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5 tháng 5 2022

\(=\left(40^1_4-25^1_4\right):\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\dfrac{1}{2021}\)

\(=15:0-\dfrac{1}{2021}\)

\(=0-\dfrac{1}{2021}\)

\(=\dfrac{-1}{2021}\)

 

 

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25

23 tháng 2 2023

\(1\dfrac{4}{5}+2\dfrac{5}{7}+3\dfrac{4}{5}+4\dfrac{5}{7}\)

\(\text{=}\left(1\dfrac{4}{5}+3\dfrac{4}{5}\right)+\left(2\dfrac{5}{7}+4\dfrac{5}{7}\right)\)

\(\text{=}1+3+\left(\dfrac{4}{5}+\dfrac{4}{5}\right)+2+4+\left(\dfrac{5}{7}+\dfrac{5}{7}\right)\)

\(\text{=}10+\dfrac{8}{5}+\dfrac{10}{7}\text{=}131\dfrac{1}{35}\)

8 tháng 3 2022

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8 tháng 3 2022

\(C=\dfrac{-5}{7}+\dfrac{-2}{7}+\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{-1}{5}=-1+1-\dfrac{1}{5}=\dfrac{-1}{5}\)

4 tháng 12 2017

a)\(\left|-0.75\right|+\dfrac{1}{4}-2\dfrac{1}{2}\)

=0.75+0.25-2.5

=1-2.5=-1.5

b)\(15.\dfrac{1}{5}:\left(\dfrac{-5}{7}\right)-2\dfrac{1}{5}.\left(\dfrac{-7}{5}\right)\)

=3.(-1.4)+3.08

=-4.2+3.08=-1.12

c)\(\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{20}{12}+\dfrac{7}{9}+\dfrac{12}{17}\)

=\(\dfrac{49}{51}-\dfrac{5}{3}+\dfrac{7}{9}+\dfrac{12}{17}\)

=\(\dfrac{-12}{17}+\dfrac{7}{9}+\dfrac{12}{17}\)

=\(\dfrac{11}{153}+\dfrac{12}{17}\)

=\(\dfrac{7}{9}\)

d)\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\)

=\(\dfrac{67}{75}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)

=-0.44+\(\dfrac{127}{175}\)

=\(\dfrac{2}{7}\)

5 tháng 5 2018

A,

\(\left(7\dfrac{4}{9}+3\dfrac{7}{11}\right)-3\dfrac{4}{9}=7\dfrac{4}{9}+3\dfrac{7}{11}-3\dfrac{4}{9}\)

\(=7\dfrac{4}{9}-3\dfrac{4}{9}+3\dfrac{7}{11}=4+3\dfrac{7}{11}=7\dfrac{7}{11}\)

5 tháng 5 2018

B,

\(5\dfrac{2}{7}.\dfrac{8}{11}+5\dfrac{2}{7}.\dfrac{5}{11}-5\dfrac{2}{7}.\dfrac{2}{11}=5\dfrac{2}{7}.\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)\)

\(=5\dfrac{2}{7}.1=5\dfrac{2}{7}\)

AH
Akai Haruma
Giáo viên
22 tháng 4 2023

Lời giải:
$S=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+...+\frac{69}{7^{70}}$

$7S=\frac{1}{7}+\frac{2}{7^2}+\frac{3}{7^3}+...+\frac{69}{7^{69}}$

$6S=7S-S=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{69}}-\frac{69}{7^{70}}$

$42S=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{68}}-\frac{69}{7^{69}}$

$\Rightarrow 42S-6S=(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{68}}-\frac{69}{7^{69}})-(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{69}}-\frac{69}{7^{70}})$

$\Rightarrow 36S=1-\frac{69}{7^{69}}-\frac{1}{7^{69}}+\frac{69}{7^{70}}$

Hay $36S=1-\frac{69.7-7-69}{7^{70}}=1-\frac{407}{7^{70}}$

$\Rightarrow S=\frac{1}{36}(1-\frac{407}{7^{70}})$

a) \(x=\dfrac{-2}{7}+\dfrac{9}{7}=1\) 

b) \(\dfrac{x}{3}=\dfrac{2}{5}+\dfrac{-4}{3}\) 

     \(\dfrac{x}{3}=\dfrac{-14}{15}\) 

\(\Rightarrow x=\dfrac{3.-14}{15}=\dfrac{-14}{5}\)

\(x=\dfrac{-2}{7}+\dfrac{9}{7}\) 

\(x=1\)

28 tháng 4 2023

Đây nha bạn:

=7−55.7+12−77.12+19−1212.19+28−1919.28+39−2828.39+40−3939.40

=15−17+17−112+112−119+119−128+128−139+139−140

=15−140=740