phân tích các đa thức sau thành nhân tử:
a) (ab-1)^2 + (a+b)^2
b) x^3+2x^2+2x+1
c) x^3-4x^2+12x-27
d)x^4-2x^3+2x-1
e)x^4+2x^3+2x^2+2x+1
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b, \(x^3+2x^2+2x+1=\left(x^2+x+1\right)\left(x+1\right)\)
c, \(x^3-4x^2+12x-27=\left(x^2-x+9\right)\left(x-3\right)\)
d, \(x^4-2x^3+2x-1=\left(x-1\right)^3\left(x+1\right)\)
e, sai đề
a, \(\left(ab-1\right)^2+\left(a+b\right)^2=\left(a^2+1\right)\left(b^2+1\right)\)
b, \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2+x+1\right)\)
c, \(x^3-4x^2+12x-27=\left(x-3\right)\left(x^2-x+9\right)\)
d, \(x^4-2x^3+2x-1=\left(x-1\right)^3\left(x+1\right)\)
e, cho mình sửa đề xíu
\(x^4+2x^3+2x^2+2x+1=\left(x+1\right)^2\left(x^2+1\right)\)
a) \(4x\left(a-b\right)+6xy\left(b-a\right)\)
\(=4x\left(a-b\right)-6xy\left(a-b\right)\)
\(=\left(4x-6xy\right)\left(a-b\right)\)
\(=2x\left(2-3y\right)\left(a-b\right)\)
a: \(x^3-2x+4\)
\(=x^3+2x^2-2x^2-4x+2x+4\)
\(=\left(x+2\right)\left(x^2-2x+2\right)\)
b: \(x^3-4x^2+12x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
c: \(x^3+2x^2+2x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
\(x^4+2x^3+2x^2+2x+1\)
\(=\left(x^4+2x^3+x^2\right)+\left(x^2+2x+1\right)\)
\(=\left(x^2+x\right)^2+\left(x+1\right)^2\)
\(=x^2\left(x+1\right)^2+\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left(x^2+1\right)\)
mấy cái này chỉ cần dùng hằng đẳng thức thui mà ..tự lm nha
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)
phân tích các đa thức sau thành nhân tử:
a) \(\left(ab-1\right)^2+\left(a+b\right)^2\)
Ta thấy \(\left\{{}\begin{matrix}\left(ab-1\right)^2\ge0\\\left(a+b\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(ab-1\right)^2+\left(a+b\right)^2>0\) nên k phân tích thành nhân tử đc.
b) \(x^3+2x^2+2x+1\)
= \(x^3+x^2+x^2+x+x+1\)
= \(x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+x+1\right)\)
c) \(x^3-4x^2+12x-27\)
= \(x^3-3x^2-x^2+3x+9x-27\)
= \(x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
= \(\left(x-3\right)\left(x^2-x+9\right)\)
d) \(x^4+2x^3+2x^2+2x+1\)
= \(x^4+x^3+x^3+x^2+x^2+x+x+1\)
= \(x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
= \(\left(x+1\right)\left(x^3+x^2+x+1\right)\)
= \(\left(x+1\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\)
= \(\left(x+1\right).\left(x+1\right)\left(x^2+1\right)\)
= \(\left(x+1\right)^2\left(x^2+1\right)\)
a, \(\left(ab-1\right)^2+\left(a+b\right)^2\)
\(=a^2b^2-2ab+1+a^2+2ab+b^2\)
\(=a^2b^2+a^2+b^2+1=a^2.\left(b^2+1\right)+\left(b^2+1\right)\)
\(=\left(b^2+1\right).\left(a^2+1\right)\)
b, \(x^3+2x^2+2x+1\)
\(=x^3+x^2+x^2+x+x+1\)
\(=x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2+x+1\right)\)
c, \(x^3-4x^2+12x-27\)
\(=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2.\left(x-3\right)-x.\left(x-3\right)+9.\left(x-3\right)\)
\(=\left(x-3\right).\left(x^2-x+9\right)\)
d, \(x^4-2x^3+2x-1=x^4-x^3-x^3+x^2-x^2+x+x-1\)
\(=x^3.\left(x-1\right)-x^2.\left(x-1\right)-x.\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right).\left(x^3-x^2-x+1\right)\)
\(=\left(x-1\right).\left[x^2.\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right).\left(x-1\right).\left(x^2-1\right)=\left(x-1\right)^2\left(x^2-1\right)\)
e, \(x^4+2x^3+2x^2+2x+1\)
\(=x^4+x^3+x^3+x^2+x^2+x+x+1\)
\(=x^3.\left(x+1\right)+x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^3+x^2+x+1\right)\)
\(=\left(x+1\right).\left[x^2.\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(x+1\right).\left(x+1\right).\left(x^2+1\right)=\left(x+1\right)^2\left(x^2+1\right)\)
Chúc bạn học tốt!!!