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\(A=\frac{\frac{1}{3}+\frac{1}{9}-\frac{1}{27}}{\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}\)
\(\Leftrightarrow\frac{\frac{11}{27}}{\frac{55}{27}}\)
\(=\frac{11}{27}.\frac{27}{55}=\frac{1}{5}\)
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\(x-\frac{10}{3}=\frac{7}{15}.\frac{3}{5}\)
\(x-\frac{10}{3}=\frac{7}{25}\)
\(x=\frac{7}{25}+\frac{10}{3}=\frac{271}{75}\)
Vậy x = 271/75
\(x-\frac{10}{3}\)\(=\)\(\frac{7}{15}\)\(.\)\(\frac{3}{5}\)
\(x-\frac{10}{3}\)\(=\) \(\frac{7}{25}\)
\(x\) \(=\)\(\frac{7}{25}\)\(+\)\(\frac{10}{3}\)
\(x\) \(=\) \(\frac{271}{75}\)
a, 1/2 x ( 2/9 + 3/7 - 5/27 )
= 1/2 x ( 41/63 - 5/27 )
= 1/2 x 88/189
= 44/189
b, (-5/8 + 1,75 + 8/35) : (-3 và 9/10)
= (-5/8 + 7/4 + 8/35) : (-39/10)
= (9/8 + 8/35) : (-39/10)
=379/280 : (-39/10)
= -379/1092
c, 1/3 x 5/7 - 5/27 x 36/14
= 5/21 - 10/21
= -5/21
Chúc bạn học tốt !
a) 1/2 x (\(\frac{42}{189}-\frac{81}{189}-\frac{35}{189}\))=1/2 x 88/189 = 44/189
b) ( \(\frac{-5}{8}+\frac{7}{4}+\frac{8}{35}\)) : \(-3\frac{9}{10}\)
= ( \(\frac{-175}{280}+\frac{490}{280}+\frac{64}{280}\)) : \(\frac{-39}{10}\)
=\(\frac{379}{280}\) x \(\frac{-10}{39}\)=\(\frac{-379}{1092}\)
c) \(\frac{5}{21}\)- \(\frac{10}{21}\)= \(\frac{-5}{21}\)
\(\frac{9}{27}=-\frac{5}{a}\)
=>-5x27=a x 9
a=-5x27/9
a=-15
-5/15=b/-18
=>-18x5/15=b
b=-6
a) Ta có: \(\dfrac{-3}{5}x+\dfrac{-7}{4}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{-3}{5}x=\dfrac{3}{10}+\dfrac{7}{4}=\dfrac{41}{20}\)
\(\Leftrightarrow x=\dfrac{41}{20}:\dfrac{-3}{5}=\dfrac{41}{20}\cdot\dfrac{-5}{3}\)
hay \(x=-\dfrac{41}{12}\)
Vậy: \(x=-\dfrac{41}{12}\)
\(\frac{\frac{1}{3}+\frac{1}{9}-\frac{1}{27}}{\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}=\frac{\frac{9}{27}+\frac{3}{27}-\frac{1}{27}}{\frac{45}{27}+\frac{15}{27}-\frac{5}{27}}=\frac{\frac{9+3-1}{27}}{\frac{45+15-5}{27}}=\frac{\frac{11}{27}}{\frac{55}{27}}=\frac{11}{55}=\frac{1}{5}.\)