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\(=\dfrac{1}{97\cdot99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)

\(=\dfrac{1}{97\cdot99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{97\cdot99}-\dfrac{1}{2}\cdot\dfrac{96}{97}\)

\(=\dfrac{1}{97\cdot99}-\dfrac{48}{97}=\dfrac{1-48\cdot99}{97\cdot99}=\dfrac{-4751}{9603}\)

14 tháng 1 2020

\(=\frac{1}{99.97}-\frac{1}{2}\left(\frac{1}{95}-\frac{1}{97}+\frac{1}{93}-\frac{1}{95}+...+\frac{1}{3}-\frac{1}{5}+1-\frac{1}{3}\right)\)

\(=\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}=\frac{1}{99.97}-\frac{48}{97}=-\frac{4751}{99.97}\)

16 tháng 8 2017

\(-\dfrac{1}{30}-\dfrac{1}{15.13}-\dfrac{1}{13.11}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

Đặt \(A=-\dfrac{1}{30}-\dfrac{1}{15.13}-\dfrac{1}{13.11}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(\Rightarrow2A=-\dfrac{1}{15}-\left(\dfrac{2}{15.13}+\dfrac{2}{13.11}+...+\dfrac{2}{5.3}+\dfrac{2}{3.1}\right)\)

\(\Rightarrow2A=-\dfrac{1}{15}-\left(\dfrac{1}{15}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{11}+...+\dfrac{1}{5}-\dfrac{1}{3}+\dfrac{1}{3}-1\right)\)

\(\Rightarrow2A=-\dfrac{1}{15}-\left(\dfrac{1}{15}-1\right)=-\dfrac{1}{15}-\dfrac{1}{15}+1\)

\(=\dfrac{13}{15}\)

Vậy...................

Chúc bạn học tốt!!!

20 tháng 8 2019

\(9^{x+1}-5.3^{2x}=72\)

\(\rightarrow9^x.9-5.\left(3^2\right)^x=72\)

\(\rightarrow9^x.9-5.9^x=72\)

\(\rightarrow9^x\left(9-5\right)=72\)

\(\rightarrow4.9^x=72\)

\(\rightarrow9^x=18\)

\(\Rightarrow x=1,315...\)

Vậy \(x=1,315...\)

a: \(7\cdot3^x=5\cdot3^7+2\cdot3^7\)

\(\Leftrightarrow7\cdot3^x=7\cdot3^7\)

=>3x=37

hay x=7

b: \(4^{x+3}-3\cdot4^{x+1}=13\cdot4^{11}\)

\(\Leftrightarrow4^{x+1}\left(4^2-3\right)=13\cdot4^{11}\)

=>x+1=11

hay x=10

d: \(\left(x-1\right)^{13}=\left(x-1\right)^{12}\)

\(\Leftrightarrow\left(x-1\right)^{12}\left(x-2\right)=0\)

hay \(x\in\left\{1;2\right\}\)

23 tháng 12 2019

.-. ... anh dùng mấy tính tính được mà

23 tháng 12 2019

6,1 (làm tròn)

8 tháng 7 2019

\(\left\{x^2-\left[6^2-\left(8^2-9\cdot7\right)^3-7\cdot5\right]^3-5\cdot3\right\}^3=1\\ \Rightarrow x^2-\left[36-\left(64-63\right)^3-35\right]^3-15=1\\ \Rightarrow x^2-\left[36-35-1^3\right]^3=16\\ \Rightarrow x^2-0^3=16\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)Vậy \(x=\pm4\)

17 tháng 7 2019

{\(x\)2-[62-(82-9.7)3-7.5]3-5.3}3=1

{\(x\)2-[36-(64-63)3-35]3-15}3=1

{\(x\)2-[36-1-35]3-15}3=1

{\(x\)2-0-15}3=13

\(x\)2-0=1+15

\(x\)2-0=16

\(x\)2=16+0

\(x\)2=16

\(x\)2=16⇒42

\(x\)2=4

0
23 tháng 5 2016

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

 

 

23 tháng 5 2016

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1}X2+\frac{1}{2}X3+\frac{1}{3}X4+\frac{1}{4}X5+\frac{1}{5}X6+\frac{1}{6}X7\)

\(=\) \(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)