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Ta có : \(\frac{y-1}{2}=\frac{5-2y}{3}\)
=> \(3\left(y-1\right)=2\left(5-2y\right)\)
=> 3y - 3 = 10 - 4y
=> 3y + 4y = 10 + 3
=> 7y = 13
=> y = 13/7
![](https://rs.olm.vn/images/avt/0.png?1311)
2 (x-1) - 5 (x+2) = -10
2x-2 - 5x+10 = -10
2x-5x-2+10=-10
2x-5x=-10-10+2
-3x=-18
x=6
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a) \(\left|x-2\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3+2=5\\x=-3+2=-1\end{cases}}\)
Vậy x=5 hoặc x=-1
Ta có :\(\left|x-2\right|=3\)
\(=>\orbr{\begin{cases}x-2=-3\\x-2=3\end{cases}}\)
\(=>\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
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\(\frac{x-1}{4}=\frac{2x+1}{5}\)
\(\Rightarrow5\left(x-1\right)=4\left(2x+1\right)\)
\(\Rightarrow5x-5=8x+4\)
\(\Rightarrow5x-8x=4+5\)
\(\Rightarrow-3x=9\)
\(\Rightarrow x=-3\)
vậy_
\(\frac{x+2}{x-1}=\frac{x-3}{x+1}\)
\(\Rightarrow\left(x+2\right)\left(x+1\right)=\left(x-1\right)\left(x-3\right)\)
\(\Rightarrow x^2+x+2x+2=x^2-3x-x+3\)
\(\Rightarrow x^2+x+2x-x^2+3x+x=3-2\)
\(\Rightarrow7x=1\)
\(\Rightarrow x=\frac{1}{7}\)
vậy_
\(x:\dfrac{4}{x}=\dfrac{5}{2}\)
\(\Leftrightarrow x^2:4=\dfrac{5}{2}\)
=>x2=10
hay \(x=\pm\sqrt{10}\)