a: Ta có: \(\dfrac{x+2}{5}=\dfrac{1}{x-2}\)

\(\Leftrightarrow x^2-4=5\)

hay \(x\in\left\{3;-3\right\}\)

d: Ta có: \(\dfrac{x-3}{x+1}=\dfrac{x+2}{x-5}\)

\(\Leftrightarrow x^2-8x+15=x^2+3x+2\)

\(\Leftrightarrow-11x=-13\)

hay \(x=\dfrac{13}{11}\)

a) ĐKXĐ: \(x\ne2\)

\(\Rightarrow\left(x+2\right)\left(x-2\right)=5.1\)

\(\Rightarrow x^2-4=5\Rightarrow x^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)

b) ĐKXĐ: \(x\ne-1\)

\(\Rightarrow\left(x+1\right)^2=2.8=16\)

\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)

c) giống câu a

d) ĐKXĐ: \(x\ne5,x\ne-1\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)=\left(x-3\right)\left(x-5\right)\)

\(\Rightarrow x^2+3x+2=x^2-8x+15\)

\(\Rightarrow11x=13\)

\(\Rightarrow x=\dfrac{13}{11}\left(tm\right)\)

\(\left(x+2\right)-2=0\)

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

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

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\Leftrightarrow\left(4-x\right)\left(4-x\right)=-5\times-5\)
\(\Rightarrow\left(4-x\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)

\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)

\(\Rightarrow\left(4-x\right).\left(4-x\right)=\left(-5\right).\left(-5\right)\)

\(\Rightarrow\left(4-x\right)^2=25\)

\(\Rightarrow\left(4-x\right)^2=5^2\)

\(\Rightarrow4-x=\pm5\)

\(\Rightarrow\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4-5\\x=4-\left(-5\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)

=>(4-x)^2=25

=>(x-4)^2=25

=>x-4=5 hoặc x-4=-5

=>x=-1 hoặc x=9

\(d,\dfrac{6}{5}-3:\dfrac{15}{4}=\dfrac{6}{5}-3\times\dfrac{4}{15}=\dfrac{6}{5}-\dfrac{4}{5}=\dfrac{2}{5}\)

\(b,\dfrac{2}{5}+\dfrac{4}{5}:4=\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{1}{4}=\dfrac{2}{5}+\dfrac{1}{5}=\dfrac{3}{5}\)

\(a,\dfrac{5}{8}+\dfrac{4}{3}=\dfrac{15}{24}+\dfrac{32}{24}=\dfrac{47}{24}\)

x/y=3/4

=>x/3=y/4

=>x/15=y/20

y/z=5/7

=>y/5=z/7

=>y/20=z/28

=>x/15=y/20=z/28=(2x+3y-z)/(2*15+3*20-28)=186/62=3

=>x=45; y=60; z=84

a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)

b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)

\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)

chắc b