a=78/35

b=22/12

c=1/1

d=40202090/4040090

e=1,24025667172...

f=871,82

ko biết đúng ko [0_0'] hihi

`a, 3-(x+5/7 )=9/21`

`=>x+5/7= 3-9/21`

`=>x+5/7= 63/21-9/21`

`=>x+5/7= 54/21`

`=>x= 54/21-5/7`

`=>x= 54/21 - 15/21`

`=>x= 39/21`

`=>x= 13/7`

`b, x/2+ x/5 = 17/10`

`=> (5x)/10 + (2x)/10=17/10`

`=> 7x/10=17/10`

`=> 7x.10=10.17`

`=>7x.10=170`

`=>7x=170:10`

`=>7x=17`

`=>x=17/7`

`c, 1/2x + 1/3 -1= 3 1/3`

`=>  1/2x + 1/3 -1=  10/3`

`=>   1/2x + 1/3=10/3+1`

`=>   1/2x + 1/3=10/3 + 3/3`

`=>   1/2x + 1/3=13/3`

`=>1/2 x= 13/3 -1/3`

`=> 1/2x= 12/3`

`=> 1/2x= 4`

`=>x= 4 :1/2`

`=>x= 4 xx 2`

`=>x=8`

\(a,3-\left(x+\dfrac{5}{7}\right)=\dfrac{9}{21}\\ x+\dfrac{5}{7}=3-\dfrac{9}{21}\\ x+\dfrac{5}{7}=\dfrac{18}{7}\\ x=\dfrac{18}{7}-\dfrac{5}{7}\\ x=\dfrac{13}{7}\\ b,\dfrac{x}{2}+\dfrac{x}{5}=\dfrac{17}{10}\\ \dfrac{5x}{10}+\dfrac{2x}{10}=\dfrac{17}{10}\\ \dfrac{7x}{10}=\dfrac{17}{10}\\ 7x=17\\ x=\dfrac{17}{7}\\ c,\dfrac{1}{2}x+\dfrac{1}{3}-1=3\dfrac{1}{3}\\ \dfrac{1}{2}x+\dfrac{1}{3}-1=\dfrac{10}{3}\\ \dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{10}{3}+1\\ \dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{13}{3}\\ \dfrac{1}{2}x=\dfrac{13}{3}-\dfrac{1}{3}\\ \dfrac{1}{2}x=4\\ x=4:\dfrac{1}{2}\\ x=10\)

a: =-1/3+1/3=0

b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)

c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)

d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)

a: =-1/3+1/3=0

b: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167

d: 

12+3/7-11+3/7 tính bằng cách thuận tiện nhé

a: \(\Leftrightarrow\dfrac{5}{3}+\dfrac{4}{3}< x< 3+\dfrac{1}{5}+1+\dfrac{4}{5}\)

=>3<x<5

=>x=4

b: \(\Leftrightarrow\dfrac{1}{3}:2x=-5+\dfrac{1}{4}=-\dfrac{19}{4}\)

=>\(2x=\dfrac{1}{3}:\dfrac{-19}{4}=\dfrac{1}{3}\cdot\dfrac{-4}{19}=\dfrac{-4}{57}\)

=>x=-2/57

c: \(\Leftrightarrow x\cdot\dfrac{-3}{2}=\dfrac{10}{3}-\dfrac{6}{7}=\dfrac{70-18}{21}=\dfrac{52}{21}\)

=>\(x=\dfrac{-52}{21}:\dfrac{3}{2}=\dfrac{-52}{21}\cdot\dfrac{2}{3}=\dfrac{-104}{63}\)

d: \(\Leftrightarrow70+18< x< 120+70\)

=>88<x<190

hay \(x\in\left\{89;90;...;188;189\right\}\)

\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)

\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)

\(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{2010}{2012}:2\)

\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{1005}{2012}\)

\(\Rightarrow\dfrac{1}{\left(x+1\right)}=\dfrac{1}{4}-\dfrac{1005}{2012}\)

\(\dfrac{1}{\left(x+1\right)}=\dfrac{-251}{1006}\)

\(\Rightarrow1:\left(x+1\right)=\dfrac{-251}{1006}\)

\(\left(x+1\right)=1:\dfrac{-251}{1006}\)

\(x+1=\dfrac{-1006}{251}\)

\(x=\dfrac{-1006}{251}-1\)

\(x=\dfrac{-1257}{251}\)

Nếu bạn tìm \(x\in Z\) hay \(x\in N\) thì \(x=\varnothing\) (không có x thoả mãn)

+) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2^{10}}=\dfrac{2^{10}-1}{2^{10}}\)

Vậy \(A=\dfrac{2^{10}-1}{2^{10}}\)

+) \(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)

\(\Rightarrow\dfrac{1}{2}F=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{380}\)

\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=\dfrac{1}{5}-\dfrac{1}{20}=\dfrac{3}{20}\Rightarrow F=\dfrac{3}{20}:\dfrac{1}{2}=\dfrac{3}{10}\)

Vậy \(F=\dfrac{3}{10}\)

+) \(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)

\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}=\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{25.28}\)

\(=\dfrac{4}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{25.28}\right)\)

\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)

Vậy \(G=\dfrac{2}{7}\)

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(A=1-\dfrac{1}{2^{10}}=\dfrac{1024-1}{1024}=\dfrac{1023}{1024}\)

\(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)

\(=\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{380}\)

\(=\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{19.20}\)

\(=2\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{20}\right)=2.\dfrac{3}{20}=\dfrac{3}{10}\)

\(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)

\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}\)

\(=\dfrac{4}{4.7}+\dfrac{4}{7.10}+\dfrac{4}{10.13}+...+\dfrac{4}{25.28}\)

\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)

Cần đáp án thoi

A

\(a,x-\dfrac{5}{7}=\dfrac{19}{21}\\ x=\dfrac{34}{21}\\ b,\dfrac{5}{3}-\left|x-\dfrac{1}{5}\right|=\dfrac{1}{3}\\ \left|x-\dfrac{1}{5}\right|=\dfrac{4}{3}\\ TH1:x-\dfrac{1}{5}=\dfrac{4}{3}\\ x=\dfrac{23}{15}\\ TH2:x-\dfrac{1}{5}=-\dfrac{4}{3}\\ x=-\dfrac{17}{15}\\ c,x-\dfrac{2}{5}=\dfrac{1}{4}\\ x=\dfrac{13}{20}\\ d,5\sqrt{x}-30=15\\ 5\sqrt{x}=45\\ \sqrt{x}=9\\ x=9^2=81\)

https://hoc24.vn/question/246430.html

\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)

\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{2010}{2012}:2\)

\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{1005}{2012}\)

\(\dfrac{1}{\left(x+1\right)}=\dfrac{1}{4}-\dfrac{1005}{2012}\)

\(\dfrac{1}{\left(x+1\right)}=\dfrac{-251}{1006}\)

\(\Rightarrow1:\left(x+1\right)=\dfrac{-251}{1006}\)

\(\left(x+1\right)=1:\dfrac{-251}{1006}\)

\(x+1=\dfrac{-1006}{251}\)

\(x=\dfrac{-1006}{251}-1\)

\(x=\dfrac{-1257}{251}\)

Vì \(x\in N\) nên \(x=\varnothing\) (không có giá trị nào của x thoả mãn)

Ta có :

\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+.........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2010}{2012}\)

\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+..........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2010}{2012}\)

\(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{7.8}+.......+\dfrac{2}{x\left(x+1\right)}=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{7.8}+...........+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.........+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{x+1}\right)=\dfrac{2010}{2012}\)

\(\dfrac{1}{4}-\dfrac{1}{x+1}=\dfrac{2010}{2012}:2\)

\(\dfrac{1}{4}-\dfrac{1}{x+1}=\dfrac{1005}{2012}\)

\(\dfrac{1}{x+1}=\dfrac{1}{4}-\dfrac{1005}{2012}\)

\(\dfrac{1}{x+1}=\dfrac{-215}{1006}\)

\(\Rightarrow1.1006=\left(x+1\right).\left(-215\right)\)

\(1006=\left(x+1\right).\left(-215\right)\)

\(x+1=1006:\left(-215\right)\)

\(x+1=\dfrac{-1006}{215}\)

\(x=\dfrac{-1006}{215}-1\)

\(x=\dfrac{-1221}{215}\)(ko thỏa mãn \(x\in N\))

Vậy ko tìm dc giá trị của x thỏa mãn theo yêu cầu

~ Học tốt ~