8^x+2^3x+1=24

8^x+8^x*2=24

8^x(1+2)=24

8^x*3=24

8^x=24/3

8^x=8=8¹

=>x=1

`1/8+(2x+5)/24=(3x+5)/2+x/6`

`=>3+2x+5=12(3x+5)+4x`

`=>2x+8=36x+60+4x`

`=>2x+8=40x+60`

`=>38x=-52`

`=>x=-26/19`

Vậy `x=-26/19`

\(a,\Rightarrow\dfrac{1}{2}x=-2\Rightarrow x=-4\\ b,3x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x+y}{5+3}=\dfrac{24}{8}=3\\ \Rightarrow\left\{{}\begin{matrix}x=15\\y=9\end{matrix}\right.\)

\(a,\Rightarrow12x-91=101\\ \Rightarrow12x=192\\ \Rightarrow x=16\\ b,\Rightarrow x:23+45=133\\ \Rightarrow x:23=88\\ \Rightarrow x=\dfrac{88}{23}\\ c,\Rightarrow\left(6x-39\right):7=3\\ \Rightarrow6x-39=21\\ \Rightarrow6x=60\\ \Rightarrow x=10\\ d,\Rightarrow3x-24=\dfrac{148}{73}\\ \Rightarrow3x=\dfrac{1900}{73}\\ \Rightarrow x=\dfrac{1900}{219}\\ e,\Rightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\\ f,\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ d,\left(9-x\right)^3=64=4^3\\ \Rightarrow9-x=4\\ \Rightarrow x=5\\ h,\Rightarrow x=27\\ i,\Rightarrow6x=312\cdot12=624\cdot6\\ \Rightarrow x=624\\ j,\Rightarrow\left(19x+104\right):14=25-42=-17\\ \Rightarrow19x+104=-238\\ \Rightarrow19x=-342\\ \Rightarrow x=-18\)

\(PT\Leftrightarrow\sqrt{x^2+24}-\sqrt{x^2+8}=3x-1\)

Mà \(\sqrt{x^2+24}>\sqrt{x^2+8}\) nên \(3x-1>0\Leftrightarrow x>\frac{1}{3}\)

Ta có : \(PT\Leftrightarrow\left(\sqrt{x^2+24}-5\right)-\left(\sqrt{x^2+8}-3\right)-3x+3=0\)

\(\Leftrightarrow\frac{x^2+24-25}{\sqrt{x^2+24}+5}-\frac{x^2+8-9}{\sqrt{x^2+8}+3}-3\left(x-1\right)=0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+24}+5}-\frac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+8}+3}-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\frac{\left(x+1\right)}{\sqrt{x^2+24}+5}-\frac{\left(x+1\right)}{\sqrt{x^2+8}+3}-3\right]=0\)

Suy luận dựa \(ĐK\) ta được \(x=1\)

Yêu cầu của đề là gì nhỉ ? 

Nhầm

(⁠-⁠_⁠-⁠;⁠)⁠・⁠・⁠・

a: \(8\cdot3^x=24\)

=>\(3^x=3\)

=>x=1

b: \(3\cdot2^{x+1}-5=19\)

=>\(3\cdot2^{x+1}=24\)

=>\(2^{x+1}=8\)

=>x+1=3

=>x=2

c: \(5^x+17=42\)

=>\(5^x=25\)

=>x=2