GIẢI PT:
\(\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)x+\frac{1}{4.5^{99}.x}=\frac{1}{50}+\frac{1}{150}+\frac{1}{300}+...+\frac{1}{9500}\)
giải PT: a, (4x-5)2 (2x-3)(x-1)=9
b,\(\frac{5}{x-8}+1=\frac{23}{x^2-5x-24}+\frac{2}{x+3}\)
c,(\(\left(\frac{x-1}{99}+\frac{x-99}{1}\right)+\left(\frac{x-3}{97}+\frac{x+97}{3}\right)+\left(\frac{x-5}{93}+\frac{x-95}{5}\right)=6\)
c, Trừ hai vế cho 6
Vế trái thì lấy từng số hạng trừ 1 là được
Giaỉ PT
a) \(\left(\frac{x-1}{99}+x-99\right)+\left(\frac{x-3}{97}+\frac{x-7}{93}\right)+\left(\frac{x-5}{95}+\frac{x-95}{5}\right)=6\)
b) \(\left(4x-5\right)^2\left(2x-3\right)\left(x-1\right)=9\)
c) \(\frac{5}{x-8}+1=\frac{23}{x^2-5x-24}-\frac{2}{x+3}\)
Help !!
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 \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
tìm x :
\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{x.\left(x+1\right)}=\frac{99}{20}\)
Ta có : \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)
\(\Rightarrow5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{20}.\frac{1}{5}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x + 1 = 100
=> x = 99
Ta có \(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)
=> \(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)
=> \(5\left(1-\frac{1}{x+1}\right)=\frac{99}{20}\)
=> \(1-\frac{1}{x+1}=\frac{99}{100}\)
=> \(\frac{1}{x+1}=\frac{1}{100}\)
=> x+1 = 100
=> x = 99
Bài 1:CMR:\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{15}<2\)
Bài 2: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right)}=\frac{99}{101}\)
Bài 3:\(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2449}{2500}\)
Bài 4:CMR:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Tính:
a, A=\(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
b, B=\(\left(\frac{1}{2}-\frac{1}{3}\right).\left(\frac{1}{2}-\frac{1}{5}\right).\left(\frac{1}{2}-\frac{1}{7}\right)....\left(\frac{1}{2}-\frac{1}{99}\right)\)
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+\frac{1}{5.995}+......+\frac{1}{999.1}}\)
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+......+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+4.5+......+98.99}\)
\(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+......+\frac{1}{100.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+......+\frac{1}{299.400}}\)
\(D=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{97}-......-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+......+\frac{1}{500}}\)
Giup tui voi !!!!!!!!!!!!!!!!!!!!!!!!!!! Mai phai nop roi !!!!!!!!!!!!!!!!!!!
a) \(A=\left(\frac{1}{2}-\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{5}\right)\left(\frac{1}{2}-\frac{1}{7}\right)...\left(\frac{1}{2}-\frac{1}{99}\right)\)
b) \(B=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
Giúp mình nhanh nhé, 3 tick đấy!
Tính:
a.A = \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b. B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c. C = \(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)