Question

[Trâm Nguyễn]

tìm x:

a) x-2/-4  =  -9/5-x

b) x-5/3  = -12/5-x

c) (2x-1)^2 - (2x-1) = 0

Answer 1

[HT.Phong (9A5)]

a) 

\(x-\dfrac{2}{-4}=\dfrac{-9}{5}-x\\ x+x=\dfrac{-9}{5}+\dfrac{2}{-4}\\ 2x=\dfrac{-9}{5}-\dfrac{1}{2}\\ 2x=\dfrac{-23}{10}\\ x=\dfrac{-23}{20}\)

b) 

\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\\ \dfrac{x-5}{3}=\dfrac{12}{x-5}\\ \left(x-5\right)^2=3\cdot12\\ \left(x-5\right)^2=36\\ \left(x-5\right)^2=6^2\)

TH1: x - 5 = 6 => x = 6 + 5 = 11

TH2: x - 5 = -6 => x = -6 + 5 = -1

c) 

\(\left(2x-1\right)^2-\left(2x-1\right)=0\\ \Rightarrow\left(2x-1\right)\left(2x-1-1\right)=0\\ \Rightarrow\left(2x-1\right)\left(2x-2\right)=0\)

TH1: 2x - 1=0 => 2x = 1 => x =1/2

TH2: 2x - 2=0 => 2x = 2 => x = 1 

Answer 2

[Nguyễn Lê Phước Thịnh]

a:

ĐKXĐ: x<>5

 \(\dfrac{x-2}{-4}=\dfrac{-9}{5-x}\)

=>\(\dfrac{\left(x-2\right)}{-4}=\dfrac{9}{x-5}\)

=>\(\left(x-2\right)\left(x-5\right)=-4\cdot9=-36\)

=>\(x^2-7x+10+36=0\)

=>\(x^2-7x+46=0\)

\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot46=49-184=-135< 0\)

=>Phương trình vô nghiệm

b: ĐKXĐ: x<>5

\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\)

=>\(\dfrac{x-5}{3}=\dfrac{12}{x-5}\)

=>\(\left(x-5\right)\left(x-5\right)=3\cdot12=36\)

=>\(\left(x-5\right)^2=36\)

=>\(\left[{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

c: \(\left(2x-1\right)^2-\left(2x-1\right)=0\)

=>(2x-1)(2x-1-1)=0

=>(2x-1)(2x-2)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)