Ta có: \(x^4=24x+32\)

\(\Leftrightarrow x^4+4x^2+4=24x+32+4x^2+4\)

\(\Leftrightarrow\left(x^2+2\right)^2=4x^2+24x+36\)

\(\Leftrightarrow\left(x^2+2\right)^2=4\left(x+3\right)^2\)

\(\Leftrightarrow\left(x^2+2\right)^2=\left(2x+6\right)^2\)

\(\Leftrightarrow\left(x^2+2\right)^2-\left(2x+6\right)^2=0\)

\(\Leftrightarrow\left(x^2+2-2x-6\right)\left(x^2+2+2x+6\right)=0\)

\(\Leftrightarrow\left(x^2-2x-4\right)\left(x^2+2x+8\right)=0\)(1)

Ta có: \(x^2+2x+8\)

\(=x^2+2x+1+7\)

\(=\left(x+1\right)^2+7\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+7\ge7\forall x\)

hay \(x^2+2x+8>0\forall x\)(2)

Từ (1) và (2) suy ra \(x^2-2x-4=0\)

\(\Leftrightarrow x^2-2x+1-5=0\)

\(\Leftrightarrow\left(x-1\right)^2-5=0\)

\(\Leftrightarrow\left(x-1\right)^2=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{5}\\x-1=-\sqrt{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+1\\x=-\sqrt{5}+1\end{matrix}\right.\)

Vậy: \(x\in\left\{\sqrt{5}+1;-\sqrt{5}+1\right\}\)

\(\Leftrightarrow x^4+4x^2+4=4\left(x^2+6x+9\right)\)

\(\Leftrightarrow\left(x^2+2\right)^2=4\left(x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2=2\left(x+3\right)\\x^2+2=-2\left(x+3\right)\end{matrix}\right.\)..

Giải vô tư

a) \(x^4-24x+32=0\)

\(\Leftrightarrow x^4-2x^3+2x^3-4x^2+4x^2-8x-16x+32=0\)

\(\Leftrightarrow x^3\left(x-2\right)+2x^2\left(x-2\right)+4x\left(x-2\right)-16\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2+4x-16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+2x^2+4x-16=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x\approx1,62\end{matrix}\right.\)

b) \(x^4-8x\sqrt{2}+12=0\)

\(\Leftrightarrow x^4-\sqrt{2}x^3+\sqrt{2}x^3-2x^2+2x^2-2\sqrt{2}x-6\sqrt{2}x+12=0\)

\(\Leftrightarrow x^3\left(x-\sqrt{2}\right)+\sqrt{2}x^2\left(x-\sqrt{2}\right)+2x\left(x-\sqrt{2}\right)-6\sqrt{2}\left(x-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x^3+\sqrt{2}x^2+2x-6\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x\approx1,4142135...\end{matrix}\right.\)

b)

<=>x^3+3x^2-3x+1=0

<=> (x-1)^3=0

<=> x=1.

c)<=>x^4=x^2-2x+1

<=>x^4=(x-1)^2

<=>(x^2-x+1)(x^2+x-1)=0

Do x^2-x+1>0

=> x^2+x-1=0

<=> x=(-1+căn 5)/2; (-1-căn 5)/2

1000001011101001010101101110111111111111111111110000010010010010010000100101100101001000011110100000001001000010100101010010100100000011010100000101101101010111110100100101

a/ \(\Leftrightarrow x^4-2x^3-4x^2+2x^3-4x^2-8x+8x^2-16x-32=0\)

\(\Leftrightarrow x^2\left(x^2-2x-4\right)+2x\left(x^2-2x-4\right)+8\left(x^2-2x-4\right)=0\)

\(\Leftrightarrow\left(x^2-2x-4\right)\left(x^2+2x+8\right)=0\)

\(\Leftrightarrow x^2-2x-4=0\Rightarrow x=1\pm\sqrt{5}\)

b/ \(2x^3=x^3-3x^2+3x-1\)

\(\Leftrightarrow2x^3=\left(x-1\right)^3\)

\(\Leftrightarrow x\sqrt[3]{2}=x-1\)

\(\Rightarrow x=\frac{1}{1-\sqrt[3]{2}}\)

c/ \(x^4-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow x^2+x-1=0\Rightarrow x=\frac{-1\pm\sqrt{5}}{2}\)

Ta có: \(x^4-8x^3+21x^2-24x+9=0\)

\(\Leftrightarrow x^4-5x^3+3x^2-3x^3+15x^2-9x+3x^2-5x+9=0\)

\(\Leftrightarrow\left(x^2-5x+3\right)\left(x^2-3x+3\right)=0\)

\(\Leftrightarrow x^2-5x+3=0\)

\(\text{Δ}=\left(-5\right)^2-4\cdot1\cdot3=25-12=13\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{5-\sqrt{13}}{2}\\x_2=\dfrac{5+\sqrt{13}}{2}\end{matrix}\right.\)

\(sin^23x-cos^24x=sin^25x-cos^26x\)

\(\Leftrightarrow2sin^23x-2cos^24x=2sin^25x-2cos^26x\)

\(\Leftrightarrow2sin^23x-1+1-2cos^24x=2sin^25x-1+1-2cos^26x\)

\(\Leftrightarrow-cos6x-cos8x=-cos10x-cos12x\)

\(\Leftrightarrow cos6x-cos12x+cos8x-cos10x=0\)

\(\Leftrightarrow sin9x.sin6x+sin9x.sin4x=0\)

\(\Leftrightarrow sin9x.\left(sin6x+sin4x\right)=0\)

\(\Leftrightarrow2sin9x.sin5x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin9x=0\\sin5x=0\\cosx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{9}\\x=\dfrac{k\pi}{5}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)