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\(\left(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}\right)-\left(3-\frac{1}{2}-\frac{35}{11}\right)+\left(\frac{11}{4}-\frac{2}{5}\right)\)
= \(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}-3+\frac{1}{2}+\frac{35}{11}+\frac{11}{4}-\frac{2}{5}\)
= \(\left(\frac{3}{4}+\frac{11}{4}+\frac{1}{2}\right)\left(-\frac{13}{11}+\frac{35}{11}\right)+\left(\frac{7}{5}-\frac{2}{5}\right)-3\)
= \(8+2+1-3\)
= \(8\)
#)Giải :
\(\left(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}\right)-\left(3-\frac{1}{2}-\frac{35}{11}\right)+\left(\frac{11}{4}-\frac{2}{5}\right)\)
\(=\frac{3}{4}-\frac{13}{11}+\frac{7}{5}-3+\frac{1}{2}+\frac{35}{11}+\frac{11}{4}-\frac{2}{5}\)
\(=\left(\frac{3}{4}+\frac{11}{4}\right)+\left(-\frac{13}{11}+\frac{35}{11}\right)+\left(\frac{7}{5}-\frac{2}{5}\right)-3+\frac{1}{2}\)
\(=\frac{7}{2}+2+1-3+\frac{1}{2}\)
\(=\frac{7}{2}+\frac{1}{2}\)
\(=4\)
a)\(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=\dfrac{-5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
b) \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
a) Ta có: \(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=-\dfrac{5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
b) Ta có: \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
a)
x=3 y=0
b)x=1 y=1
c)x=0 y=7 mình không biết có đúng ko nữa bạn suy nghĩ xem nhé #kết bạn với mk nha# cho hỏi người lạ minhf trả lời thế có k ko <3
\(\left(2x-1\right)^5=x^5\)
\(\Leftrightarrow2x-1=x\)
\(\Leftrightarrow x=1\)
(2x-1)5=x5
2x-1=x
2x-x=1
x=1