(5/2-1/2):2+5/9×3/4
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1: =1/8*9/4=9/32
2: =8/27*243/32=9/4
3: =(5/4*4/5)^5*(4/5)^2=16/25
4: \(=\left(-\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^2\cdot\left(\dfrac{6}{5}\right)^2=\dfrac{36}{25}\)
5: \(=\left(-\dfrac{4}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^{10}=\left(-1\right)\left(\dfrac{3}{4}\right)^7=-\left(\dfrac{3}{4}\right)^7\)
6: \(=\left(\dfrac{1}{3}\cdot\dfrac{-9}{2}\right)^4\left(-\dfrac{9}{2}\right)^2=\left(-\dfrac{3}{2}\right)^4\cdot\dfrac{81}{4}=\dfrac{9}{4}\cdot\dfrac{81}{4}=\dfrac{729}{16}\)
8: =(0,2*5)^4*5^2=25
10: =-0,5^5*2^10
=-0,5^5*2^5*2^5
=-32
13: =(0,5*2)^2*2^2=4
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
1; = ( -4/10 + 3/10 ) : ( -2/5 + 2/3 ) = -1/10 : ( -6/15 + 10/15 ) = -1/10 : 4/15 = -1/10 . 15/4 = -15/40 = -3/8
2; = 25/2 . -5/7 + 39/4 + -3/2 . 5/7 = -125/14 + 39/4 + -15/14 = ( -125/14 + -15/14 ) + 39/4 = -10 + 39/4 = -40/4 + 39/4 = -1/4
3; = 5/52 + 35/52 + 40/52 = 40/52 + 40/52 = 80/52 = 20/13
4; = ( -39/52 + 20/52 ) . 7/2 - ( 117/52 + 32/52 ) . 7/2 = -19/52 . 7/2 - 149/52 . 7/2 = ( -19/52 + -149/52 ) . 7/2 = -168/52 .7/2 = -147/13
5; = ( 36/12 + -9/12 + 8/12 ) - ( -12/6 + -8/6 + -9/6 ) - ( 6/6 - 14/6 - 27/6 ) = 35/12 + 10/12 + 70/12 = 115/12
6; = -1/3 + -8/35 +-2/9 + -1/135 +4/5 +-4/9 +3/7 = (-1/3 + -2/9 + -4/9 ) + ( -8/35 + 4/5 + 3/7 ) + -1/135 = ( -1/3 + -2/3 ) + ( -8/35 + 28/35 + 15/35 ) + -1/135 = -1 + 1 + -1/135 = -1/135
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
= \(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(\frac{10}{11}\)
A= \(\frac{10}{11}:\frac{2}{3}\)
A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)
d) giả tương tự câu c kết quả \(\frac{25}{11}\)
tổng đặc biệt đó bạn
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(1-\frac{1}{10}=\frac{9}{10}\)
những câu sau cũng áp dụng như vậy nhé
1) Ta có: \(\frac{-4}{7}-\frac{11}{19}+\frac{13}{19}\cdot\frac{-3}{7}+\frac{2}{19}:\frac{-7}{4}\)
\(=\frac{-4}{7}-\frac{11}{19}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-76}{133}-\frac{77}{133}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-200}{133}\)
2) Ta có: \(\left(\frac{-4}{9}+\frac{3}{5}\right):\frac{1}{\frac{1}{5}}+\left(\frac{1}{5}-\frac{5}{9}\right):\frac{1}{\frac{1}{5}}\)
\(=\left(\frac{-4}{9}+\frac{3}{5}\right)\cdot\frac{1}{5}+\left(\frac{1}{5}-\frac{5}{9}\right)\cdot\frac{1}{5}\)
\(=\frac{1}{5}\left(\frac{-4}{9}+\frac{3}{5}+\frac{1}{5}-\frac{5}{9}\right)\)
\(=\frac{1}{5}\left(-1+\frac{4}{5}\right)\)
\(=\frac{1}{5}\cdot\frac{-1}{5}=\frac{-1}{25}\)
3) Ta có: \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\)
\(=\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\)
\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\)
\(=\frac{27}{70}\)
4) Ta có: \(\frac{2}{7}-\left(-\frac{13}{15}+\frac{4}{9}\right)-\left(\frac{5}{9}-\frac{2}{15}\right)\)
\(=\frac{2}{7}+\frac{13}{15}-\frac{4}{9}-\frac{5}{9}+\frac{2}{15}\)
\(=\frac{2}{7}+1-1=\frac{2}{7}\)
a: =91/105+60/105-101/105
=50/105=10/21
c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)
d: =2-2/9
=18/9-2/9
=16/9
e: =24/36-9/36+8/36
=23/36
g: =5/2+1/2
=3
\(\left(\frac{5}{2}-\frac{1}{2}\right):2+\frac{5}{9}\times\frac{3}{4}\)
\(=\frac{4}{2}:2+\frac{5}{9}\times\frac{3}{4}\)
\(=1+\frac{5}{9}\times\frac{3}{4}\)
\(=1+\frac{15}{36}\)
\(=1+\frac{5}{12}\)
\(=\frac{17}{12}\)
\(\left(\dfrac{5}{2}-\dfrac{1}{2}\right):2+\dfrac{5}{9}\times\dfrac{3}{4}\\ =2:2+\dfrac{15}{36}\\ =1+\dfrac{5}{12}\\ =\dfrac{12}{12}+\dfrac{5}{12}\\ =\dfrac{17}{12}\)