Phân tích thành nhân tử
(3x+2)^2-(x-6)^2
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\(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=8\left(x+4\right)\left(x-1\right)\)
\(\left(3x+2\right)^2-\left(x-6\right)^2\)
\(=\left(3x+2+x-6\right)\left(3x+2-x+6\right)\)
\(=\left(4x-4\right)\left(2x+8\right)\)
\(=8\left(x-1\right)\left(x+4\right)\)
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:
\(=\left(x^2+3x+1+3\right)\left(x^2+3x+1-2\right)\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)
\(\left(5x-10\right)\left(x^2-1\right)-\left(3x-6\right)\left(x^2-2x+1\right)\)
\(=\left(5x-10\right)\left(x-1\right)\left(x+1\right)-\left(3x-6\right)\left(x-1\right)^2\)
\(=\left(x-1\right)\left[\left(5x-10\right)\left(x+1\right)-\left(3x-6\right)\left(x-1\right)\right]\)
\(=\left(x-1\right)\left[5\left(x-2\right)\left(x+1\right)-3\left(x-2\right)\left(x-1\right)\right]\)
\(=\left(x-1\right)\left[\left(x-2\right)\left(5x+5-3x+3\right)\right]\)
\(=\left(x-1\right)\left[\left(x-2\right)\left(2x+8\right)\right]\)
\(=\left(x-1\right)\left(x-2\right)\left(2x+8\right)\)
\(a,x^2-5x+6\\=x^2-3x-2x+6\\=x(x-3)-2(x-3)\\=(x-3)(x-2)\\---\\b,3x^2+9x-30\\=3x^2-6x+15x-30\\=3x(x-2)+15(x-2)\\=(x-2)(3x+15)\\=3(x-2)(x+5)\\---\)
\(c,x^2-3x+2\\=x^2-x-2x+2\\=x(x-1)-2(x-1)\\=(x-1)(x-2)\\---\\d,3x^2-5x-2\\=3x^2-6x+x-2\\=3x(x-2)+(x-2)\\=(x-2)(3x+1)\\Toru\)
\(3x\cdot\left(x-y\right)^2-6\cdot\left(y-x\right)\)
\(=3x\left(x-y\right)^2+6\left(x-y\right)\)
\(=\left(x-y\right)\left[3x\left(x-y\right)+6\right]\)
\(=\left(x-y\right)\left(3x^2-3xy+6\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
1) \(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=8\left(x+4\right)\left(x-1\right)\)
2) \(A=x^2+2y^2+2xy-2y+2021=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+2020=\left(x+y\right)^2+\left(y-1\right)^2+2020\ge2020\)
\(minA=2020\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
\(x.\left(x^2-4\right)-3x+6\)
\(=x.\left(x+2\right).\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x^2+2x\right).\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2+2x-3\right)\)
\(=\left(x-2\right).\left(x^2-x+3x-3\right)\)
\(=\left(x-2\right).[x.\left(x-1\right)+3.\left(x-1\right)]\)
\(=\left(x-2\right).\left(x-1\right).\left(x+3\right)\)
(3x+2)2-(x-6)2=(3x+2-x+6)(3x+2+x-6)=(2x+8)(4x-4)=8(x+4)(x-1)
\((3x+2)^2-(x-6)^2=(3x+2-x+6)(3x+2+x-6) =(2x+8)(4x-4)=2.4(x+4)(x-1)=8(x+4)(x-1)\)