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\(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)\)
\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)
Đặt \(x^2+3x+1=t\)
\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)
\(=t^2+t-6=\left(t^2-2t\right)+\left(3t-6\right)\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
\(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1=a\)ta có :
\(a\left(a+1\right)-6\)
\(=a^2+a-6\)
\(=a^2+6a-a-6\)
\(=\left(a^2+6a\right)-\left(a+6\right)\)
\(=a\left(a+6\right)-\left(a+6\right)\)
\(=\left(a+6\right)\left(a-1\right)\)
Thay \(a=x^2+3x+1\)vào A ta có :
\(A=\left(x^2+3x+1+6\right)\left(x^2+3x+1-1\right)\)
\(=\left(x^2+3x+7\right)\left(x^2+3x\right)\)
\(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(\left(x^2+3x+1\right)=a\), ta được:
\(a\left(a+1\right)-6\)\(=a^2+a-6\)\(=\left(a^2+3a\right)-\left(2a+6\right)\)\(=a\left(a+3\right)-2\left(a+3\right)\)
\(=\left(a+3\right)\left(a-2\right)\)
Thay \(a=\left(x^2+3x+1\right)\), ta được:
\(=\left(x^2+3x+1+3\right)\left(x^2+3x+1-2\right)\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)
1) \(x\left(x-1\right)+\left(1-x\right)^2\)
\(=x\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(x-1\right)\left(x+x-1\right)\)
\(=\left(x-1\right)\left(2x-1\right)\)
2) \(2x\left(x-2\right)-\left(x-2\right)^2\)
\(=\left(x-2\right)\left[2x-\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(2x-x+2\right)\)
\(=\left(x-2\right)\left(x+2\right)\)
3) \(3x\left(x-1\right)^2-\left(1-x\right)^3\)
\(=3x\left(x-1\right)^2+\left(x-1\right)^3\)
\(=\left(x-1\right)^2\left(3x+x-1\right)\)
\(=\left(x-1\right)^2\left(4x-1\right)\)
4) \(3x\left(x+2\right)-5\left(x+2\right)^2\)
\(=\left(x+2\right)\left[3x-5\left(x+2\right)\right]\)
\(=\left(x+2\right)\left(3x-5x-10\right)\)
\(=\left(x+2\right)\left(-2x-10\right)\)
\(=-2\left(x+2\right)\left(x+5\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(\left(3x+1\right)^2-\left(3x-1\right)^2\)
\(=\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\)
\(=2\cdot6x\)
\(=12x\)
_________
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=2x\cdot2y\)
\(=4xy\)
\(\left(x+y\right)^3+\left(x-y\right)^3\)
\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2x\cdot\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)
\(=2x\cdot\left(x^2+3y^2\right)\)
______
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3+3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2-xy-xz-yz\right)\)
\(x^3+y^3-3x^2+3x-1\\=(x^3-3x^2+3x-1)+y^3\\=(x-1)^3+y^3\\=(x-1+y)[(x-1)^2-(x-1)y+y^2]\\=(x+y-1)(x^2-2x+1-xy+y+y^2)\)
(-x-1)2-(3x-4)2=(x2+2x+1)-(9x2-24x+16)=-8x2+26x-15=\(-8\left(x-\dfrac{5}{2}\right)\left(x-\dfrac{3}{4}\right)\)
\(\left(x^2-1+x\right)\left(x^2-1+3x\right)+x^2\)
Đặt \(a=x^2+2x-1\)
Phương trình trở thành \(\left(a-x\right)\left(a+x\right)+x^2\)
\(=a^2-x^2+x^2\)
\(=a^2=(x^2+2x-1)^2\)