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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 \(x\in\left\{\frac{11}{4};-\frac{25}{12}\right\}\).
\(\frac{x+2}{3}=\frac{2x-1}{5}\)
=> \(\left(x+2\right)\cdot5=3\left(2x-1\right)\)
=> \(5x+10=6x-3\)
=> \(6x-5x=10+3\)
=> \(x=13\)
\(\frac{-x}{4}=\frac{-9}{x}\)
=> \(-x^2=4\cdot\left(-9\right)\)
=> \(-x^2=-36\)
=> \(x^2=36\)
=> \(\orbr{\begin{cases}x^2=6^2\\x^2=\left(-6\right)^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
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 : \(x+3=5\Leftrightarrow x=2\)
\(\frac{1}{2}\left(x+1\right):\frac{3}{7}=\frac{32}{135}\)
\(\frac{1}{2}\left(x+1\right)=\frac{32}{135}.\frac{3}{7}\)
\(\frac{1}{2}\left(x+1\right)=\frac{32}{315}\)
\(x+1=\frac{32}{315}:\frac{1}{2}\)
\(x+1=\frac{64}{315}\)
\(x=\frac{64}{315}-1=-\frac{251}{315}\)
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 ~
X=2 NHA