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Ta có:
\(x^3+2x^2+x+2\)
\(=x^2\left(x+2\right)+\left(x+2\right)\)
\(=\left(x^2+1\right)\left(x+2\right)\)
\(a.x^3+3x^2+4x+2\)
\(=x^3+x^2+2x^2+2x+2\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+2\right)\)
\(b.6x^4-x^3-7x^2+x+1\)
\(=6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1\)
\(=6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(6x^3+5x^2-2x-1\right)\)
\(=\left(x-1\right)\left(6x^3+6x^2-x^2-x-x-1\right)\)
\(=\left(x-1\right)\left[6x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(6x^2-x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(6x^2-3x+2x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left[3x\left(2x-1\right)+\left(2x-1\right)\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(3x+1\right)\)
k giùm cái cho đỡ buồn!
a, \(x^4-x^2-2=0\Leftrightarrow x^4-2x^2+x^2-2=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\Leftrightarrow\left(x^2+1>0\right)\left(x^2-2\right)=0\Leftrightarrow x=\pm\sqrt{2}\)
b, \(\Leftrightarrow x^2\left(x^2+2x+1\right)=0\Leftrightarrow x^2\left(x+1\right)^2=0\Leftrightarrow x=0;x=-1\)
c, \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1>0\right)=0\Leftrightarrow x=1\)
d, \(\Leftrightarrow6x^2-3x-4x+2=0\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\Leftrightarrow x=\dfrac{2}{3};x=\dfrac{1}{2}\)
a)
/ \(x^4+x^2-2=0\)
\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
a) \(-y^2+\dfrac{1}{9}\)
\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)
\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)
b) \(4^4-256\)
\(=4^4-4^4\)
\(=0\)
\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
1.\(x^3+6x^2+12xy+8=x^3+3.2x^2+3.2^2x+2^3=\left(x+2\right)^3\)
3.\(x^4+2x^3+x^2-y^2=\left(x^2\right)^2+2x^2.x+x^2-y^2\)\(=\left(x^2+x\right)^2-y^2=\left(x^2+x-y\right)\left(x^2+x+y\right)\)
k mình nha bn !!!!!!! cái 2 bn xem lại đề đi, rồi mình giải cho
cau so 2 ne ban
x4 - 4x3- 8x2+ 8x