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Ta có :\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
=> \(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
=> \(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
=> \(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
=> \(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
=> \(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 = 18
=> x = 17
Vậy x = 17
Bài 1:
a) \(x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=1\)
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 =18
x = 17
bài 2 ko bk lm, xl nha
đối với câu a :
nhân 2 ở tử số còn mẫu số giữ nguyên:
a)1/21 + 1/28 + 1/36 + ... + 2/x(x+1) = 2/9
=> 2/42 + 2/56 + 2/72 + ... + 2/x(x+1) = 2/9
=> 2 (1/42 + 1/56 + 1/72 + ... + 1/x(x+1) =2/9
=> 2 (1/6*7 + 1/7*8 + 1/8*9 + ... + 1/x(x+1) = 2/9
=> 2 ( 1/6 - 1/7 + 1/7 - 1/8 + ... +1/x - 1/x+1 = 2/9
=> 2 ( 1/6 - 1/x+1 ) = 2/9
=> 1/6 - 1/x+1 = 1/9
=> 1/x+1 = 1/6 - 1/9
=> 1/x+1 = 1/18
=> x+1 =18
=> x= 17
chúc các bn học tốt!
\(A=\frac{99}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)
\(A=\frac{99}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{99}{100}-\frac{99}{100}\)
\(A=\frac{99-99}{100}=0\)
Bài 2
\(\left(3x+5\right).\left(2x-4\right)=0\)
\(TH1:3x+5=0\)
\(3x=-5\)
\(x=-\frac{5}{3}\)
\(TH2:2x-4=0\)
\(2x=4\)
\(x=2\)
\(\left(x^2-1\right).\left(x+3\right)=0\)
\(\Rightarrow x^2-1=0\)
\(x^2=1\)
\(\Rightarrow x=1\)
\(x+3=0\)
\(x=-3\)
\(5x^2-\frac{1}{2}x=0\)
\(\Rightarrow5x^2-\frac{x}{2}=0\)
\(\Rightarrow5x^2=\frac{5x^2}{1}=\frac{5x^2.2}{2}\)
\(10x^2-x=x.\left(10x-1\right)\)
\(\frac{x.\left(10x-1\right)}{2}=0\)
\(\frac{x.\left(10x-1\right)}{2}.2=0.2\)
\(10x-1=0\)
\(x=\frac{1}{10}=0.100\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}=0.100\\x=0\end{cases}}\)
\(\frac{x}{4}-\frac{1}{2}=\frac{3}{4}\)
\(\frac{x}{4}=\frac{3}{4}+\frac{1}{2}\)
\(\frac{x}{4}=\frac{5}{4}\)
\(\Rightarrow x=5\)
\(\frac{1}{8}+\frac{7}{8}:x=\frac{3}{4}\)
\(\frac{7}{8}:x=\frac{3}{4}-\frac{1}{8}\)
\(x=\frac{7}{8}:\frac{5}{8}\)
\(x=\frac{56}{40}=\frac{28}{20}=\frac{14}{10}=\frac{7}{5}\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
a)=\(\frac{1}{9}.\frac{2}{145}-\frac{13}{3}.\frac{2}{145}+\frac{2}{145}\)
=\(\frac{2}{145}\left(\frac{1}{9}-\frac{13}{3}+1\right)\)
=\(\frac{2}{145}.\frac{-29}{9}\)
=\(\frac{-2}{45}\)
Học tốt nha!!!^^