1/7:5/17+1/7:5/18
help me
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1/ \(4\left(\dfrac{-1}{2}\right)^3+\dfrac{1}{2}:5\)
\(=4.\dfrac{-1}{8}+\dfrac{1}{2}.\dfrac{1}{5}\)
\(=\dfrac{-1}{2}+\dfrac{1}{10}\)
\(=\dfrac{-5}{10}+\dfrac{1}{10}\)
\(=\dfrac{-4}{10}\)
\(=\dfrac{-2}{5}\)
2/ \(17\dfrac{1}{5}:\left(-\dfrac{5}{7}\right)-2\dfrac{1}{5}.\left(-\dfrac{7}{5}\right)\)
\(=\dfrac{86}{5}.\left(\dfrac{-7}{5}\right)-\dfrac{11}{5}.\left(\dfrac{-7}{5}\right)\)
\(=\dfrac{-7}{5}.\left(\dfrac{86}{5}-\dfrac{11}{5}\right)\)
\(=\dfrac{-7}{5}.15\)
\(=-21\)
\(B=\dfrac{12}{17}.\dfrac{5}{7}-\dfrac{-12}{17}.\dfrac{1}{7}+\dfrac{1}{17.7}.12\\ =\dfrac{12}{17}.\dfrac{5}{7}+\dfrac{12}{17}.\dfrac{1}{7}+\dfrac{12}{17}.\dfrac{1}{7}\\ =\dfrac{12}{17}\left(\dfrac{5}{7}+\dfrac{1}{7}+\dfrac{1}{7}\right)\\ =\dfrac{15}{17}.1=\dfrac{15}{17}\)
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 ...
a: =(5/7+2/7)+(4/3+5/3)=3+1=4
b: =(17/12+7/12)+(29/7-8/7)
=2+3=5
c: =(2/5+3/5)+(6/9+1/3)+(7/4+1/4)
=1+2+1
=4
d: =(1/3+2/3)+(13/17+4/17)+(29/11+4/11)
=1+1+3=5
a)=\(\frac{-2.4}{5.7}+\frac{-3.2}{5.7}+\frac{-3}{5}\)
=\(\frac{-2.4}{5.7}+\frac{-2.3}{5.7}+\frac{-3}{5}\)
=\(\frac{-2}{5}\left(\frac{4+3}{7}\right)+\frac{-3}{5}\)
=\(\frac{-2}{5}.1+\frac{-3}{5}\)
=-1
b)
4: \(=\dfrac{2+3}{7}+\dfrac{1+6}{9}-\dfrac{5}{6}=\dfrac{5}{7}+\dfrac{7}{9}-\dfrac{5}{6}=\dfrac{83}{126}\)
5: \(=\dfrac{-5-2}{7}+\dfrac{3+1}{4}-\dfrac{1}{5}=-\dfrac{1}{5}\)
6: \(=\dfrac{-3-28}{31}+\dfrac{-6-1}{17}+\dfrac{1-5}{25}=-1-\dfrac{7}{17}-\dfrac{4}{25}=-\dfrac{668}{425}\)
4. \(\dfrac{2}{7}+\dfrac{1}{9}+\dfrac{3}{7}+\dfrac{5}{9}+\dfrac{-5}{6}\)
\(=\left(\dfrac{2}{7}+\dfrac{3}{7}\right)+\left(\dfrac{1}{9}+\dfrac{5}{9}\right)+\dfrac{-5}{6}\)
\(=\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{-5}{6}\)
\(=\dfrac{30+28+\left(-35\right)}{42}=\dfrac{23}{42}\)
5. \(\dfrac{-5}{7}+\dfrac{3}{4}+\dfrac{-1}{5}+\dfrac{-2}{7}+\dfrac{1}{4}\)
\(=\left(\dfrac{-5}{7}+\dfrac{-2}{7}\right)+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\dfrac{-1}{5}\)
\(=\dfrac{-7}{7}+\dfrac{4}{4}+\dfrac{-1}{5}\)
\(=-1+1+\dfrac{-1}{5}\)
\(=-\dfrac{1}{5}\)
6. \(\dfrac{-3}{31}+\dfrac{-6}{17}+\dfrac{1}{25}+\dfrac{-28}{31}+\dfrac{-1}{17}+\dfrac{-1}{5}\)
\(=\left(\dfrac{-3}{31}+\dfrac{-28}{31}\right)+\left(\dfrac{-6}{17}+\dfrac{-1}{17}\right)+\dfrac{1}{25}+\dfrac{-1}{5}\)
\(=\dfrac{-31}{31}+\dfrac{-7}{17}+\dfrac{1}{25}+\dfrac{-1}{5}\)
\(=-1+\dfrac{-7}{17}+\dfrac{1}{25}+\dfrac{-1}{5}\)
\(=\dfrac{-425+\left(-175\right)+17+\left(-85\right)}{425}=\dfrac{-668}{425}\)
1/7:(5/17+5/18)
1/7:175/306
1/7.306/175
25/306
1/7 : 5/17 + 1/7 : 5/18
= 1/7 x 17/5 + 1/7 x 18/5
= 1/7 x ( 17/5 + 18/5 )
= 1/7 x 35/5
= 1/7 x 7
= 1