Phân tích đa thức thành nhân tử: A)x³-16x; B)3x²-3y²-6xy-12; C)x²+6x+5; D)x⁴+x³+2x²+x+1. Tìm x: A)(x+6)²=144 B)x³+27+(x+3)(x-9)=0; C)2x²-x-6=0. Giúp mình với.Gấp lắm r.Thanks
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\(=\left(x-3\right)\left(8x^3-16x^2\right)=8x^2\left(x-2\right)\left(x-3\right)\)
\(8x^3\left(x-3\right)+16x^2\left(3-x\right)\)
\(=8x^3\left(x-3\right)-16x^2\left(x-3\right)\)
\(=8x^2\left(x-3\right)\left(x-2\right)\)
= ( x3 + 2x2y + xy2 ) - 16x
= x (x2 + 2xy + y2) - 16x
= x( x + y)2 - 16x
= x [ ( x + y)2 - 16 ]
= x ( x + y +4) ( x + y - 4)
https://h7.net/hoi-dap/toan-8/phan-h-da-thuc-x-4-16-thanh-nhan-tu-faq324398.html
\(A=x^3-15x^4+16x^3-29x^2\)
\(A=\left(x^3+16x^3\right)-15x^4-29x^2\)
\(A=17x^3-15x^4-29x^2\)
\(A=-15x^4+17x^3-29x^2\)
\(A=-x^2\left(15x^2-17x+29\right)\)
\(=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
\(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
Bài 1:
a) \(x^3-16x=x\left(x-4\right)\left(x+4\right)\)
b) \(3x^2+3y^2-6xy-12=3\left(x^2-2xy+y^2-4\right)=3\left(x-y-2\right)\left(x-y+2\right)\)
c) \(x^2+6x+5=\left(x+1\right)\left(x+5\right)\)
d) \(x^4+x^3+2x^2+x+1=\left(x^2+x+1\right)\left(x^2+1\right)\)
Bài 2:
a) Ta có: \(\left(x+6\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=12\\x+6=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-18\end{matrix}\right.\)
b) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
c) Ta có: \(2x^2-x-6=0\)
\(\Leftrightarrow2x^2-4x+3x-6=0\)
\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)