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\(\frac{1}{a+1}-\frac{1}{a+7}=\frac{a+7}{\left(a+1\right)\left(a+7\right)}-\frac{a+1}{\left(a+1\right)\left(a+7\right)}=\frac{6}{\left(a+1\right)\left(a+7\right)}\)
=>x+7=6
=>x=6-7
=>x=-1
vậy x=-1
\(\frac{1}{a+1}-\frac{1}{a+7}=\frac{\left(a+7\right)-\left(a+1\right)}{\left(a+1\right)\left(a+7\right)}=\frac{6}{\left(a+1\right)\left(a+7\right)}\)=> x + 7 = 6 => x = -1
a)Ta có:
\(\frac{x-1}{x+2}=\frac{4}{5}\Leftrightarrow5\left(x-1\right)=4\left(x+2\right)\)
\(\Leftrightarrow5x-5=4x+8\)
\(\Leftrightarrow5x-4x=8+5\)
\(\Leftrightarrow x=13\)
b)Ta có:
\(2^{2x+1}+4^{x+3}=2^{2x+1}+2^{2x+6}=2^{2x+1}\left(1+2^5\right)=2^{2x+1}.33=264\Leftrightarrow2^{2x+1}=8=2^3\)\(\Rightarrow2x+1=3\Leftrightarrow2x=2\Leftrightarrow x=1\)
c)Ta có:
\(\frac{x^2}{-8}=\frac{27}{x}\Leftrightarrow x^3=-8.27=-216\Leftrightarrow x=-6\)
d)Ta có:
\(\frac{x+7}{-20}=\frac{-5}{x+7}\Leftrightarrow\left(x+7\right)^2=\left(-20\right)\left(-5\right)=100\Leftrightarrow\left[{}\begin{matrix}x+7=10\\x+7=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-17\end{matrix}\right.\)e)Ta có:
\(\frac{x}{-8}=\frac{2}{-x^3}\Leftrightarrow x.\left(-x^3\right)=-8.2\)
\(\Leftrightarrow-x^4=-16\Leftrightarrow x^4=16\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(a.\frac{x-1}{x+2}=\frac{4}{5}\)
\(\Rightarrow\frac{x+2-3}{x+2}=\frac{4}{5}\)
\(\Rightarrow1-\frac{3}{x+2}=\frac{4}{5}\)
\(\Rightarrow\frac{3}{x+2}=1-\frac{4}{5}\)
\(\Rightarrow\frac{3}{x+2}=\frac{1}{5}\)
\(\Rightarrow\frac{3}{x+2}=\frac{3}{15}\Rightarrow x+2=15\)
\(\Rightarrow x=13\)( thỏa mãn )
\(\Rightarrow x+7=a+7-\left(a+1\right)\)
\(\Rightarrow x+7=a+7-a-1\)
\(\Rightarrow x+7=6\)
\(\Rightarrow x=-1\)