2 1/3 + 5 - 1 1/3 + 30% + 2,75 - 5
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1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42
2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57
3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165
4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315
5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5
\(-2\dfrac{3}{4}.\left(-0,4\right)+1\dfrac{3}{5}.2,75-1,2:\dfrac{4}{11}\)
\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}:\dfrac{4}{11}\)
\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}.\dfrac{11}{4}\)
\(=\dfrac{11}{4}\left(\dfrac{-2}{5}+\dfrac{8}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{11}{4}.0\)
\(=0\)
a . ( 2,75 + 125% ) x 5/2 - 0,4 x 2 : 8% b . ( 1/3 - 25% ) : 50% + ( 1/3 + 0,25 ) x 1/7
= ( 2,75 + 1,25 ) x 2,5 - 0,4 x 2 : 0,08 = ( 1/3 - 1/4 ) : 1/2 + ( 1/3 + 1/4 ) x 1/7
= 4 x 2,5 - 0,4 x 2 : 0,8 = 1/12 : 1/2 + 1/12 x 1/7
= 10 - 0,8 : 0,8 = 1/12 x 2 + 1/12 x 1/7
= 10 - 1 = 1/12 x ( 2 + 1/7 )
= 9 = 1/12 x 15/7 = 5/28
\(-2\dfrac{3}{4}.\left(-0,4\right)+1\dfrac{2}{5}.2,75-\left(-1,2\right):\dfrac{4}{11}\)
\(=\dfrac{11}{4}.\dfrac{2}{5}+\dfrac{7}{5}.\dfrac{11}{4}+\dfrac{6}{5}.\dfrac{11}{4}\)
\(=\dfrac{11}{4}.\left(\dfrac{2}{5}+\dfrac{7}{5}+\dfrac{6}{5}\right)\)
\(=\dfrac{11}{4}.\dfrac{15}{5}\)
\(=\dfrac{11}{4}.3\)
\(=\dfrac{33}{4}\)
a) \(2\frac{3}{4}\cdot\left(-0,4\right)-1\frac{3}{5}\cdot2,75+1,2:\frac{4}{11}\)
\(=2\frac{3}{4}\cdot\left(-\frac{2}{5}\right)-1\frac{3}{5}\cdot\frac{11}{4}+\frac{6}{5}:\frac{4}{11}\)
\(=\frac{11}{4}\cdot\left(-\frac{2}{5}\right)-1\frac{3}{5}\cdot\frac{11}{4}+\frac{6}{5}\cdot\frac{11}{4}\)
\(=\frac{11}{4}\left(-\frac{2}{5}-1\frac{3}{5}+\frac{6}{5}\right)\)
\(=\frac{11}{4}\left(-\frac{2}{5}-\frac{8}{5}+\frac{6}{5}\right)\)
\(=\frac{11}{4}\cdot\left(-\frac{4}{5}\right)=\frac{11}{1}\cdot\left(-\frac{1}{5}\right)=-\frac{11}{5}\)
b) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)....\left(\frac{1}{31}+1\right)\)
\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{31}+\frac{31}{31}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{32}{31}\)
\(=\frac{3\cdot4\cdot5\cdot...\cdot32}{2\cdot3\cdot4\cdot...\cdot31}=\frac{32}{2}=16\)
c) Đặt \(C=1+2+3+...+30\)
Số số hạng là : \(\left(30-1\right):1+1=30\)(số)
Tổng của dãy số là : \(\frac{\left(1+30\right)\cdot30}{2}=465\)
Do đó : \(\frac{930}{C}=\frac{930}{465}=2\)
\(=\dfrac{11}{4}\cdot\dfrac{2}{5}-\dfrac{8}{5}\cdot\dfrac{11}{4}+\dfrac{-6}{5}\cdot\dfrac{11}{4}\)
=11/4x(-12/5)=-132/20=-33/5
=\(\frac{81}{20}=4,05\)