2-3x+2\right)-8\left(x-2\right)\)
\(=x^2\left(x-2\right)\left(x+2\right)-4x\left(x-2\right)\left(x-1\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left[x^2\left(x+2\right)-4x\left(x-1\right)-8\right]=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)
\(=\left(x-2\right)\left[\left(x^3-8\right)-\left(2x^2-4x\right)\right]=\left(x-2\right)\left[\left(x-2\right)\left(x^2+2x+4\right)-2x\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(x-2\right)\left(x^2+2x+4-2x\right)=\left(x-2\right)^2\left(x^2+4\right)\)
\(x^2\left(1-x^2\right)-4-4x^2\)
\(\text{ Phân tích thành nhân tử}\)
\(\left(-\left(x^2-x+2\right)\right)\left(x^2+x+2\right)\)
a, \(11x+11y+x^2+xy=\left(11x+11y\right)+\left(x^2+xy\right)=11\left(x+y\right)+x\left(x+y\right)=\left(x+y\right)\left(x+11\right)\)
b. \(255-4x^2-4xy-y^2=255-\left(4x^2+4xy+y^2\right)=255-\left(2x+y\right)^2=\left(15+2x+y\right)\left(15-2x-y\right)\)
Bài 2:
\(x^2-y^2-4x+4=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2=\left(x-2-y\right)\left(x-2+y\right)\)
\(=\left(72-2\right)\left(102-2\right)=70.100=7000\) ( x+y=102, x-y=72 )
a) \(\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\)
\(=\left(y+z\right)\left(xy+yz+zx\right)+x^2y+x^2z+xyz-xyz\)
\(=\left(y+z\right)\left(xy+yz+zx\right)+x^2\left(y+z\right)\)
\(=\left(y+z\right)\left(xy+yz+zx+x^2\right)\)
\(=\left(y+z\right)\left[y\left(x+z\right)+x\left(z+x\right)\right]\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
b) \(\left(x^2+y^2+5\right)^2-4x^2y^2-16xy-16\)
\(=\left(x^2+y^2+5\right)^2-\left(4x^2y^2+16xy+16\right)\)
\(=\left(x^2+y^2+5\right)^2-\left(2xy+4\right)^2\)
\(=\left(x^2+y^2+5-2xy-4\right)\left(x^2+y^2+5+2yx+4\right)\)
\(=\left(x^2+y^2+5-2xy-4\right)\left(x^2+y^2+5+2yx+4\right)\)
c)sai đề.
đặt \(x^2+x+1=t\)
\(\Rightarrow\left(x^2+x+1\right)^2+\left(x^2+x+2\right)-12\)
\(=t^2+t+1-12\)
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mình sửa đề không biết có đúng hay không nên mình chỉ nêu hướng làm thôi. bạn thông cảm.
d) \(x^2-x-2001.2002\)
\(=x\left(x+2001\right)-2002\left(x+2001\right)\)
\(=\left(x-2002\right)\left(x+2001\right)\)