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\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left(4x^3-x\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left[\left(2x\right)^2-1\right]\)
\(=x\left(x+1\right)\left(2x+1\right)\left(2x-1\right)\)
(Nhớ k cho mình với nhá!)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(x^4+4x^3+2x^2-4x+1\)
\(=x^4+2x^3-x^2+2x^3+4x^2-2x-x^2-2x+1\)
\(=x^2\left(x^2+2x-1\right)+2x\left(x^2+2x-1\right)-\left(x^2+2x-1\right)\)
\(=\left(x^2+2x-1\right)^2\)
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
=\(\left(3x^2-4x-13-4x^2+9\right)\left(3x^2-4x-13+4x^2-9\right)-\left(x+2\right)^4\)
=\(\left(-x^2-4x-4\right)\left(7x^2-4x-22\right)-\)\(\left(x+2\right)^{^{ }2.2}\)
=\(-\left(x+2\right)^2\left(7x^2-4x-22\right)-\left(x+2\right)^2\left(x+2\right)^2\)
=\(-\left(x+2\right)^2\)\(\left(7x^2-4x-22-x^2-4x-4\right)\)
\(-\left(x+2\right)^2\)(\(6x^2-8x-26\))
Phân tích thành nhân tử:
x^2 - y^2 - 4x + 4
=(x^2-4x+4)-y^2
=(x-2)^2-y^2
=(x-2+y)(x-2-y)
\(3\left(x+4\right)-x^2-4x\)
\(\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)\)
\(\Leftrightarrow\left(3-x\right)\left(x+4\right)\)