Violympic toán 7

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thanh nguyen van long

\(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}với\) x=\(-\dfrac{1}{3}\)

6)a) \(\left|\dfrac{5}{3}:x\right|=\left|-\dfrac{1}{6}\right|\)

b)\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|-\dfrac{3}{4}\right|\)

JakiNatsumi
3 tháng 10 2018 lúc 21:24

\(B=\dfrac{\dfrac{2}{10}-\dfrac{3}{8}+\dfrac{5}{11}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)

\(B=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{10}{22}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)

\(B=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}\)\(-\dfrac{1}{3}\)

\(B=\dfrac{-2}{3}-\dfrac{1}{3}=-1\)

6)a) \(\left|\dfrac{5}{3}:x\right|=\left|\dfrac{-1}{6}\right|\)

\(\left|\dfrac{5}{3}:x\right|=\dfrac{1}{6}\)

\(\dfrac{5}{3}:x=\dfrac{1}{6}\) hoặc \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)

*TH1 : \(\dfrac{5}{3}:x=\dfrac{1}{6}\)

\(x=\dfrac{5}{3}:\dfrac{1}{6}=10\)

*TH2 : \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)

\(x=\dfrac{5}{3}:\dfrac{-1}{6}=-10\)

Vậy \(x\)\(\left\{10;-10\right\}\)

\(b,\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|\dfrac{-3}{4}\right|\)

\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)

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

\(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\) hoặc \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)

TH1 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\)

\(\dfrac{3}{4}x=\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{9}{4}\)

\(x=\dfrac{9}{4}:\dfrac{3}{4}=3\)

TH2 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)

\(\dfrac{3}{4}x=\dfrac{-3}{2}+\dfrac{3}{4}=\dfrac{-3}{4}\)

\(x=\dfrac{-3}{4}:\dfrac{3}{4}=-1\)

Vậy \(x\)\(\left\{3;1\right\}\)


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