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15 tháng 7 2020

Bài làm:

Ta có: \(\frac{x-1}{3}+\frac{x-3}{4}=2\)

\(\Leftrightarrow\left(\frac{x}{3}+\frac{x}{4}\right)=2+\frac{1}{3}+\frac{3}{4}\)

\(\Leftrightarrow\frac{7}{12}x=\frac{37}{12}\)

\(\Leftrightarrow x=\frac{37}{12}\div\frac{7}{12}\)

\(\Rightarrow x=\frac{37}{7}\)

15 tháng 7 2020

\(\frac{x-1}{3}+\frac{x-3}{4}=2\)

\(\Leftrightarrow\frac{4\left(x-1\right)}{12}+\frac{3\left(x-3\right)}{12}=\frac{24}{12}\)

\(\Leftrightarrow4\left(x-1\right)+3\left(x-3\right)=24\)

\(\Leftrightarrow4x-4+3x-9=24\)

\(\Leftrightarrow7x-13=24\)

\(\Leftrightarrow7x=37\)

\(\Leftrightarrow x=\frac{37}{7}\)

16 tháng 6 2016

*\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left[\frac{1}{14}+\frac{1}{7}-\left(-\frac{3}{35}\right)\right].\frac{4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right).\frac{4}{3}}=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{4}{3}}=\frac{\frac{-1}{12}}{\frac{14}{35}}=-\frac{1}{12}.\frac{35}{14}=\frac{-35}{168}\)

*\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{63}{10}.12-21.\frac{18}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{378}{5}-\frac{378}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)

20 tháng 5 2017

\(7:\frac{5.8.x-57}{19+3}=7.\frac{22}{40x-57}=1775\)

<=> \(\frac{154}{40x-57}=1775=>x=\frac{154+1775.57}{40}\)

2: \(=\dfrac{-2}{75}+\dfrac{5}{39}=\dfrac{33}{325}\)

3: \(=\dfrac{6}{11}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{6}{11}\)

4: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-1\right)=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)

5: \(=\dfrac{2}{7}\left(\dfrac{4}{23}-\dfrac{27}{23}+1\right)=0\)

6: \(=\dfrac{3}{8}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{11}{8}=\dfrac{3}{8}+\dfrac{11}{8}=\dfrac{14}{8}=\dfrac{7}{4}\)

10 tháng 12 2019

\(4x^4-4x^3+4=4y^2\)

Ta có:

\(\left(2x^2-x-1\right)^2< 4x^4-4x^3+4=4y^2< \left(2x^4-x+3\right)^2\)

\(\Leftrightarrow\left(4x^4-4x^3+4\right)=\left(\left(2x^2-x\right)^2;\left(2x^2-x+1\right)^2;\left(2x^2-x+2\right)^2\right)\)

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