a) \(\frac{x}{x+1}=\frac{1}{2}\)
=> 2x = x + 1
=> 2x - x = 1
=> x = 1
b) \(\frac{x}{2}=\frac{x}{3}\)
=> 3x = 2x
=> 3x - 2x = 0
=> x = 0
c) \(\frac{x+1}{2}=\frac{x+1}{2017}\)
=> \(2017\left(x+1\right)=2\left(x+1\right)\)
=> 2017x + 2017 = 2x + 2
=> 2017x - 2x = 2 - 2017
=> 2015x = -2015
=> x = -2015 : 2015
=> x = -1
i) \(\frac{3}{x}=\frac{x}{2017}\)
=> x2 = 2017.3
=> x2 = 6051
=> \(\orbr{\begin{cases}x=\sqrt{6051}\\x=-\sqrt{6051}\end{cases}}\)
còn lại tự lm
\(a,\frac{x}{x+1}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow x=\frac{1}{2}x+\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow x=1\)
\(b,\frac{x}{2}=\frac{x}{3}\)
\(\Rightarrow x=\frac{x}{3}.2\)
\(\Rightarrow x=\frac{2x}{3}\)
\(\Rightarrow3x=2x\)
\(\Rightarrow x=0\)
\(c,\frac{x+1}{2}=\frac{x+1}{2017}\)
\(\Rightarrow x+1=\frac{x+1}{2017}.2\)
\(\Rightarrow x+1=\frac{2x+2}{2017}\)
\(\Rightarrow2017x+2017=2x+2\)
\(\Rightarrow2017x-2x=2-2017\)
\(\Rightarrow2015x=-2015\)
\(\Rightarrow x=-1\)
\(i,\frac{3}{x}=\frac{x}{2017}\)
\(\Rightarrow x=3:\frac{x}{2017}\)
\(\Rightarrow x=\frac{6051}{x}\)
\(\Rightarrow x^2=6051\)
\(\Rightarrow x=\sqrt{6051}\)
\(o,\frac{x}{3}=\frac{x+1}{2}\)
\(\Rightarrow x=\frac{x+1}{2}.3\)
\(\Rightarrow x=\frac{3x+3}{2}\)
\(\Rightarrow2x=3x+3\)
\(\Rightarrow-x=3\)
\(\Rightarrow x=-3\)
\(m,\frac{x+1}{2}=\frac{x+2}{3}\)
\(\Rightarrow x+1=\frac{x+2}{3}.2\)
\(\Rightarrow x+1=\frac{2x+4}{3}\)
\(\Rightarrow3x+3=2x+4\)
\(\Rightarrow x=1\)
\(p,\frac{x+1}{2}=x\)
\(\Rightarrow2x=x+1\)
\(\Rightarrow x=1\)
\(m,\frac{2}{x}=\frac{x}{8}\)
\(\Rightarrow x=2:\frac{x}{8}\)
\(\Rightarrow x=\frac{16}{x}\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=4\)
\(Q,\frac{x^2}{2}=\frac{8}{x^2}\)
\(\Rightarrow x^2=\frac{8}{x^2}.2\)
\(\Rightarrow x^2=\frac{16}{x^2}\)
\(\Rightarrow x^4=16\)
\(\Rightarrow x=2\)
\(r,\frac{x^3}{2}=\frac{32}{x}\)
\(\Rightarrow x^3=\frac{32}{x}.2\)
\(\Rightarrow x^3=\frac{64}{x}\)
\(\Rightarrow x^4=64\)
\(\Rightarrow x=\sqrt[4]{64}\)
