NN
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
GG
14 tháng 4 2019
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
NT
17 tháng 8 2016
\(a\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\\ =>\left(x-\frac{1}{2}\right)=\frac{1}{3}\\ =>x=\frac{1}{3}+\frac{1}{2}\\ =>x=\frac{5}{6}\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\\ =>\left(x+\frac{1}{2}\right)=\frac{2}{5}\\ =>x=\frac{-1}{10}\)
d) (2x+3)2016=(2x+3)2018 khi 2x+3=0 hoặc 1
Nếu 2x+3=0
=2x=-3 ( loại )
Nếu 2x+3=1
=>2x=-2
=>x=-1 ( thỏa )
VL
3
F
16 tháng 4 2019
\(\left(75\%+\frac{1}{2}\right)\text{ : }\left|-1\frac{1}{2}\right|-\frac{5}{27}\cdot1,8\cdot2012^0\)
\(=\left(\frac{3}{4}+\frac{1}{2}\right):\left|-\frac{3}{2}\right|-\frac{5}{27}\cdot\frac{9}{5}\)
\(=\frac{5}{4}\text{ : }\frac{3}{2}-\frac{1}{3}\)
\(=\frac{5}{6}-\frac{1}{3}\)
\(=\frac{1}{2}\)
16 tháng 4 2019
\(\left(75\%+\frac{1}{2}\right)-\left|-1\frac{1}{2}\right|-\frac{5}{27}\cdot1,8\cdot2012^0\)
\(=\left(\frac{75}{100}+\frac{1}{2}\right)-\left|-\frac{3}{2}\right|-\frac{5}{27}\cdot\frac{9}{5}\cdot1\)
\(=\left(\frac{3}{4}+\frac{1}{2}\right)-\frac{3}{2}-\frac{5}{27}\cdot\frac{9}{5}\)
\(=\left(\frac{3}{4}+\frac{2}{4}\right)-\frac{3}{2}\cdot\frac{1}{3}\)
\(=\frac{5}{4}-\frac{1}{2}=\frac{5}{4}-\frac{2}{4}=\frac{3}{4}\)
Ez mà :)))
NG
18 tháng 6 2020
a)
\(\frac{8}{11}.\frac{14}{23}+\frac{9}{23}:\frac{11}{8}-\frac{8}{11} \)
\(=\frac{8}{11}.\frac{14}{23}+\frac{9}{23}.\frac{8}{11}-\frac{8}{11}\)
\(=\frac{8}{23}.(\frac{14}{23}+\frac{9}{23}-1)\)
\(=\frac{8}{23}.0\)
=0
b)
\(1,8.\frac{-20}{27}\)+(75%\(-\frac{5}{16}):3\frac{1}{2}\)
=\(\frac{9}{5}.\frac{-20}{27}+(\frac{3}{4}-\frac{5}{16}):\frac{7}{2}\)
=\(\frac{-4}{3}+\frac{1}{8}\)
=\(\frac{-29}{24}\)
7 tháng 8 2018
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~