Đặt \(A=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)

\(A=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{93.95}+\frac{1}{95.97}\right)\)

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

\(A=\frac{1}{99.97}-\left(\frac{1}{2}.\left(1-\frac{1}{97}\right)\right)=\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}=\frac{1}{99.97}-\frac{48}{97}=-\frac{4751}{9603}\)

\(=\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)

\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)

\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=\dfrac{1-48\cdot99}{97\cdot99}=\dfrac{-4751}{9603}\)

bn tách  1/ 97 .95 = 1/2 . ( 1/95 -1/97) nha! rồi sử dụng phương pháp khử liên tiếp ! 

123987564210

bài này có thể sai đề, viết lại

Áp dụng tc của dãy tỉ số bằng nhau :

\(\frac{2.\left(x-1\right)}{4}=\frac{3.\left(y-2\right)}{9}=\frac{z-3}{4}=\frac{2x+3x-z-2-6+3}{4+9-4}=\frac{90}{9}=10\)

\(=>\hept{\begin{cases}\frac{x-1}{2}=10=>x-1=20=>x=21\\\frac{y-2}{3}=10=>y-2=30=>y=32\\\frac{z-3}{4}=10=>z-3=40=>z=43\end{cases}}\)

Vậy ...

Trả lời:

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)

Đặt\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)

\(\Rightarrow\hept{\begin{cases}x-1=2k\\y-2=3k\\z-3=4k\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2k+1\\y=3k+2\\z=4k+3\end{cases}}\)

Mà\(2x+3y-z=95\)

\(\Rightarrow2\left(2k+1\right)+3\left(3k+2\right)-\left(4k+3\right)=95\)

\(\Leftrightarrow4k+2+9k+6-4k-3=95\)

\(\Leftrightarrow9k+5=95\)

\(\Leftrightarrow9k=90\)

\(\Leftrightarrow k=10\)

\(\Rightarrow\hept{\begin{cases}x=2.10+1=21\\y=3.10+2=32\\z=4.10+3=43\end{cases}}\)(Thỏa mãn)

Vậy\(\hept{\begin{cases}x=21\\y=32\\z=43\end{cases}}\)

Hok tốt!

Good girl

Sửa đề: \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{3.1}\)

\(=\dfrac{1}{97.99}-\left(\dfrac{1}{1.3}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(\dfrac{2}{1.3}+...+\dfrac{2}{93.95}+\dfrac{2}{95.97}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+...+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{97.99}\right)-\dfrac{1}{2}.\dfrac{96}{97}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{48}{97}\)

.........................

Đề đúng r mà sửa j

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=10\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=10\\\dfrac{y-2}{3}=10\\\dfrac{z-3}{4}=10\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=21\\y=32\\z=43\end{matrix}\right.\)

a) \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)

\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{19}-\frac{1}{19}+\frac{1}{23}-\frac{1}{23}+\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)

\(=\frac{1}{2}-\frac{2}{27}\)

\(=\frac{23}{54}\)

b) \(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)

\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)

\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+\frac{1}{20}-...-\frac{1}{95}-\frac{1}{100}\right)\)

\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)

\(=1-\frac{1}{5}.\frac{19}{100}\)

\(=1-\frac{19}{500}\)

\(=\frac{481}{500}\)

Dãy tỉ số bằng nhau ạ!

Áp dụng tính chất của dayxc tỉ số bằng nhau ta có :

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-2}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{2x-2+3y-6-z+2}{4+9-4}=\frac{89}{9}.\)

Tù đó rồi 

=> z , y , z nha easy quá còn j nx