giúp em bài này ạ em thank trc ạ
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20 tháng 4
\(\dfrac{x+1}{3}=\dfrac{-3}{-9}\)
=>\(\dfrac{x+1}{3}=\dfrac{1}{3}\)
=>x+1=1
=>x=0
DN
1
20 tháng 4
\(\dfrac{x+1}{3}=\dfrac{-3}{-9}\)
=>\(\dfrac{x+1}{3}=\dfrac{1}{3}\)
=>x+1=1
=>x=0
TH
0
20 tháng 4
\(B=\left(3-\dfrac{1}{5}\right)\left(3-\dfrac{2}{5}\right)\cdot...\cdot\left(3-\dfrac{29}{5}\right)\)
\(=\left(3-\dfrac{15}{5}\right)\left(3-\dfrac{1}{5}\right)\cdot...\cdot\left(3-\dfrac{29}{5}\right)\)
\(=\left(3-3\right)\left(3-\dfrac{1}{5}\right)\cdot...\cdot\left(3-\dfrac{29}{5}\right)=0\)
a: \(\left(\dfrac{1}{4}+\dfrac{-3}{11}\right)+\left(\dfrac{2}{11}+\dfrac{-8}{11}+\dfrac{3}{4}\right)\)
\(=\dfrac{1}{4}+\dfrac{-3}{11}+\dfrac{2}{11}+\dfrac{-8}{11}+\dfrac{3}{4}\)
\(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)-\dfrac{9}{11}=1-\dfrac{9}{11}=\dfrac{2}{11}\)
b: \(\left(-1\dfrac{1}{3}+1\dfrac{1}{2}\right)+2\dfrac{1}{2}\)
\(=-1-\dfrac{1}{3}+1+\dfrac{1}{2}+2+\dfrac{1}{2}\)
\(=2+1-\dfrac{1}{3}=3-\dfrac{1}{3}=\dfrac{8}{3}\)
c: \(\left(-2\dfrac{1}{4}\right)+\dfrac{5}{2}:3\dfrac{3}{4}\)
\(=-\dfrac{9}{4}+\dfrac{5}{2}:\dfrac{15}{4}\)
\(=-\dfrac{9}{4}+\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{-9}{4}+\dfrac{2}{3}\)
\(=\dfrac{-27+8}{12}=-\dfrac{19}{12}\)
d: \(\dfrac{5}{8}:0,25-\left(0,6-2\dfrac{1}{4}\right):2\dfrac{1}{5}\)
\(=\dfrac{5}{8}:\dfrac{1}{4}-\left(\dfrac{3}{5}-\dfrac{9}{4}\right):\dfrac{11}{5}\)
\(=\dfrac{5}{2}-\dfrac{12-45}{20}\cdot\dfrac{5}{11}\)
\(=\dfrac{5}{2}-\dfrac{5}{20}\cdot\dfrac{-33}{11}=\dfrac{5}{2}+\dfrac{3}{4}=\dfrac{13}{4}\)
e: \(1\dfrac{1}{2}\cdot1\dfrac{1}{3}\cdot...\cdot1\dfrac{1}{999}\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{1000}{999}\)
\(=\dfrac{1000}{2}=500\)
f: \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)