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b)\(\dfrac{1}{7}B=\dfrac{1}{10.18}+\dfrac{1}{18.26}+\dfrac{1}{26.34}+...+\dfrac{1}{802.810}\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{8}{10.18}+\dfrac{8}{18.26}+\dfrac{8}{26.34}+...+\dfrac{8}{802.810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{34}+...+\dfrac{1}{802}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}.\dfrac{8}{81}\)
\(\dfrac{1}{7}B=\dfrac{1.8}{8.81}\)
\(\dfrac{1}{7}B=\dfrac{1}{81}\)
\(B=\dfrac{1}{81}:\dfrac{1}{7}\)
\(B=\dfrac{7}{81}\)
`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
a) \(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{6-3+2}{6}=\dfrac{1}{6}\)
\(b.\) \(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{6+10}{15}=\dfrac{16}{15}\)
\(c.\) \(\dfrac{7}{11}.\dfrac{3}{4}+\dfrac{7}{11}.\dfrac{1}{4}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{16}{44}=\dfrac{21+7+16}{44}=\dfrac{44}{44}=1\)
a/\(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}\)
b/\(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{16}{15}\)
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
a: =>x-3=9
=>x=12
b: =>10-x=-26
=>x=36
c: =>x:4-1=2
=>x:4=3
=>x=12
d: =>x^2=4
=>x=2 hoặc x=-2
e: =>(x-2)^2=100
=>x-2=10 hoặc x-2=-10
=>x=12 hoặc x=-8
a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=3+\dfrac{2}{3}-\dfrac{1}{3}\)
\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)
b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)
\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)
\(=-10\cdot\dfrac{1}{5}=-2\)
c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)
\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)
\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)
\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)
\(=\dfrac{16}{16}-\dfrac{25}{25}\)
\(=1-1=0\)
e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)
\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)
\(=\dfrac{5}{6}+\dfrac{4}{3}\)
\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)
chữ đẹp hem:
chữ đẹp ạ:> cảm ơn bạn UwU nếu đc thì bạn có thể giúp mk câu D đó đc ko ạ?