Question

Giải các phương trình:

a)\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2}-25\)

b) x²+ 6x+9= 144

c) \(\frac{2-x}{2016}-1=\frac{1-x}{2017}-\frac{x}{2018}\)

( mink đag cần gấp)

Answer 1

b) \(x^2+6x+9=144\)

\(\Leftrightarrow\left(x+3\right)^2=12^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)

Answer 2

b, Ta có : \(x^2+6x+9=144\)

=> \(\left(x+3\right)^2=12^2\)

=> \(\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{9,-15\right\}\)

c, Ta có : \(\frac{2-x}{2016}-1=\frac{1-x}{2017}-\frac{x}{2018}\)

=> \(\frac{2-x}{2016}-1=\frac{1-x}{2017}+\frac{-x}{2018}\)

=> \(\frac{2-x}{2016}+1=\frac{1-x}{2017}+1+\frac{-x}{2018}+1\)

=> \(\frac{2-x}{2016}+\frac{2016}{2016}=\frac{1-x}{2017}+\frac{2017}{2017}+\frac{-x}{2018}+\frac{2018}{2018}\)

=> \(\frac{2018-x}{2016}=\frac{2018-x}{2017}+\frac{2018-x}{2018}\)

=> \(\frac{2018-x}{2016}-\frac{2018-x}{2017}-\frac{2018-x}{2018}=0\)

=> \(\left(2018-x\right)\left(\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)

=> \(2018-x=0\)

=> \(x=2018\)

Vậy phương trình có tập nghiệm là \(S=\left\{2018\right\}\)