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\(=\dfrac{13}{12}\cdot\dfrac{27}{5}\cdot2\cdot\dfrac{34}{9}\cdot2\cdot\dfrac{2}{17}=4\cdot\dfrac{4}{9}\cdot\dfrac{13}{5}\cdot\dfrac{9}{4}=4\cdot\dfrac{13}{5}=\dfrac{52}{5}\)
\(\left(1+\dfrac{2}{24}\right)\cdot5=\dfrac{13}{12}\cdot5=\dfrac{65}{12}=\dfrac{195}{36}\)
\(2.5\cdot2\cdot3\dfrac{7}{9}\cdot2\cdot\dfrac{2}{17}\)
\(=10\cdot\dfrac{2}{17}\cdot\dfrac{34}{9}=\dfrac{40}{9}=\dfrac{160}{36}\)
Do đó: 195/36>160/36
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
`Answer:`
a. \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{41}{4}+\frac{3}{4}\\2x=-\frac{41}{4}+\frac{3}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=11\\2x=-\frac{19}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=11:2\\x=-\frac{19}{2}:2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{19}{4}\end{cases}}\)
b. \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}-\frac{1}{5}\\x=-\frac{3}{5}-\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)
c. \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}-\left(-\frac{24}{27}\right)\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Leftrightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Leftrightarrow3x=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{1}{9}:3\)
\(\Leftrightarrow x=\frac{1}{27}\)
1,
\(\dfrac{6}{7}\) + \(\dfrac{-5}{8}\) : (-5) -\(\dfrac{3}{16}\) . \(\left(-2\right)^2\)
= \(\dfrac{6}{7}\) + \(\dfrac{-5}{8}\) . \(\dfrac{-1}{5}\) - \(\dfrac{3}{16}\) . 4
= \(\dfrac{6}{7}\) + \(\dfrac{1}{8}\) - \(\dfrac{3}{4}\)
= \(\dfrac{48}{56}\) + \(\dfrac{7}{56}\) - \(\dfrac{42}{56}\)
= \(\dfrac{13}{56}\)
3,
a,\(\dfrac{3}{5}\)x - \(\dfrac{7}{10}\)x = \(\dfrac{-1}{2}\)
x.(\(\dfrac{3}{5}\) - \(\dfrac{7}{10}\)) = \(\dfrac{-1}{2}\)
x.\(\dfrac{-1}{10}\)= \(\dfrac{-1}{2}\)
x =\(\dfrac{-1}{2}\):\(\dfrac{-1}{10}\)
x = 5
c, (2,5x - 3,6) : \(\dfrac{15}{7}\) = -1
( \(\dfrac{5}{2}\)x - \(\dfrac{18}{5}\)) : \(\dfrac{15}{7}\) = -1
\(\dfrac{5}{2}\)x - \(\dfrac{18}{5}\) = -1 . \(\dfrac{15}{7}\)
\(\dfrac{5}{2}\)x - \(\dfrac{18}{5}\) = \(\dfrac{-15}{7}\)
\(\dfrac{5}{2}\)x = \(\dfrac{-15}{7}\) + \(\dfrac{18}{5}\)
\(\dfrac{5}{2}\)x = \(\dfrac{51}{35}\)
x =\(\dfrac{51}{35}\) : \(\dfrac{5}{2}\)
x = \(\dfrac{102}{175}\)
Mình không biết có chỗ nào sai không nữa.
