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\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
=> \(\frac{2}{3}x=-\frac{29}{70}\)
=> \(x=-\frac{29}{70}:\frac{2}{3}\)
=> \(x=-\frac{29}{70}.\frac{3}{2}\)
=> \(x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
=> \(-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
=> \(-\frac{21}{13}x=-\frac{3}{3}\)
=> \(-\frac{21}{13}x=1\)
=> \(x=1:\left(-\frac{21}{13}\right)\)
=> \(x=-\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
=> \(\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2+1,5\\x=-2+1,5\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)(T/M)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
=> \(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
=> \(=>\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}\\x=-\frac{1}{2}-\frac{3}{4}\end{matrix}\right.=>\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{matrix}\right.\)(T/M)
HỌC TỐT
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
\(\Leftrightarrow\frac{2}{3}x=-\frac{29}{70}\)
\(\Leftrightarrow x=-\frac{29}{70}:\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-1\)
\(\Leftrightarrow x=-1:\left(-\frac{21}{13}\right)\)
\(\Leftrightarrow x=\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{5}{4}\end{matrix}\right.\)
a) Ta có: \(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)=\frac{2}{3}\)
⇒\(\frac{2}{3}:\left(x-1\right)=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
⇒\(x-1=\frac{2}{3}:\frac{1}{6}=\frac{2}{3}\cdot6=4\)
hay x=5
Vậy: x=5
b) \(5,4-3\left[x-120\%\right]=\frac{3}{10}\)
⇔\(\frac{27}{5}-3\cdot\left(x-\frac{6}{5}\right)=\frac{3}{10}\)
⇔\(3\left(x-\frac{6}{5}\right)=\frac{27}{5}-\frac{3}{10}=\frac{51}{10}\)
hay \(x-\frac{6}{5}=\frac{51}{10}\cdot\frac{1}{3}=\frac{17}{10}\)
⇔\(x=\frac{17}{10}+\frac{6}{5}=\frac{29}{10}\)
Vậy: \(x=\frac{29}{10}\)
c) \(10\cdot3^{x+2}-3^x=89\)
\(\Leftrightarrow10\cdot3^2\cdot3^x-3^x=89\)
\(\Leftrightarrow3^x\left(90-1\right)=89\)
\(\Leftrightarrow3^x=1\)
hay x=0
Vậy: x=0
d) \(5\cdot\left(x-0,2\right)=3x+\left(\frac{-2}{3}\right)^3\)
⇒\(5\cdot\left(x-\frac{1}{5}\right)=3x+\frac{-8}{27}\)
\(\Leftrightarrow5x-1-3x-\frac{-8}{27}=0\)
\(\Leftrightarrow2x-\frac{19}{27}=0\)
\(\Leftrightarrow2x=\frac{19}{27}\)
hay \(x=\frac{\frac{19}{27}}{2}=\frac{19}{27}\cdot\frac{1}{2}=\frac{19}{54}\)
Vậy: \(x=\frac{19}{54}\)
e) \(\left(2x+\frac{3}{4}\right)^2-1,5=2\frac{1}{2}\)
\(\Leftrightarrow\left(2x+\frac{3}{4}\right)^2=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{3}{2}=-2\\2x+\frac{3}{2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2-\frac{3}{2}\\2x=2-\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{7}{2}\\2x=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{2}\cdot\frac{1}{2}\\x=\frac{1}{2}\cdot\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{7}{4};\frac{1}{4}\right\}\)
Câu 2 đây:
\(|x^2+|x-1||=x^2+2\)
\(\Rightarrow\orbr{\begin{cases}x^2+\left|x-1\right|=x^2+2\\x^2+\left|x-1\right|=-x^2-2\left(l\right)\end{cases}}\)
\(\Rightarrow\left|x-1\right|=2\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
a) \(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+0,5}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{2}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2012}{2013}\)
\(=\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\right):\frac{2012}{2013}\)
\(=\left(\frac{2}{7}-\frac{2}{7}\right):\frac{2012}{2013}\)
\(=0\)
a ) \(\left|x-2\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-2=2\\x-2=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2+2\\x=-2+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)
Vậy \(x=4\) hoặc \(x=0\)
b ) \(\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}+\frac{4}{5}\\x=-\frac{3}{4}+\frac{4}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{15}{20}+\frac{16}{20}\\-\frac{15}{20}+\frac{16}{20}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{31}{20}\\x=\frac{1}{20}\end{cases}}\)
Vậy \(x=\frac{31}{20}\) hoặc \(x=\frac{1}{20}\)
c ) \(6-\left|\frac{1}{2}-x\right|=\frac{2}{5}\)
\(\left|\frac{1}{2}-x\right|=6-\frac{2}{5}\)
\(\left|\frac{1}{2}-x\right|=\frac{28}{5}\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}-x=\frac{28}{5}\\\frac{1}{2}-x=-\frac{28}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}-\frac{28}{5}\\x=\frac{1}{2}-\left(-\frac{28}{5}\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{51}{10}\\x=\frac{61}{10}\end{cases}}\)
Vậy \(x=-\frac{51}{10}\) hoặc \(x=\frac{61}{10}\)
d ) \(\frac{1}{5}-\left|\frac{1}{5}.x\right|=\frac{1}{5}\)
\(\left|\frac{1}{5}.x\right|=\frac{1}{5}-\frac{1}{5}\)
\(\left|\frac{1}{5}.x\right|=0\)
\(\Rightarrow\frac{1}{5}.x=0\)
\(x=0:\frac{1}{5}\)
\(x=0\)
Vậy \(x=0\)
e ) \(-2,5+\left|3x+5\right|=-1,5\)
\(\left|3x+5\right|=-1,5-\left(-2,5\right)\)
\(\left|3x+5\right|=1\)
\(\Rightarrow\orbr{\begin{cases}3x+5=1\\3x+5=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=1-5\\3x=-1-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=-4\\3x=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4:3\\x=-6:3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{4}{3}\\x=-2\end{cases}}\)
Vậy \(x=-\frac{4}{3}\) hoặc \(x=-2\)
g ) \(\left(\frac{2}{3}x-1\right)+\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\left(\frac{2}{3}x-1\right)=0\\\left(\frac{3}{4}x+\frac{1}{2}\right)=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-1=0\\\frac{3}{4}x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x=0+1\\\frac{3}{4}x=0-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x=1\\\frac{3}{4}x=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1:\frac{2}{3}\\x=-\frac{1}{2}:\frac{3}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{2}{3}\end{cases}}\)
Vậy \(x=\frac{3}{2};x=-\frac{2}{3}\)
Bạn bạn hơi dài nhưng đã xong
Chúc bạn học tốt !!!
Trả lời:
a, | x - 2 | = 2
=> x - 2 = 2 hoặc x - 2 = -2
=> x = 4 x = 0
Vậy x = 4; x = 0
b, \(\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
\(\Rightarrow x-\frac{4}{5}=\frac{3}{4}\)hoặc \(x-\frac{4}{5}=\frac{-3}{4}\)
\(\Rightarrow x=\frac{31}{20}\)hoặc \(x=\frac{1}{20}\)
Vậy \(x=\frac{31}{20};x=\frac{1}{20}\)
c, \(6-\left|\frac{1}{2}-x\right|=\frac{2}{5}\)
\(\Rightarrow\left|\frac{1}{2}-x\right|=\frac{28}{5}\)
\(\Rightarrow\frac{1}{2}-x=\frac{28}{5}\)hoặc \(\frac{1}{2}-x=\frac{-28}{5}\)
\(\Rightarrow x=\frac{-51}{10}\)hoặc \(x=\frac{61}{10}\)
Vậy \(x=\frac{-51}{10};x=\frac{61}{10}\)
d, \(\frac{1}{5}-\left|\frac{1}{5}\cdot x\right|=\frac{1}{5}\)
\(\Rightarrow\left|\frac{1}{5}\cdot x\right|=0\)
\(\Rightarrow\frac{1}{5}\cdot x=0\)
\(\Rightarrow x=0\)
Vậy x = 0
e, \(-2,5+\left|3x+5\right|=-1,5\)
\(\Rightarrow\left|3x+5\right|=1\)
\(\Rightarrow3x+5=1\)hoặc \(3x+5=-1\)
\(\Rightarrow3x=-4\)hoặc \(3x=-6\)
\(\Rightarrow x=\frac{-4}{3}\)hoặc \(x=-2\)
Vậy \(x=\frac{-4}{3};x=-2\)
g, \(\left(\frac{2}{3}x-1\right)+\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\frac{2}{3}x-1+\frac{3}{4}x+\frac{1}{2}=0\)
\(\Rightarrow\frac{17}{12}x-\frac{1}{2}=0\)
\(\Rightarrow\frac{17}{12}x=\frac{1}{2}\)
\(\Rightarrow x=\frac{6}{17}\)
Vậy \(x=\frac{6}{17}\)
a) x=4/7 - 1/3=19/21
b) /x-5/=7 -->x-5=7 hoặc x-5=-7
--> x=12 hoặc x= -2
a, \(\left(2x-1\right)=-8\)
\(2x=-8+1\)
\(2x=-7\)
\(x=-7:2\)
\(x=-3,5\)
a) (2x - 1) = -8
⇒ 2x = -8 + 1
⇒ 2x = -7
b) (3x - 2)\(^2\) = \(\frac{1}{49}\)
Ta có: \(\frac{1}{49}\) = \(\frac{1}{7}\). \(\frac{1}{7}\) hoặc \(\frac{1}{49}\) = \(\frac{-1}{7}\). \(\frac{-1}{7}\)
TH1: 3x - 2 = \(\frac{1}{7}\) TH2: 3x - 2 = \(\frac{-1}{7}\)
⇒ 3x = \(\frac{1}{7}\)+2 ⇒ 3x = \(\frac{-1}{7}\)+2
⇒ 3x = \(\frac{15}{7}\) ⇒ 3x = \(\frac{13}{7}\)
⇒ x = \(\frac{5}{7}\) ⇒ x = \(\frac{13}{21}\)
Vậy: x = \(\frac{5}{7}\) hoặc x = \(\frac{13}{21}\)
\(\frac{x+2}{x+6}=\frac{3}{x+1}\)
\(\Rightarrow\left(x+2\right)\left(x+1\right)=3\left(x+6\right)\)
\(\Rightarrow x^2+x+2x+2=3x+18\)
\(\Rightarrow x^2+x+2x-3x=18-2\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=\pm4\)
các phần còn lại tương tự :)
a)\(\frac{x+2}{x+6}\) =\(\frac{3}{x+1}\)
<=>\(\frac{\left(x+2\right)\left(x+1\right)}{\left(x+6\right)\left(x+1\right)}\) =\(\frac{3\left(x+6\right)}{\left(x+1\right)\left(x+6\right)}\)
=> ( x+2) ( x+1) = 3(x+6)
<=> x2 +3x +3 = 3x +18
<=> x2 +3x -3x = 18 -3
<=> x2 = 15
=> x = \(\sqrt{15}\)
Vậy x=\(\sqrt{15}\)
b)
a) Quy đồng lên đi.
b) \(\frac{x+2}{0.5}=\frac{2x+1}{2}\Leftrightarrow\frac{x+2}{\left(\frac{1}{2}\right)}=\frac{2x+1}{2}\)
\(\Leftrightarrow2x+4=\frac{2x+1}{2}\Leftrightarrow4x+8=2x+1\)
\(\Leftrightarrow x=-\frac{7}{2}\)
c) \(\Leftrightarrow\left|x+\frac{1}{5}\right|=6\). VỚi x >= -1/5 thì:
\(x+\frac{1}{5}=6\Leftrightarrow x=\frac{29}{5}\left(TM\right)\)
Với x < -1/5 thì \(-x-\frac{1}{5}=6\Leftrightarrow x=-\frac{31}{5}\left(TM\right)\)
d) TƯơng tự ý a, quy đồng lên thôi (mẫu chung là 24 thì phải)
c) \(\left|x+\frac{1}{5}\right|-4=2\)
=> \(\left|x+\frac{1}{5}\right|=2+4\)
=> \(\left|x+\frac{1}{5}\right|=6\)
=> \(\left\{{}\begin{matrix}x+\frac{1}{5}=6\\x+\frac{1}{5}=-6\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=6-\frac{1}{5}\\x=\left(-6\right)-\frac{1}{5}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{29}{5}\\x=-\frac{31}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{29}{5};-\frac{31}{5}\right\}\).
Mình chỉ làm câu c) thôi nhé.
Chúc bạn học tốt!
\(\frac{-2}{1,5}=\frac{-2.2}{1,5.2}=\frac{-4}{3}=\frac{x-1}{3}\)
\(\Rightarrow x-1=-4\)\(\Rightarrow x=-3\)
Vậy \(x=-3\)