-_-......................um hà 

Bà như ......................

bài này áp dụng công thức lớp 4 là ra mà

\(\text{Đặt }\frac{x}{2}=\frac{y}{3}=k\)

\(\Rightarrow x=2k;y=3k\)

Từ \(\frac{x+2}{x-2}=\frac{2k+2}{2k-2}=\frac{2\left(k+1\right)}{2\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)

\(;\frac{y+3}{y-3}=\frac{3k+3}{3k-3}=\frac{3\left(k+1\right)}{3\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\)

\(\text{Từ (1) và (2) }\Rightarrow\frac{x+2}{x-2}=\frac{x+3}{x-3}\left(\text{đpcm}\right)\)

Đặt \(\frac{x}{2}\) và \(\frac{y}{3}=k\)

Từ đó ta có : \(\frac{x+2}{x-2}=\frac{2k+2}{2k-2}=\frac{k+1}{k-1}\left(1\right)\) 

\(\frac{y+3}{y-3}=\frac{3k+3}{3k-3}=\frac{k+1}{k-1}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\frac{x}{2}=\frac{y}{3}\)

Trả lời

\(A=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{2.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}{\frac{7}{6}-\frac{7}{8}-\frac{7}{10}}\right):\left(1^2+2^2+...+2015^2\right).\)

\(A=\left(\frac{2}{7}-\frac{2}{7}\right):\left(1^2+2^2+3^2+...+2015^2\right)\)

\(A=0:\left(1^2+2^2+3^2+.....+2015^2\right)\)

A=0

Study well 

\(A=...\)

\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\left(1^2+2^2+...+2015^2\right)\)

\(=\left[\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\right]:\left(1^2+2^2+...+2015^2\right)\)

\(=\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right):\left(1^2+2^2+...+2015^2\right)\)

\(=\left(\frac{2}{7}-\frac{2}{7}\right):\left(1^2+2^2+...+2015^2\right)\)

\(=0:\left(1^2+2^2+...+2015^2\right)=0\)

A = \(\left(\frac{1}{8}+\frac{1}{8.15}+\frac{1}{15.22}+...+\frac{1}{43.50}\right)\cdot\frac{4-3-5-7-...-49}{217}\)

A = \(\frac{1}{7}.\left(\frac{7}{1.8}+\frac{7}{8.15}+\frac{7}{15.22}+...+\frac{7}{43.50}\right)\cdot\frac{4-\left(3+5+7+...+49\right)}{217}\)

A = \(\frac{1}{7}.\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+...+\frac{1}{43}-\frac{1}{50}\right)\cdot\frac{4-\left(49+3\right)\left[\left(49-3\right):2+1\right]:2}{217}\)

A = \(\frac{1}{7}\cdot\left(1-\frac{1}{50}\right)\cdot\frac{4-52.24:2}{217}\)

A = \(\frac{1}{7}\cdot\frac{49}{50}\cdot\frac{4-624}{217}\)

A = \(\frac{7}{50}\cdot\frac{-620}{217}=-\frac{2}{5}\)

a) \(A=3+3^2+...+3^{2019}\)

\(\Rightarrow3A=3^2+3^3+...+3^{2020}\)

Lấy 3A trừ A theo vế ta có : 

\(3A-A=\left(3^2+3^3+...+3^{2020}\right)-\left(3+3^2+3^3+...+3^{2019}\right)\)

\(2A=3^{2020}-3\)

\(A=\frac{3^{2020}-3}{2}\)

b) Ta có : \(2A=3^{2020}-3\)

\(=3^{505.4}-3\)

\(=\left(3^4\right)^{505}-3\)

\(=81^{505}-3\)

\(=\overline{....1}-3\)

\(=...8\)

\(\Rightarrow A=...4\)

Vậy chữ số tận cùng của A là 4

a) 2x-5=3+2x-7x

2x-2x+7x=3+5

7x=8

  x=8/7

vậy x=8/7

a) 2x - 5 = 3 + 2x - 7x

=> 2x - 2x + 7x = 3 +5 

=> 7x = 8

=> x = 8/7

b) \(\left(2x-1\right)^2=\left(2x-1\right)^5\)

=> \(\left(2x-1\right)^2-\left(2x-1\right)^5=0\)

=> \(\left(2x-1\right)^2\left[1-\left(2x-1\right)^3\right]=0\)

=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)^3=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^3=1\end{cases}}\)

=> \(\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)

\(1^2:0,25^2=16;2^2:0,5^2=16;....;10^2:2,5^2=16\Rightarrow B=\frac{385}{16}\)

Gọi A=\(\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

A= -(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\))

A=-(1-\(\frac{1}{100}\))

A=-(\(\frac{99}{100}\))

A=-99/100

\(\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Leftrightarrow-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(\Leftrightarrow\)\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Leftrightarrow-\left(1-\frac{1}{100}\right)\)

\(\Leftrightarrow-\left(\frac{99}{100}\right)\)

\(=-\frac{99}{100}\)