Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
a.
\(-\frac{2}{3}-\frac{1}{3}\times\left(2x-5\right)=\frac{3}{2}\)
\(-\frac{2}{3}-\frac{2}{3}x+\frac{5}{3}=\frac{3}{2}\)
\(\left(-\frac{2}{3}+\frac{5}{3}\right)-\frac{2}{3}x=\frac{3}{2}\)
\(\frac{3}{3}-\frac{2}{3}x=\frac{3}{2}\)
\(1-\frac{2}{3}x=\frac{3}{2}\)
\(\frac{2}{3}x=1-\frac{3}{2}\)
\(\frac{2}{3}x=\frac{2}{2}-\frac{3}{2}\)
\(\frac{2}{3}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\frac{2}{3}\)
\(x=-\frac{1}{2}\times\frac{3}{2}\)
\(x=-\frac{3}{4}\)
b.
\(\frac{1}{3}x+\frac{2}{5}\times\left(x-1\right)=0\)
\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(x\times\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)
\(x\times\left(\frac{5}{15}+\frac{6}{15}\right)=\frac{2}{5}\)
\(x\times\frac{11}{15}=\frac{2}{5}\)
\(x=\frac{2}{5}\div\frac{11}{15}\)
\(x=\frac{2}{5}\times\frac{15}{11}\)
\(x=\frac{6}{11}\)
Chúc bạn học tốt
a ) \(-\frac{2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=-\frac{2}{3}-\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=\frac{-13}{6}\)
\(\left(2x-5\right)=-\frac{13}{6}:\frac{1}{3}\)
\(\left(2x-5\right)=-\frac{13}{6}.\frac{3}{1}\)
\(\left(2x-5\right)=-\frac{13}{2}\)
\(2x=-\frac{13}{2}+5\)
\(2x=-\frac{3}{2}\)
\(\Rightarrow x=-\frac{3}{2}:2\)
\(\Rightarrow x=-\frac{3}{2}.\frac{1}{2}\)
\(\Rightarrow x=-\frac{3}{4}\)
7/4.x+3/2=-4/5
7/4.x=-4/5-3/2
7/4.x=-23/10
x=-23/10:7/4
x=-46/35
vậy x=-46/35
1/4+3/4.x=3/4
1.x=3/4
x=3/4:1
x=3/4
vậy x=3/4
x.(1/4+1/5)-(1/7+1/8)=0
x.9/20-15/56=0
x.51/280=0
x=0:51/280
x=0
vậy x=0
3/35-(3/5+x)=2/7
(3/5+x)=3/35-2/7
(3/35+x)=-1/5
x=-1/5-3/5
x=-4/5
vậy x=-4/5
\(a,1\frac{3}{4}.x+1\frac{1}{2}=\frac{4}{5}\)
\(\frac{7}{4}.x=\frac{4}{5}-\frac{3}{2}\)
\(\frac{7}{4}.x=\frac{-7}{10}\)
\(x=\frac{-7}{10}:\frac{7}{4}\)
\(x=\frac{-2}{5}\)
\(b,\frac{1}{4}+\frac{3}{4}.x=\frac{3}{4}\)
\(\frac{3}{4}.x=\frac{3}{4}-\frac{1}{4}\)
\(\frac{3}{4}.x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{3}{4}\)
\(x=\frac{2}{3}\)
\(c,x.\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)
\(x.\frac{9}{20}-\frac{15}{56}=0\)
\(x.\frac{9}{20}=\frac{15}{56}\)
\(x=\frac{15}{56}:\frac{9}{20}\)
\(x=\frac{25}{42}\)
\(d,\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{-1}{5}\)
\(x=\frac{-1}{5}-\frac{3}{5}\)
\(x=\frac{-4}{5}\)
Học tốt
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) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)\(5\)
=> \(\frac{2}{3}-\left(\frac{1}{3}x-\frac{1}{2}\right)-\left(x+\frac{1}{2}\right)=5\)
=>\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
=>\(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5\)
=>\(\frac{2}{3}-\frac{4}{3}x=5\)
=>\(\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)
=>\(x=-\frac{13}{3}:\frac{4}{3}=-\frac{13}{4}\)
b)\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
=>\(4x-x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)
=> \(3x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)
=>\(x=-\left(-\frac{9}{2}\right)+\frac{1}{2}=5\)
\(\text{a) }\left|x-\frac{1}{3}\right|=\left|2x-3\right|\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2x-3\\x-\frac{1}{3}=3-2x\end{cases}\Rightarrow\orbr{\begin{cases}x-2x=-3+\frac{1}{3}\\x+2x=3+\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}-x=-\frac{8}{3}\\3x=\frac{10}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{8}{3}\\x=\frac{10}{9}\end{cases}}}\)
\(\text{Vậy }S=\left\{\frac{8}{3};\frac{10}{9}\right\}\)
\(\text{b) }\left(3\frac{5}{7}.x-1\frac{5}{7}.x\right)-\frac{1}{3}=\frac{2}{3}\)
\(\Leftrightarrow\left(\frac{26}{7}.x-\frac{12}{7}.x\right)=\frac{2}{3}+\frac{1}{3}\)
\(\Leftrightarrow x.\left(\frac{26}{7}-\frac{12}{7}\right)=1\)
\(\Leftrightarrow x.2=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
\(\text{Vậy }S=\left\{\frac{1}{2}\right\}\)
a, \(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}=-\frac{1}{5}\)
\(x=-\frac{1}{5}-\frac{3}{5}\)
\(x=-\frac{4}{5}\)
b,\(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
=> \(\left(5x-1\right)=0\) hoặc \(\left(2x-\frac{1}{3}\right)=0\)
=> \(5x=1\) hoặc \(2x=\frac{1}{3}\)
=> \(x=\frac{1}{5}\) hoặc \(x=\frac{1}{6}\)