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11 tháng 10 2021

\(a,\Rightarrow\left|x+\dfrac{4}{9}\right|=\dfrac{3}{2}+\dfrac{1}{2}=2\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{9}=2\\x+\dfrac{4}{9}=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{14}{9}\\x=-\dfrac{22}{9}\end{matrix}\right.\\ b,\Rightarrow\left\{{}\begin{matrix}x-\dfrac{4}{11}=0\\5+y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{11}\\y=-5\end{matrix}\right.\)

11 tháng 10 2021

a) \(\left|x+\dfrac{4}{9}\right|-\dfrac{1}{2}=\dfrac{3}{2}\)

\(\Rightarrow\left|x+\dfrac{4}{9}\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{9}=2\\x+\dfrac{4}{9}=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{14}{9}\\x=-\dfrac{22}{9}\end{matrix}\right.\)

b) \(\left|x-\dfrac{4}{11}\right|+\left|5+y\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{4}{11}=0\\5+y=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{11}\\y=-5\end{matrix}\right.\)

17 tháng 10 2021

Ta có: \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}-\dfrac{3}{7}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{3}{7}x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{7}{6}\end{matrix}\right.\)

a: \(\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(y-x\right)^2}\)

\(=\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(x-y\right)^2}\)

\(=3\left(x-y\right)^2+2\left(x-y\right)-5\)

b: \(\dfrac{\left(x-2y\right)^3}{x^2-4xy+4y^2}\)

\(=\dfrac{\left(x-2y\right)^3}{\left(x-2y\right)^2}\)

=x-2y

c: \(\dfrac{x^3+y^3}{x+y}\)

\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x+y}\)

\(=x^2-xy+y^2\)

8 tháng 6 2020

Bài làm

a) \(\left(\frac{3}{8}+\frac{-1}{4}+\frac{5}{12}\right):\frac{2}{3}\)

\(=\left(\frac{9}{24}+\frac{-6}{24}+\frac{10}{24}\right).\frac{3}{2}\)

\(=\frac{13}{24}.\frac{3}{2}\)

\(=\frac{39}{48}\)

b) \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)

\(=-\frac{5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)

\(=-\frac{5}{7}.1+\frac{12}{7}\)

\(=\frac{7}{7}\)

\(=1\)

c) \(0,25:\left(10,3-9,8\right)-\frac{3}{4}\)

\(=\frac{25}{100}:\left(0,5\right)-\frac{3}{4}\)

\(=\frac{5}{4}:\frac{1}{2}-\frac{3}{4}\)

\(=\frac{5}{4}.2-\frac{3}{4}\)

\(=\frac{5}{2}-\frac{3}{4}\)

\(=\frac{10}{4}-\frac{3}{4}=\frac{7}{4}\)

d) \(-\frac{5}{9}.\frac{13}{28}-\frac{13}{28}.\frac{4}{9}\)

\(=-\frac{13}{28}\left(\frac{5}{9}+\frac{4}{9}\right)\)

\(=-\frac{13}{28}.\frac{9}{9}\)

\(=-\frac{13}{28}.1\)

\(=-\frac{13}{28}\)

14 tháng 9 2021

c)\(\left(xy^2-1\right)\left(x^2y+5\right)\)

\(=x^3y^3+5xy^2-x^2y-5\)

d)\(4\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x^2+1\right)\)

\(=4\left(x^2-\dfrac{1}{4}\right)\left(4x^2+1\right)\)

\(=4\left(4x^4+x^2-x-\dfrac{1}{4}\right)\)

\(=16x^4+4x^2-4x-1\)

14 tháng 9 2021

Bài 9

a)\(\left(x+3\right)\left(x+4\right)\)                               b)\(\left(x-4\right)\left(x^2+4x+16\right)\)

\(=x^2+4x+3x+12\)                         \(=\left(x-4\right)\left(x^2+x.4+4^2\right)\)

\(=x^2+7x+12\)                                  \(=x^3-4^3=x^3-64\)

3 tháng 5 2017

\(\dfrac{-7}{12}+\dfrac{11}{8}-\dfrac{5}{9}=\dfrac{-7}{12}+\dfrac{59}{72}=\dfrac{17}{72}\) \(\dfrac{1}{7}-\dfrac{8}{7}:8-3:\dfrac{3}{4}.\left(-2\right)^2=\dfrac{1}{7}-\dfrac{1}{7}-4.4=\left(-16\right)\) \(\dfrac{1}{4}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}=\dfrac{15}{49}-\dfrac{22}{15}:\dfrac{11}{5}=\dfrac{15}{49}-\dfrac{2}{3}=\dfrac{-53}{147}\)

3 tháng 5 2017

\(\dfrac{-11x}{12}+\dfrac{3}{4}=\dfrac{1}{6}\Leftrightarrow\dfrac{-11x}{12}=\dfrac{1}{6}-\dfrac{3}{4}\Leftrightarrow\dfrac{-11x}{12}=\dfrac{-7}{12}\Leftrightarrow-11x=-7\Leftrightarrow x=\dfrac{7}{11}\)