1) Tìm x:
a) \(-\frac{2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)
b) \(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
CÁC BẠN GIÚP MIK 2 BÀI NÀY NHÉ
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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)\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
=>\(4x-x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)
=> \(3x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)
=>\(x=-\left(-\frac{9}{2}\right)+\frac{1}{2}=5\)
a.
\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)
TH1:
\(x+\frac{1}{2}=0\)
\(x=-\frac{1}{2}\)
TH2:
\(x-\frac{3}{4}=0\)
\(x=\frac{3}{4}\)
Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)
b.
\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)
TH1:
\(\frac{1}{2}x-3=0\)
\(\frac{1}{2}x=3\)
\(x=3\div\frac{1}{2}\)
\(x=3\times2\)
\(x=6\)
TH2:
\(\frac{2}{3}x+\frac{1}{2}=0\)
\(\frac{2}{3}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\frac{2}{3}\)
\(x=-\frac{1}{2}\times\frac{3}{2}\)
\(x=-\frac{3}{4}\)
Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)
c.
\(\frac{2}{3}-\frac{1}{3}\times\left(x-\frac{3}{2}\right)-\frac{1}{2}\times\left(2x+1\right)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5-\frac{2}{3}\)
\(-\frac{4}{3}x=\frac{13}{3}\)
\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)
\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)
\(x=-\frac{13}{4}\)
d.
\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)
\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)
\(x=5\)
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
( 3x - 1/2 ) + ( 1/2y + 3/5 ) = 0
=> ( 3 x - 1/2 ) = 0
3x = 0+1/2
3x = 1/2
x = 1/2 : 3
x = 1/6
=> ( 1/2 y + 3/5 ) = 0
1/2y = 0 - 3/5
1/2 y = -3/5
y = -3/5 : 1/2
y = -6/5
a) 4/3 - x = 3/5 + 1/2
=> 4/3 - x= 0,8
=> x = 4/3 + 0/8
=> x = 5/8
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 ) \(-\frac{2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=-\frac{2}{3}-\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=\frac{-13}{6}\)
\(\left(2x-5\right)=-\frac{13}{6}:\frac{1}{3}\)
\(\left(2x-5\right)=-\frac{13}{6}.\frac{3}{1}\)
\(\left(2x-5\right)=-\frac{13}{2}\)
\(2x=-\frac{13}{2}+5\)
\(2x=-\frac{3}{2}\)
\(\Rightarrow x=-\frac{3}{2}:2\)
\(\Rightarrow x=-\frac{3}{2}.\frac{1}{2}\)
\(\Rightarrow x=-\frac{3}{4}\)