Tìm x:
a) \(x^4+5x^3-8x-40=0\)
\(\Leftrightarrow x^3\left(x+5\right)-8\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^3-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x^3=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: ...
b) \(x^4+4x^3-16x-16=0\)
\(\Leftrightarrow x^4-2x^3+6x^3-12x^2+12x^2-24x+8x-16=0\)
\(\Leftrightarrow x^3\left(x-2\right)+6x^2\left(x-2\right)+12x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+6x^2+12x+8\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)^3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: ...
a) $9x^2-y^2+10yz-25z^2$
$=9x^2-(y^2-10yz+25z^2)$
$=(3x)^2-(y-5z)^2$
$=(3x-y+5z)(3x+y-5z)$
$---$
a) $x^4+5x^3-8x-40=0$
$\Leftrightarrow (x^4-8x)+(5x^3-40)=0$
$\Leftrightarrow x(x^3-8)+5(x^3-8)=0$
$\Leftrightarrow (x^3-8)(x+5)=0$
\(\Leftrightarrow \left[\begin{array}{} x^3-8=0\\ x+5=0 \end{array} \right. \Leftrightarrow \left[\begin{array}{} x^3=2^3\\ x=-5 \end{array} \right.\\ \Leftrightarrow \left[\begin{array}{} x=2\\ x=-5 \end{array} \right.\)
Vậy $x\in\{2;-5\}$.
b) $x^4+4x^3-16x-16=0$
$\Leftrightarrow (x^4-16)+(4x^3-16x)=0$
$\Leftrightarrow (x^2-4)(x^2+4)+4x(x^2-4)=0$
$\Leftrightarrow (x^2-4)(x^2+4+4x)=0$
$\Leftrightarrow (x-2)(x+2)^3=0$
\(\Leftrightarrow \left[\begin{array}{} x-2=0\\ x+2=0 \end{array} \right. \Leftrightarrow \left[\begin{array}{} x=2\\ x=-2 \end{array} \right.\)
Vậy $x\in\{2;-2\}$.
$\text{#}Toru$
1:
a: \(9x^2-y^2+10yz-25z^2\)
\(=9x^2-\left(y^2-10yz+25z^2\right)\)
\(=\left(3x\right)^2-\left(y-5z\right)^2\)
\(=\left(3x-y+5z\right)\left(3x+y-5z\right)\)
2:
a: \(x^4+5x^3-8x-40=0\)
=>\(x^3\left(x+5\right)-8\left(x+5\right)=0\)
=>\(\left(x+5\right)\left(x^3-8\right)=0\)
=>\(\left(x+5\right)\left(x-2\right)\left(x^2+2x+4\right)=0\)
=>(x+5)(x-2)=0
=>\(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b: \(x^4+4x^3-16x-16=0\)
=>\(x^4-16+4x\left(x^2-4\right)=0\)
=>\(\left(x^2-4\right)\left(x^2+4\right)+4x\cdot\left(x^2-4\right)=0\)
=>\(\left(x^2-4\right)\left(x^2+4x+4\right)=0\)
=>\(\left(x-2\right)\left(x+2\right)^3=0\)
=>\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Phân tích đa thức thành nhân tử
\(9x^2-y^2+10yz-25z^2\)
\(=9x^2-\left(y^2-10yz+25z^2\right)\)
\(=\left(3x\right)^2-\left(y-5z\right)^2\)
\(=\left(3x-y+5z\right)\left(3x+y-5z\right)\)
