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ND
1
DP
7 tháng 7 2017
\(\frac{11}{15}-\left(\frac{7}{9}+x\right)\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)
\(\frac{11}{15}-\frac{7}{24}-\frac{3}{8}x=\frac{61}{90}+\frac{x}{3}\)
\(\frac{x}{3}+\frac{3}{8}x=\frac{11}{15}-\frac{7}{24}-\frac{61}{90}\)
\(\frac{17}{24}x=-\frac{17}{72}\)
\(\Rightarrow x=-\frac{1}{3}\)
17 tháng 10 2020
a) \(\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
=> \(\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x+\frac{1}{3}=0\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}x\right)+\left(-\frac{2}{5}+\frac{1}{3}\right)=0\)
=> \(\frac{1}{6}x-\frac{1}{15}=0\Rightarrow\frac{1}{6}x=\frac{1}{15}\Rightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{2}{5}\)
Vậy x = 2/5
b) \(\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\)
=> \(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
=> \(\frac{11}{15}x+\frac{2}{5}=0\Rightarrow\frac{11}{15}x=-\frac{2}{5}\)
=> \(x=\left(-\frac{2}{5}\right):\frac{11}{15}=\left(-\frac{2}{5}\right)\cdot\frac{15}{11}=-\frac{6}{11}\)
Vậy x = -6/11
c) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
=> \(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
=> \(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{1}{3}x-x\right)=5\)
=> \(\frac{2}{3}-\frac{4}{3}x=5\)
=> \(\frac{4}{3}x=-\frac{13}{3}\Rightarrow x=\left(-\frac{13}{3}\right):\frac{4}{3}=\left(-\frac{13}{3}\right)\cdot\frac{3}{4}=-\frac{13}{4}\)
Vậy x = -13/4
d) \(\frac{11}{5}-\left(\frac{7}{9}-x\right)\cdot\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)
=> \(\frac{11}{5}-\frac{3}{8}\left(\frac{7}{9}-x\right)=\frac{61}{90}+\frac{30x}{90}\)
=> \(\frac{11}{5}-\frac{7}{24}+\frac{3}{8}x=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{3}{8}x=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{3x}{8}=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{45x}{120}=\frac{61+30x}{90}\)
=> \(\frac{229+45x}{120}=\frac{61+30x}{90}\)
=> \(\frac{3\left(229+45x\right)}{360}=\frac{4\left(61+30x\right)}{360}\)
=> \(3\left(229+45x\right)=4\left(61+30x\right)\)
=> \(687+135x=244+120x\)
=> \(687+135x-244-120x=0\)
=> \(\left(687-244\right)+\left(135x-120x\right)=0\)
=> \(443+15x=0\)
=> \(15x=-443\Rightarrow x=-\frac{443}{15}\)
Vậy x = -443/15
29 tháng 6 2019
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
29 tháng 6 2019
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
DT
2
\(\frac{11}{15}-\left(\frac{7}{9}+x\right).\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)(. là dấu nhân)
\(\Leftrightarrow\frac{11}{15}-\frac{7}{24}-\frac{3x}{8}-\frac{61}{90}-\frac{x}{3}=0\)
\(\Leftrightarrow\frac{-17}{72}-\left(\frac{3x}{8}+\frac{x}{3}\right)=0\)
\(\Leftrightarrow\frac{17x}{24}=-\frac{17}{72}\)
\(\Leftrightarrow17x.72=-17.24\)
\(\Leftrightarrow x=-\frac{1}{3}\)
\(\frac{11}{15}-\left(\frac{7}{9}+x\right).\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)
\(\Rightarrow\frac{11}{15}-\frac{7}{9}.\frac{3}{8}-\frac{3x}{8}=\frac{61}{90}+\frac{x}{3}\)
\(\Rightarrow\frac{11}{15}-\frac{7}{24}-\frac{3x}{8}=\frac{61}{90}+\frac{x}{3}\)
\(\Rightarrow\frac{53}{120}-\frac{3x}{8}=\frac{61}{90}+\frac{x}{3}\)
\(\Rightarrow\frac{53}{120}-\frac{61}{90}=\frac{8x+9x}{24}\)
\(\Rightarrow-\frac{17}{72}=\frac{17x}{24}\)
\(\Rightarrow\frac{-17}{72}.24=17x\)
\(\Rightarrow-17.3=17x\Leftrightarrow x=-3\)