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5 tháng 8 2022
2: =>2x-1/4=5/6-1/2x
=>5/2x=5/6+1/4=13/12
=>x=13/30
3: =>3x-5/6=2/3-1/2x
=>3,5x=2/3+5/6=4/6+5/6=9/6=3,2
hay x=32/35
3 tháng 7 2018
a) \(\frac{2x-1}{2}=\frac{x+5}{4}\Leftrightarrow4\left(2x-1\right)=2\left(x+5\right)\)
\(\Leftrightarrow8x-4=2x+10\)
\(\Leftrightarrow6x=14\)
\(\Leftrightarrow x=\frac{7}{3}\)
b) \(6x+3\left(x-\frac{1}{2}\right)=5-\frac{1}{2}x\)
\(\Rightarrow6x+3x-\frac{3}{2}=5-\frac{1}{2}x\)
\(\Rightarrow6x+3x+\frac{1}{2}x=5+\frac{3}{2}\)
\(\Leftrightarrow9,5x=6,5\)
\(\Leftrightarrow x=\frac{13}{19}\)
c) \(3\left(x-2\right)=\frac{1}{2}\left(4-2x\right)\)
\(\Rightarrow3x-6=2-x\)
\(\Rightarrow4x=8\Rightarrow x=2\)
26 tháng 9 2018
\(a)\)\(\left(\frac{3}{5}\right)^{2x+1}=\frac{81}{625}\)
\(\Leftrightarrow\)\(\left(\frac{3}{5}\right)^{2x+1}=\left(\frac{3}{5}\right)^4\)
\(\Leftrightarrow\)\(2x+1=4\)
\(\Leftrightarrow\)\(x=\frac{3}{2}\)
Vậy \(x=\frac{3}{2}\)
\(b)\)\(\left(\frac{2}{3}\right)^x.\left(\frac{2}{3}\right)^3=\frac{32}{243}\)
\(\Leftrightarrow\)\(\left(\frac{2}{3}\right)^{x+3}=\left(\frac{2}{3}\right)^5\)
\(\Leftrightarrow\)\(x+3=5\)
\(\Leftrightarrow\)\(x=2\)
Vậy \(x=2\)
\(c)\)\(\left(2x-1\right)^2=\left(2x-1\right)^3\)
\(\Leftrightarrow\)\(\left(2x-1\right)^3-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\)\(\left(2x-1\right)^2\left(2x-1-1\right)=0\)
\(\Leftrightarrow\)\(\left(2x-1\right)^2\left(2x-2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(2x-1\right)^2=0\\2x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}}\)
Vậy \(x=\frac{1}{2}\) hoặc \(x=1\)
Chúc bạn học tốt ~
DH
8 tháng 10 2019
a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)
\(\frac{1}{2}-x=\frac{57}{28}\)
\(x=-\frac{43}{28}\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
DH
8 tháng 10 2019
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy ...
VM
8 tháng 10 2019
a) \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=\frac{-11}{4}\)
\(\Rightarrow\left(\frac{1}{2}-x\right)=\left(-\frac{5}{7}\right)+\frac{11}{4}\)
\(\Rightarrow\frac{1}{2}-x=\frac{57}{28}\)
\(\Rightarrow x=\frac{1}{2}-\frac{57}{28}\)
\(\Rightarrow x=-\frac{43}{28}\)
Vậy \(x=-\frac{43}{28}.\)
b) \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=20+5\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5+1=6\\2x=\left(-5\right)+1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6:2\\x=\left(-4\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{3;-2\right\}.\)
d) \(\frac{x-6}{4}=\frac{4}{x-6}\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=4.4\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=16\)
\(\Rightarrow\left(x-6\right)^2=16\)
\(\Rightarrow x-6=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+6\\x=\left(-4\right)+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{10;2\right\}.\)
Chúc bạn học tốt!
VM
18 tháng 9 2019
1) \(\frac{1}{3}x-\frac{2}{5}=\frac{1}{3}\)
⇒ \(\frac{1}{3}x=\frac{1}{3}+\frac{2}{5}\)
⇒ \(\frac{1}{3}x=\frac{11}{15}\)
⇒ \(x=\frac{11}{15}:\frac{1}{3}\)
⇒ \(x=\frac{11}{5}\)
Vậy \(x=\frac{11}{5}.\)
2) \(2,5:7,5=x:\frac{3}{5}\)
⇒ \(\frac{5}{2}:\frac{15}{2}=x:\frac{3}{5}\)
⇒ \(\frac{1}{3}=x:\frac{3}{5}\)
⇒ \(x=\frac{1}{3}.\frac{3}{5}\)
⇒ \(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}.\)
4) \(\left|x\right|+\left|x+2\right|=0\)
Có: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{matrix}\right.\forall x.\)
⇒ \(\left|x\right|+\left|x+2\right|=0\)
⇒ \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=0-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vô lí vì \(x\) không thể nhận cùng lúc 2 giá trị khác nhau.
⇒ \(x\in\varnothing\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
10) \(5-\left|1-2x\right|=3\)
⇒ \(\left|1-2x\right|=5-3\)
⇒ \(\left|1-2x\right|=2\)
⇒ \(\left[{}\begin{matrix}1-2x=2\\1-2x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=1-2=-1\\2x=1+2=3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-1\right):2\\x=3:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
DH
18 tháng 9 2019
9, \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
\(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)
\(10=26:\left(2x-1\right)\)
\(2x-1=26:10\)
\(2x-1=2,6\)
\(2x=2,6+1\)
\(2x=3,6\)
\(x=3,6:2\)
\(x=1,8\)
KN
25 tháng 8 2020
a. \(\frac{2x+3}{15}=\frac{7}{5}\)
\(\Leftrightarrow5\left(2x+3\right)=15.7\)
\(\Leftrightarrow10x+15=105\)
\(\Leftrightarrow10x=90\)
\(\Leftrightarrow x=9\)
b. \(\frac{x-2}{9}=\frac{8}{3}\)
\(\Leftrightarrow3\left(x-2\right)=9.8\)
\(\Leftrightarrow3x-6=72\)
\(\Leftrightarrow3x=78\)
\(\Leftrightarrow x=26\)
c. \(\frac{-8}{x}=\frac{-x}{18}\)
\(\Leftrightarrow-x^2=-144\)
\(\Leftrightarrow x^2=12^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
Mấy câu kia tương tự
A
25 tháng 8 2020
d, \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=6x-12\Leftrightarrow4x=-27\Leftrightarrow x=-\frac{27}{4}\)
e, \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x=132\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f, \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x=10\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x+2\right)\left(2x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{5}{2}\end{cases}}\)
g, \(\left(2x-1\right)\left(2x+1\right)=63\Leftrightarrow4x^2+2x-2x-1=63\Leftrightarrow4x^2-64=0\)
\(\Leftrightarrow x^2=16\Leftrightarrow x=\pm4\)
h, \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow\left(10x+5\right)\left(x+1\right)=30\Leftrightarrow10x^2+10x+5x+5=30\)
\(\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(2x+5\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=1\end{cases}}\)
KN
5 tháng 6 2019
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)