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a: Để 2x+1/5=2
thì 2x+1=10
=>2x=9
hay x=9/2
Để (2x+1)/5=-2
thì 2x+1=-10
=>2x=-11
hay x=-11/2
Để (2x+1)/5=0 thì 2x+1=0
hay x=-1/2
Để (2x+1)/5=4 thì 2x+1=20
=>2x=19
hay x=19/2
b: Để (x+1)/7=0 thì x+1=0
hay x=-1
Để (3x+3)/5=0 thì 3x+3=0
hay x=-1
a: =>(3x+6)(x+5)<0
=>(x+2)(x+5)<0
=>-5<x<-2
b: \(\Leftrightarrow\dfrac{x+2}{x+1}>0\)
=>x>-1 hoặc x<-2
c: \(\Leftrightarrow\dfrac{x-1}{2x+5}-1>0\)
\(\Leftrightarrow\dfrac{x-1-2x-5}{2x+5}>0\)
\(\Leftrightarrow\dfrac{x+6}{2x+5}< 0\)
=>x>-5/2 hoặc x<-6
Để : \(\left(2x+\frac{1}{8}\right)\left(3x-\frac{4}{5}\right)>0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{1}{8}>0;3x-\frac{4}{5}>0\\2x+\frac{1}{8}< 0;3x-\frac{4}{5}< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x>-\frac{1}{8};3x>\frac{4}{5}\\2x< \frac{-1}{8};3x< \frac{4}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x>-\frac{1}{4};x>\frac{12}{5}\Rightarrow x>\frac{12}{5}\\x< -\frac{1}{4};x< \frac{12}{5}\Rightarrow x< -\frac{1}{4}\end{cases}}\)
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
a, Vì \(\left|3x-2y\right|\ge0;\left|3y-4z\right|\ge0\Rightarrow\left|3x-2y\right|+\left|3y-4z\right|\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3x-2y=0\\3y-4z=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2y\\3y=4z\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{4}=\frac{z}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{x}{8}=\frac{y}{12}\\\frac{y}{12}=\frac{z}{9}\end{cases}\Leftrightarrow}\frac{x}{8}=\frac{y}{12}=\frac{z}{9}}\)
\(\Leftrightarrow\frac{x}{8}=\frac{2y}{24}=\frac{3z}{27}=\frac{x-2y+3z}{8-24+27}=\frac{5}{11}\)
từ đây tìm x,y,z
b,Ta có: \(\frac{2x+3}{2}=\frac{3x-6}{5}\Rightarrow5\left(2x+3\right)=2\left(3x-6\right)\Rightarrow10x+15=6x-12\Rightarrow4x=-27\Rightarrow x=\frac{-27}{4}\)
Thay x=-27/4 vào \(\frac{3x-6}{5}=\frac{3x+3y+1}{3x}\), ta được:
\(\frac{3\cdot\left(\frac{-27}{4}\right)-6}{5}=\frac{3.\left(\frac{-27}{4}\right)+3y+1}{3.\left(\frac{-27}{4}\right)}\)
\(\Rightarrow\frac{-21}{4}=\frac{\frac{-77}{4}+3y}{\frac{-81}{4}}\Rightarrow\frac{-77}{4}+3y=\frac{1701}{16}\Rightarrow3y=\frac{2009}{16}\Rightarrow y=\frac{2009}{48}\)
Vậy x=-27/4,y=2009/48
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự