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a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)
Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)
b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)
=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=> \(\frac{27x}{4}=\frac{27}{40}\)
\(27x.40=27.4\)
\(1080.x=108\)
\(x=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
c) \(\left|x-1\right|+4=6\)
\(\left|x-1\right|=6-4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy \(x=\left[3,-1\right]\)
d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)
e) \(\left(x^2-3\right)^2=16\)
\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)
\(x^2=7=>x=\sqrt{7}\)
Vậy \(x=\sqrt{7}\)
f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\)
\(\frac{2}{5}x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)
Vậy \(x=-\frac{2}{3}\)
g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)
Vậy \(x=-3\)
k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)
\(\frac{2}{5}x=\frac{4}{15}\)
\(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)
Vậy \(x=\frac{2}{15}\)
I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)
\(\frac{3}{5}x=\frac{5}{14}\)
\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}\)
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
a) Ta có: \(\frac{x}{12}=\frac{y}{3}.\)
=> \(\frac{x}{12}=\frac{y}{3}\) và \(x-y=36.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4.\)
\(\left\{{}\begin{matrix}\frac{x}{12}=4=>x=4.12=48\\\frac{y}{3}=4=>y=4.3=12\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(48;12\right).\)
b)
\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
⇒ \(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
⇒ \(\frac{5}{3}x=\frac{1}{21}\)
⇒ \(x=\frac{1}{21}:\frac{5}{3}\)
⇒ \(x=\frac{1}{35}\)
Vậy \(x=\frac{1}{35}.\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
⇒ \(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
⇒ \(x-\frac{1}{2}=\frac{1}{3}\)
⇒ \(x=\frac{1}{3}+\frac{1}{2}\)
⇒ \(x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}.\)
Có 1 câu bạn đăng mình làm ở dưới rồi mà.
Chúc bạn học tốt!
a)áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4\)
\(\)x/12=4 suy ra x=12.4=48
y/3=4 suy ra y=3.4 =12
b)\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
\(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
\(\frac{5}{3}x=\frac{1}{21}\)
\(x=\frac{1}{21}:\frac{5}{3}\)
\(x=\frac{1}{35}\)
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}\)
\(\frac{2}{5}+x=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{2}{5}\)
\(x=\frac{-3}{20}\)
\(\left|x-\frac{2}{5}\right|+\frac{3}{4}=\frac{11}{4}\)
\(\left|x-\frac{2}{5}\right|=\frac{11}{4}-\frac{3}{4}\)
\(\left|x-\frac{2}{5}\right|=2\)
suy ra x-2/5=2 hoac x-2/5=-2
\(x-\frac{2}{5}=2\)
\(x=\frac{12}{5}\)
\(x-\frac{2}{5}=-2\)
\(x=\frac{-8}{5}\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
a) \(\frac{2}{3x}-\frac{3}{12}=\frac{4}{5}-\left(\frac{7}{x}-2\right)\)
\(\frac{2}{3x}+\left(\frac{7}{x}-2\right)=\frac{4}{5}+\frac{3}{12}\)
\(\frac{2}{3x}+\frac{7}{x}-2=\frac{21}{20}\)
\(\frac{2}{3x}+\frac{7}{x}=\frac{61}{20}\)
\(\frac{2}{3x}+\frac{21}{3x}=\frac{61}{20}\)
\(\frac{23}{3x}=\frac{61}{20}\)
\(3x=\frac{460}{61}\)
\(x=\frac{460}{183}\)
b) \(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
\(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
\(\frac{19}{10}:x=2\)
\(x=\frac{19}{20}\)
Thanks bạn nhìu