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a,x4-10x2+9=0
=>(x-1)(x3+x2-9x-9)=0
=> (x-1)(x+1)(x-3)(x+3)=0
=>\(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\)hoặc\(\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\pm1\\x=\pm3\end{cases}}\)
Vậy tập nghiệm cuả pt là S={\(\pm1,\pm3\)}
a/ \(\left(x+1\right)\left(x+5\right)\left(x^2+6x+19\right)=0\)
b/ \(\left(x+1\right)\left(x^2-2x+2\right)\left(x^2+x+1\right)=0\)
e/ \(\left(x+3\right)\left(x+5\right)\left(x^2+9x+19\right)=0\)
a) \(\left(x+2\right)^4+\left(x+4\right)^4=82\)
x+3=t
<=>\(\left(t-1\right)^4+\left(t+1\right)^4=82\)
<=>\(\left[\left(t-1\right)^2-\left(t+1\right)^2\right]^2=82-2\left(t-1\right)^2\left(t+1\right)^2\)
<=>\(\left[\left\{\left(t-1\right)-\left(t+1\right)\right\}\left\{\left(t-1\right)+\left(t+1\right)\right\}\right]^2=82-2\left(t^2-1\right)^2\)
<=>\(16t^2=82-2\left(t^2-1\right)^2\)
<=>\(\left(t^2-1\right)^2+8t^2-41=0\)
<=>\(\left(t^2-1\right)^2+8\left(t^2-1\right)-33=0\)
\(\Delta_{\left(t^2-1\right)}=16+33=49\)
\(\left[{}\begin{matrix}t^2-1=-4-7\left(l\right)\\t^2-1=-4+7\Leftrightarrow t^2=4\Rightarrow\left[{}\begin{matrix}t_1=2\\t_2-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=-5\\x_2=-1\end{matrix}\right.\)
a) @Cold Wind
2x^4 -x^3 -6x^2 -x+2 =0
[2 x^4 -4x^3 ]+3x^3 -6x^2 -x+2 =0
(x-2)(2x^3 +3x^2 -1) =0
(x-2)(2x^3 + 2x^2 +x^2 -1) =0
(x-2) [(x+1)(2x^2 +(x -1) ] =0
(x-2) [(x+1)(2x^2 + x - 1 ] =0
(x-2) (x+1)(x+1)(2x -1) =0
a) \(x^4-x^2+\dfrac{1}{4}-\dfrac{225}{4}=0\\ \left(x^2-\dfrac{1}{2}\right)^2-\dfrac{15}{2}^2=0\\ \left(x+7\right)\left(x-8\right)=0\\ \left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)
Vậy x = 8 hoặc x = -7
a: Ta có: \(x^4-x^2-56=0\)
\(\Leftrightarrow x^4-8x^2+7x^2-56=0\)
\(\Leftrightarrow\left(x^2-8\right)\left(x^2+7\right)=0\)
\(\Leftrightarrow x^2-8=0\)
hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)