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\(a.\)
\(\dfrac{1}{3}\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+\dfrac{3}{5}=-5x+x\)
\(\Rightarrow-4x=-\dfrac{37}{30}\)
\(\Rightarrow4x=\dfrac{37}{30}\)
\(\Rightarrow x=\dfrac{37}{120}\)
\(b.\)
\(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{3}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{3}{2}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{2}\)
\(c.\)
\(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\)
\(\Rightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\)
\(\Rightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)
\(\Rightarrow x=\dfrac{19}{21}\)
\(d.\)
\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{39}{10}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{39}{10}+1\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{49}{10}\)
\(\Rightarrow x=\dfrac{49}{5}\)
\(e.\)
\(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=\dfrac{3}{5}\\4x+3=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{12}{5}\\4x=-\dfrac{18}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{9}{10}\end{matrix}\right.\)
a) \(\dfrac{1}{3}.\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\Leftrightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)
\(\Leftrightarrow5x-x=-\dfrac{3}{5}-\dfrac{1}{6}+2\Leftrightarrow4x=\dfrac{37}{30}\Leftrightarrow x=\dfrac{\dfrac{37}{30}}{4}=\dfrac{37}{120}\)
vậy \(x=\dfrac{37}{120}\)
b) \(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\Leftrightarrow\dfrac{3}{2}x-x=\dfrac{-3}{4}+1\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}x=\dfrac{-3}{4}+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.2=\dfrac{6}{4}=\dfrac{3}{2}\) vậy \(x=\dfrac{3}{2}\)
c) \(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\Leftrightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\Leftrightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)
\(\Leftrightarrow x=\dfrac{\dfrac{19}{12}}{\dfrac{7}{4}}=\dfrac{19}{21}\) vậy \(x=\dfrac{19}{21}\)
d) \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\Leftrightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)
\(\Leftrightarrow\dfrac{3}{2}x-x=1+\dfrac{5}{2}+\dfrac{7}{5}\Leftrightarrow\dfrac{1}{2}x=\dfrac{49}{10}\Leftrightarrow x=\dfrac{49}{10}.2=\dfrac{49}{5}\)
vậy \(x=\dfrac{49}{5}\)
e) \(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\Leftrightarrow\left|4x+3\right|=\dfrac{\dfrac{6}{15}}{\dfrac{2}{3}}=\dfrac{3}{5}\)
th1 : \(4x+3\ge0\Leftrightarrow4x\ge-3\Leftrightarrow x\ge\dfrac{-3}{4}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow4x+3=\dfrac{3}{5}\Leftrightarrow4x=\dfrac{3}{5}-3=\dfrac{-12}{5}\)
\(\Leftrightarrow x=\dfrac{\dfrac{-12}{5}}{4}=\dfrac{-3}{5}\left(tmđk\right)\)
th2: \(4x+3< 0\Leftrightarrow4x< -3\Leftrightarrow x< \dfrac{-3}{4}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow-\left(4x+3\right)=\dfrac{3}{5}\Leftrightarrow-4x-3=\dfrac{3}{5}\)
\(\Leftrightarrow4x=-3-\dfrac{3}{5}=\dfrac{-18}{5}\Leftrightarrow x=\dfrac{\dfrac{-18}{5}}{4}=\dfrac{-9}{10}\left(tmđk\right)\)
vậy \(x=\dfrac{-3}{5};x=\dfrac{-9}{10}\)
a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(-\dfrac{5}{6}x=\dfrac{5}{12}\)
\(x=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)
\(3x-3.7=-\dfrac{19}{2}\)
\(3x=-5.8\)
\(x=-\dfrac{29}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)
\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)
\(\dfrac{3}{4}x=\dfrac{5}{8}\)
\(x=\dfrac{5}{6}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=-\dfrac{3}{20}\)
\(\dfrac{1}{7}=\dfrac{8}{-x}\)=> \(-x=56\)
=> \(x=56\)
2) => 18x = 18
=> x = 1
3) \(\dfrac{-4}{3}+x=\dfrac{-11}{6}\)
=> \(x=\dfrac{-11}{6}+\dfrac{4}{3}\)
=> \(x=\dfrac{-1}{2}\)
4) 45%.x =\(\dfrac{3}{5}\)
=> \(x=\dfrac{3}{5}:\dfrac{9}{20}\)
=> \(x=\dfrac{4}{3}\)
Bài 1:
a) \(\dfrac{2}{5}\cdot x-\dfrac{1}{4}=\dfrac{1}{10}\)
\(\dfrac{2}{5}\cdot x=\dfrac{1}{10}+\dfrac{1}{4}\)
\(\dfrac{2}{5}\cdot x=\dfrac{7}{20}\)
\(x=\dfrac{7}{20}:\dfrac{2}{5}\)
\(x=\dfrac{7}{8}\)
Vậy \(x=\dfrac{7}{8}\).
b) \(\dfrac{3}{5}=\dfrac{24}{x}\)
\(x=\dfrac{5\cdot24}{3}\)
\(x=40\)
Vậy \(x=40\).
c) \(\left(2x-3\right)^2=16\)
\(\left(2x-3\right)^2=4^2\)
\(\circledast\)TH1: \(2x-3=4\\ 2x=4+3\\ 2x=7\\ x=\dfrac{7}{2}\)
\(\circledast\)TH2: \(2x-3=-4\\ 2x=-4+3\\ 2x=-1\\ x=\dfrac{-1}{2}\)
Vậy \(x\in\left\{\dfrac{7}{2};\dfrac{-1}{2}\right\}\).
Bài 2:
a) \(25\%-4\dfrac{2}{5}+0.3:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}\cdot\dfrac{5}{6}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{1}{4}\)
\(=\dfrac{5}{20}-\dfrac{88}{20}+\dfrac{5}{20}\)
\(=\dfrac{5-88+5}{20}\)
\(=\dfrac{78}{20}=\dfrac{39}{10}\)
b) \(\left(\dfrac{1}{6}-\dfrac{1}{5^2}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{25}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5}{30}-\dfrac{6}{30}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5-6+1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\cdot\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\)
Bài 3:
a) \(\dfrac{4}{19}\cdot\dfrac{-3}{7}+\dfrac{-3}{7}\cdot\dfrac{15}{19}\)
\(=\dfrac{-3}{7}\left(\dfrac{4}{19}+\dfrac{15}{19}\right)\)
\(=\dfrac{-3}{7}\cdot1\)
\(=\dfrac{-3}{7}\)
b) \(7\dfrac{5}{9}-\left(2\dfrac{3}{4}+3\dfrac{5}{9}\right)\)
\(=\dfrac{68}{9}-\dfrac{11}{4}-\dfrac{32}{9}\)
\(=\dfrac{68}{9}-\dfrac{32}{9}-\dfrac{11}{4}\)
\(=4-\dfrac{11}{4}\)
\(=\dfrac{16}{4}-\dfrac{11}{4}\)
\(\dfrac{5}{4}\)
Bài 4:
\(\dfrac{4}{12\cdot14}+\dfrac{4}{14\cdot16}+\dfrac{4}{16\cdot18}+...+\dfrac{4}{58\cdot60}\)
\(=2\left(\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}+\dfrac{1}{16\cdot18}+...+\dfrac{1}{58\cdot60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{18}+...+\dfrac{1}{58}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{5}{60}-\dfrac{1}{60}\right)\)
\(=2\cdot\dfrac{1}{15}\)
\(=\dfrac{2}{15}\)
\(x:4\dfrac{1}{3}=2,5\\ x:\dfrac{13}{3}=\dfrac{5}{2}\\ x=\dfrac{5}{2}.\dfrac{13}{3}\\ x=\dfrac{65}{6}=10\dfrac{5}{6}\)
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!
6. \(\dfrac{x}{4}=\dfrac{9}{x}\)
=>x2=4.9=36
=>x\(\in\)\(\left\{-6;6\right\}\)
\((\dfrac{2x}{5}+2):\left(-4\right)=-1\dfrac{1}{2}\)
(\(\dfrac{2x}{5}+2):\left(-4\right)=-\dfrac{3}{2}\)
\(\dfrac{2x}{5}=-\dfrac{3}{2}.\left(-4\right)\)
\(\dfrac{2x}{5}=6\)
\(\dfrac{2x}{5}=\dfrac{30}{5}\)
2x = 30
x = 30 : 2 = 15
Câu 1:
a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Rightarrow-\dfrac{2}{3x}+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}x+\dfrac{2}{3}x=\dfrac{1}{6}+\dfrac{1}{3}\)
\(\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{1}{2}\)
\(\Rightarrow x.\dfrac{4}{3}=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{3}{8}\)
a: \(\Leftrightarrow x^2=\dfrac{-5}{2}\cdot\dfrac{-10}{9}=\dfrac{50}{18}=\dfrac{25}{9}\)
=>x=5/3hoặc x=-5/3
c: \(\Leftrightarrow4\left(x-\dfrac{5}{8}\right)=\dfrac{1}{4}+\dfrac{3}{4}=1\)
=>x-5/8=1/4
hay x=2/8+5/8=7/8
d: \(\Leftrightarrow\left|x-3\right|=\dfrac{2}{5}+\dfrac{3}{5}=1\)
=>x-3=1 hoặc x-3=-1
=>x=4 hoặc x=2
e: =>1-1/2x=-3
=>1/2x=4
hay x=8