Tìm x trong biểu thức sau:
\(\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{8x9}+\frac{1}{9x10}\right)x100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
Tìm y :
\(\frac{9}{10}x100-\left(\frac{5}{2}:\left(y+\frac{206}{100}\right)\right):\frac{1}{2}=89\)
\(\frac{9}{10}.100-\left(\frac{5}{2}\left(y+\frac{206}{100}\right)\right):\frac{1}{2}=89\)
\(90-\left(\frac{5}{2}\left(y+\frac{103}{50}\right)\right)=89.\frac{1}{2}\)
\(90-\left(\frac{5}{2}\left(y+\frac{103}{50}\right)\right)=\frac{89}{2}\)
\(\frac{5}{2}\left(y+\frac{103}{50}\right)=90-\frac{89}{2}\)
\(\frac{5}{2}\left(y+\frac{103}{50}\right)=\frac{180}{2}-\frac{89}{2}\)
\(\frac{5}{2}\left(y+\frac{103}{50}\right)=\frac{91}{2}\)
\(y+\frac{103}{50}=\frac{91}{2}.\frac{2}{5}\)
\(y+\frac{103}{50}=\frac{91}{5}\)
\(y=\frac{91}{5}-\frac{103}{50}\)
\(y=\frac{910}{50}-\frac{103}{50}\)
\(y=\frac{807}{50}\)
Tính :
a) \(\frac{2^{12}x3^5-4^6x9^2}{\left(2^2x3\right)^6+8^4x3^5}\)
b) \(\frac{1}{9x10}\)\(-\)\(\frac{1}{8x9}\)\(-\)\(\frac{1}{7x8}\)\(-\)\(\frac{1}{6x7}\)\(-\)\(\frac{1}{5x6}\)\(-\)\(\frac{1}{4x5}\)\(-\)\(\frac{1}{3x4}\)\(-\)\(\frac{1}{2x3}\)\(-\)\(\frac{1}{1x2}\)
c) \(\frac{1}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right):.....:\left(-1\frac{1}{100}\right)\)
Làm đc phần nào thì gúp mk với . ToT
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
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,\(\frac{1}{2}\div\left(-1\frac{1}{2}\right)\div1\frac{1}{3}:\left(-1\frac{1}{4}\right)\div...\div1\frac{1}{99}\div\left(-1\frac{1}{100}\right)\)
\(=\frac{1}{2}\div\left(-\frac{3}{2}\right)\div\frac{4}{3}\div\left(-\frac{5}{4}\right)\div...\div\frac{100}{99}\div\left(-\frac{101}{100}\right)\)
\(=\frac{1}{2}\times\left(-\frac{2}{3}\right)\times\frac{3}{4}\times...\times\left(-\frac{100}{101}\right)\)
=\(=\frac{1.\left(-2\right).3.....\left(-100\right)}{2.3.4.....101}=\frac{1}{101}\)
1, Tìm X
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right).100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \frac{9}{10}\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ 90-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=1\\ \frac{5}{2}:\left(X+\frac{206}{100}\right)=\frac{1}{2}\\ X+\frac{206}{100}=5\\ X=\frac{500}{100}-\frac{206}{100}\\ X=\frac{294}{100}=\frac{147}{50}\)
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
Tìm x, biết:
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+........+\frac{1}{Xx\left(x+1\right)}=\frac{499}{500}\)
Ai đúng cho 3tick
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{500}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{499}{500}\)
\(1-\frac{1}{x+1}=\frac{499}{500}\)
\(\frac{1}{x+1}=1-\frac{499}{500}=\frac{1}{500}\)
=> 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
Tìm \(X\), biết :
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{Xx\left(x+1\right)}=\frac{499}{500}\)
Ai giúp mk cho 5 tick
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
Chứng minh: \(\frac{3}{\left(1x2\right)}+\frac{5}{\left(2x3\right)}+...+\frac{2n+1}{\left(n\left(n+1\right)\right)^2}=\frac{n\left(n+2\right)}{\left(n+1\right)^2}\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
Bài1:Tính giá trị biểu thức sau:
A=\(\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)
Bài 2: Tính giá trị biểu thức:
B= \(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2003}\right)x\left(1-\frac{1}{2004}\right)\)
ai xong sẽ có tích , phải làm giải từng bước ra nhé!
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 1:tính giá trị biểu thức sau:
\(B=\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).\left(1-\frac{1}{36}\right)...\left(1-\frac{1}{1326}\right)\)
Bài 2 tìm x biết
\(a,1\frac{1}{30}:\left(24\frac{1}{6}-24\frac{1}{5}\right)-\frac{1\frac{1}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(-1\frac{1}{15}\right):\left(8\frac{1}{5}-8\frac{1}{3}\right)\)
\(b,\left|3x-\frac{2}{5}\right|=\frac{1}{35}+\left|\frac{-3}{7}\right|\)
\(c,-3\frac{1}{2}-\left(2x+\frac{1}{2}\right):\left(-\frac{3}{7}\right)=0\)
\(d,\left(x-4\right).\left(x+2\right)\le0\)với x thuộc z
giúp mk với