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a) \(x=-\dfrac{3}{5}\times\dfrac{9}{7}=-\dfrac{27}{35}\)
b) \(x\left(0,4-\dfrac{1}{5}\right)=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}:\dfrac{1}{5}=\dfrac{15}{4}\)
a, \(x=-3,5.\dfrac{9}{7}=-\dfrac{9}{2}\)
b, \(\dfrac{2}{5}x-\dfrac{1}{5}x=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{5}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}:\dfrac{1}{5}=\dfrac{15}{4}\)
\(10⋮\left(x+5\right)\)
\(\Rightarrow x+5\in\left\{-1;1;-2;2;-5;5;-10;10\right\}\)
\(\Rightarrow x\in\left\{-6;-4;-7;-3;-10;0;-15;5\right\}\left(x\in Z\right)\)
Để 10 ⋮ (x+5) thì (x+5) ∈ Ư(10)
⇒ (x+5) ∈ \(\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
⇒ x ∈ {5;0;-3;-4;-6;-7;-10;-15}
\(\frac{17}{2}-\left|2x-\frac{5}{2}\right|=-\frac{7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{17}{2}-\frac{-7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{51}{6}+\frac{7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{29}{3}\)
\(2x-\frac{5}{2}=\frac{29}{3}\)hoặc \(2x-\frac{5}{2}=\frac{-29}{3}\)
Trường hợp 1:
\(2x-\frac{5}{2}=\frac{29}{3}\)
\(2x=\frac{29}{3}+\frac{5}{2}\)
\(2x=\frac{73}{6}\)
\(x=\frac{73}{6}:2\)
\(x=\frac{73}{12}\)
Trường hợp 2:
\(2x-\frac{5}{2}=\frac{-29}{3}\)
\(2x=\frac{-29}{3}+\frac{5}{2}\)
\(2x=\frac{-43}{6}\)
\(x=\frac{-43}{6}:2\)
\(x=\frac{-43}{12}\)
Vậy \(x=\frac{73}{12}\)hoặc \(x=\frac{-43}{12}\)
Tính
\(\dfrac{14}{5}\times10\)
\(=\dfrac{140}{5}=28\)
Tính:
`14/5` \(\times\) 10
= `14/5` \(\times\) `10/1`
= `14xx10/5xx1`
= `140/5`
= 28.