\(\frac{9}{10}.100-\left(\frac{5}{2}\left(y+\frac{206}{100}\right)\right):\frac{1}{2}=89\)

\(90-\left(\frac{5}{2}\left(y+\frac{103}{50}\right)\right)=89.\frac{1}{2}\)

\(90-\left(\frac{5}{2}\left(y+\frac{103}{50}\right)\right)=\frac{89}{2}\)

\(\frac{5}{2}\left(y+\frac{103}{50}\right)=90-\frac{89}{2}\)

\(\frac{5}{2}\left(y+\frac{103}{50}\right)=\frac{180}{2}-\frac{89}{2}\)

\(\frac{5}{2}\left(y+\frac{103}{50}\right)=\frac{91}{2}\)

\(y+\frac{103}{50}=\frac{91}{2}.\frac{2}{5}\)

\(y+\frac{103}{50}=\frac{91}{5}\)

\(y=\frac{91}{5}-\frac{103}{50}\)

\(y=\frac{910}{50}-\frac{103}{50}\)

\(y=\frac{807}{50}\)

a) \(\frac{2^{12}x3^5-4^6.9^2}{\left(2^2x3\right)^6+8^4x3^5}=\frac{2^{12}x3^5+\left(2^2\right)^6x\left(3^2\right)^2}{2^{12}x3^6+\left(2^3\right)^4x3^5}\)

\(=\frac{2^{12}x3^5-2^{12}x3^4}{2^{12}x3^6+2^{12}x3^5}=\frac{2^{12}x3^4x\left(3-1\right)}{2^{12}x3^5x\left(3+1\right)}\)

\(=\frac{2}{3.4}=\frac{1}{3.2}=\frac{1}{6}\)

b) \(\frac{1}{9x10}-\frac{1}{8x9}-\frac{1}{7x8}-\frac{1}{6x7}-\frac{1}{5x6}-\frac{1}{4x5}-\frac{1}{3x4}-\frac{1}{2x3}-\frac{1}{1x2}\)

\(=-\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}\right)\)

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

\(=-\left(1-\frac{1}{10}\right)\)

\(=\frac{-9}{10}\)

sorry bn nha! mk ko bk lm phần c

trả lời :

a) 1/6

b)-9/10

Đặng Thị Dung cả 2 phần bạn CÔNG CHÚA ÔRI làm sai hết rồi. Nếu muốn biết sai ở đâu thì kb rồi nhắn tin mk chỉ cho

c,\(\frac{1}{2}\div\left(-1\frac{1}{2}\right)\div1\frac{1}{3}:\left(-1\frac{1}{4}\right)\div...\div1\frac{1}{99}\div\left(-1\frac{1}{100}\right)\)

\(=\frac{1}{2}\div\left(-\frac{3}{2}\right)\div\frac{4}{3}\div\left(-\frac{5}{4}\right)\div...\div\frac{100}{99}\div\left(-\frac{101}{100}\right)\)

\(=\frac{1}{2}\times\left(-\frac{2}{3}\right)\times\frac{3}{4}\times...\times\left(-\frac{100}{101}\right)\)

=\(=\frac{1.\left(-2\right).3.....\left(-100\right)}{2.3.4.....101}=\frac{1}{101}\)

\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \frac{9}{10}\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ 90-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=1\\ \frac{5}{2}:\left(X+\frac{206}{100}\right)=\frac{1}{2}\\ X+\frac{206}{100}=5\\ X=\frac{500}{100}-\frac{206}{100}\\ X=\frac{294}{100}=\frac{147}{50}\)

Vậy \(X=\frac{147}{50}\)

( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ......+ 1/9 - 1/10) . 100 - [ 5/2 : ( x + 103/50 ) ] = 89 . 1/2

( 1 - 1/10) . 100 - [ 5/2 : ( x + 103/50 ) ] = 89/2

90 - 5/2 : ( x + 103/50 ) = 89/2

5/2 : ( x + 103/50 ) = 90 - 89/2

5/2 : ( x + 103/50 ) = 91/2

x + 103/50 = 5/2 : 91/2

x + 103/50 = 5/91

x = 5/91 - 103/50

x = -9,123/4550

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{500}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{499}{500}\)

\(1-\frac{1}{x+1}=\frac{499}{500}\)

\(\frac{1}{x+1}=1-\frac{499}{500}=\frac{1}{500}\)

=> x + 1 = 500

=> x = 500 - 1

=> x = 499

Vậy x = 499

1/1.2 + 1/2.3 + 1/3.4 +...+ 1/x.(x+1)=499/500

1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 +...+ 1/x -1/(x+1) =499/500

1-1/(x+1)=499/500

=>x/(x+1)=499/500

=>x=499

Ta có: 1/1x2 + 1/2x3 + 1/3x4 +...+ 1/X x (X + 1) = 499/500

=> 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/X - 1/(X + 1) = 499/500

=> 1 - 1/(X + 1) = 499/500

=>      1/(X + 1) = 1 - 499/500

=>      1/(X + 1) = 1/500

=>          X + 1 = 500

=>          X       = 500 - 1

=>          X       = 499 

Đáp số: X = 499

1+1=3 :)))

Bài 2:

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)

\(=\frac{1}{2004}\)

Bài 2

=1/2 x 2/3 ... x 2003/2004

=1/2004

2.

1/2004

sudy well

study well