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Lời giải:
$3^{8x+4}=81^{x+3}$
$3^{4(2x+1)}=(3^4)^{x+3}$
$3^{4(2x+1)}=3^{4(x+3)}$
$\Rightarrow 4(2x+1)=4(x+3)$
$\Rightarrow 2x+1=x+3$
$\Leftrightarrow x=2$ (tm)
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a) \(x+\frac{1}{2}=\frac{3}{4}\Rightarrow x=\frac{3}{4}-\frac{1}{2}\)\(\Rightarrow x=\frac{1}{4}\)
b)\(\frac{4}{5}\cdot x=\frac{4}{7}\Rightarrow x=\frac{4}{7}:\frac{4}{5}\)\(\Rightarrow x=\frac{5}{7}\)
c) \(8x=7,8x+25\Rightarrow8x-7,8x=25\)\(\Rightarrow0,2x=25\Rightarrow x=\frac{25}{0,2}\Rightarrow x=125\)
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Ta có : \(x^4+2x^3-3x^2-8x-4=0\)
=> \(x^4-2x^3+4x^3-8x^2+5x^2-10x+2x-4=0\)
=> \(x^3\left(x-2\right)+4x^2\left(x-2\right)+5x\left(x-2\right)+2\left(x-2\right)=0\)
=> \(\left(x^3+4x^2+5x+2\right)\left(x-2\right)=0\)
=> \(\left(x^3+x^2+3x^2+3x+2x+2\right)\left(x-2\right)=0\)
=> \(\left(x^2\left(x+1\right)+3x\left(x+1\right)+2\left(x+1\right)\right)\left(x-2\right)=0\)
=> \(\left(x^2+3x+2\right)\left(x+1\right)\left(x-2\right)=0\)
=> \(\left(x^2+x+2x+2\right)\left(x+1\right)\left(x-2\right)=0\)
=> \(\left(x\left(x+1\right)+2\left(x+1\right)\right)\left(x+1\right)\left(x-2\right)=0\)
=> \(\left(x+1\right)\left(x+2\right)\left(x+1\right)\left(x-2\right)=0\)
=> \(\left(x+1\right)^2\left(x+2\right)\left(x-2\right)=0\)
=> \(\left[{}\begin{matrix}\left(x+1\right)^2=0\\x+2=0\\x-2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-1\\x=-2\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-1,-2,2\right\}\)
x4 + 2x3 - 3x2 -8x - 4 = 0
⇔ x4 + 2x3 - 2x2 - 4x - x2 - 4x - 4 = 0
⇔ x3(x + 2) - 2x(x + 2) - (x + 2)2 = 0
⇔ (x + 2)(x3 - 2x - 1) = 0
⇔ (x - 2)(x3 - x - x - 1) =
⇔ (x - 2)[x(x2 - 1) - (x + 1)] = 0
⇔ (x - 2)(x + 1)(x2 + x - 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\) (Vì x2 + x - 1 > 0)
Vậy phương trình có tập nghiệm S={2;-1}
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x+1/2=3/4
x=3/4-1/2
x=6/8--4/8
x=2/8=1/4
4/5.x=4/7
x=4/7:4/5
x=4/7.5/4
x=5/7
3^4 = 8x + x
3^4 = 9x
81 = 9x
x = 9
3^4 = 8x + x
3^4 =9x
81 = 9x
x = 9