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\(\frac{a+2}{a-2}=\frac{b+3}{b-3}\Rightarrow\left(a+2\right)\left(b-3\right)=\left(b+3\right)\left(a-2\right)\Rightarrow ab+2b-3a-6=ab+3a-2b-6.\)
\(\Rightarrow6a=4b\Rightarrow3a=2b\Rightarrow\frac{a}{2}=\frac{b}{3}\)đpcm
\(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)
\(=3+2^{x-1}=24-\left[4^2-\left(4-1\right)\right]\)
\(=3+2^{x-1}=24-\left[16-3\right]\)
\(\Rightarrow3+2^{x-1}=11\)
\(\Rightarrow2^{x-1}=11-3\)
\(\Rightarrow2^{x-1}=8\)
\(\Rightarrow2^{x-1}=2^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
\(\left(x-6\right)^2=\left(x-6\right)^3\)
\(\Leftrightarrow\left(x-6\right)^2-\left(x-6\right)^3=0\)
\(\Leftrightarrow\left(x-6\right)^2.\left(1-x+6\right)\text{=}0\)
\(\Leftrightarrow\left(x-6\right)^2.\left(7-x\right)\text{=}0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-6\right)^2\text{=}0\\7-x\text{=}0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\text{=}6\\x\text{=}7\end{matrix}\right.\)
Vậy.......
i can`t
\(\frac{20}{9}nhaban\)
h nha