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10 tháng 6 2020

c) \(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{15}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}.\frac{27}{8}\)

\(2+\frac{3}{4}x=\frac{21}{8}\)

\(\frac{3}{4}x=\frac{21}{8}-2\)

\(\frac{3}{4}x=\frac{21}{8}-\frac{16}{8}\)

\(\frac{3}{4}x=\frac{5}{8}\)

\(x=\frac{5}{8}\div\frac{3}{4}\)

\(x=\frac{5}{8}.\frac{4}{3}\)

\(x=\frac{5}{6}\)

Vậy \(x=\frac{5}{6}\).

d) \(\left|x-\frac{1}{3}\right|-\frac{3}{4}=\frac{5}{3}\)

\(\left|x-\frac{1}{3}\right|=\frac{5}{3}+\frac{3}{4}\)

\(\left|x-\frac{1}{3}\right|=\frac{20}{12}+\frac{9}{12}\)

\(\left|x-\frac{1}{3}\right|=\frac{29}{12}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{29}{12}\\x-\frac{1}{3}=-\frac{29}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{4}\\x=-\frac{25}{12}\end{cases}}\)

Vậy \(x\in\left\{\frac{11}{4};-\frac{25}{12}\right\}\).

18 tháng 5 2020

\(\frac{x+2}{3}=\frac{2x-1}{5}\)

=> \(\left(x+2\right)\cdot5=3\left(2x-1\right)\)

=> \(5x+10=6x-3\)

=> \(6x-5x=10+3\)

=> \(x=13\)

\(\frac{-x}{4}=\frac{-9}{x}\)

=> \(-x^2=4\cdot\left(-9\right)\)

=> \(-x^2=-36\)

=> \(x^2=36\)

=> \(\orbr{\begin{cases}x^2=6^2\\x^2=\left(-6\right)^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

Quỳnh ơi, chuyển 6x sang sẽ là -6x mà viết như cậu phải là -6x+5x :) 

a, \(\frac{x+2}{3}=\frac{2x-1}{5}\)

\(\Leftrightarrow\frac{5x+10}{15}=\frac{6x-3}{15}\Leftrightarrow5x+10=6x-3\Leftrightarrow-x+13=0\Leftrightarrow x=-13\)

b, \(\frac{-x}{4}=\frac{-9}{x}\)\(\Leftrightarrow x^2=36\Leftrightarrow x=\pm6\)

a, \(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)

\(\Leftrightarrow\frac{25}{9}-\frac{12}{13}x=\frac{7}{9}\Leftrightarrow\frac{12}{13}x=2\Leftrightarrow x=\frac{13}{6}\)

b, \(\frac{x-12}{4}=\frac{9-3x}{x}\)

\(\Leftrightarrow x^2-12x=36-12x\Leftrightarrow x^2-12x-36+12x=0\)

\(\Leftrightarrow x^2-36=0\Leftrightarrow x^2=36\Leftrightarrow x=\pm6\)

a, \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\Leftrightarrow\frac{x}{3}+\frac{7}{12}=0\Leftrightarrow\frac{4x}{12}+\frac{7}{12}=0\)

Khử mẫu ta đc : \(4x+7=0\Leftrightarrow4x=-7\Leftrightarrow x=-\frac{7}{4}\)

b, \(\frac{x+3}{15}=\frac{1}{3}\Leftrightarrow\frac{x+3}{15}=\frac{5}{15}\)

Khử mẫu ta đc : \(x+3=5\Leftrightarrow x=2\)

27 tháng 5 2015

\(\frac{1}{2}\left(x+1\right):\frac{3}{7}=\frac{32}{135}\)

\(\frac{1}{2}\left(x+1\right)=\frac{32}{135}.\frac{3}{7}\)

\(\frac{1}{2}\left(x+1\right)=\frac{32}{315}\)

\(x+1=\frac{32}{315}:\frac{1}{2}\)

\(x+1=\frac{64}{315}\)

\(x=\frac{64}{315}-1=-\frac{251}{315}\)

12 tháng 4 2018

Ta có : 

\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{49}+1\right)+\left(\frac{x+2}{48}+1\right)+\left(\frac{x+3}{47}+1\right)+\left(\frac{x+4}{46}+1\right)+\left(\frac{x+5}{45}+1\right)=-5+5\)

\(\Leftrightarrow\)\(\frac{x+50}{49}+\frac{x+50}{48}+\frac{x+50}{47}+\frac{x+50}{46}+\frac{x+50}{45}=0\)

\(\Leftrightarrow\)\(\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)

Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\)

Nên \(x+50=0\)

\(\Rightarrow\)\(x=-50\)

Vậy \(x=-50\)

Chúc bạn học tốt ~