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a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)
\(\Leftrightarrow2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}\right)=2.\frac{8}{17}\)
\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x\left(x+2\right)}=\frac{16}{17}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{16}{17}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{16}{17}=\frac{1}{17}\)
\(\Rightarrow x+2=17\Rightarrow x=15\)
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
Nhưng bn ơiX là x^2 hay tách biệt nếu tách biệt thì là 9/49 còn nếu là x^2 thì là 3/7 nhé
\(=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)xX=\frac{9}{7} \)\(=\left(\frac{1}{3}-\frac{1}{21}\right)xX=\frac{9}{7}\)\(=\frac{2}{7}xX=\frac{9}{7}\)
\(X=\frac{9}{7}:\frac{2}{7}\)
\(X=\frac{9}{2}\)
\(\left(x\cdot2,4-4,2\right)\div x=1\)
\(\Rightarrow x\cdot2,4-4,2=x\)
\(\Rightarrow x\cdot2,4=x+4,2\)
\(\Rightarrow\frac{12x}{5}=\frac{5x+21}{5}\)
\(\Rightarrow12x=5x+21\)
\(\Rightarrow12x-5x=21\)
\(\Rightarrow7x=21\Rightarrow x=\frac{21}{7}=3\)
Vậy x = 3
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{x\left(x+2\right)}\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\cdot\frac{x+1}{x+2}\)
\(=\frac{x+1}{2x+2}\)
1. 0,8 x 2 x 48 + 1,6 x 2 + 50 x 1,6
= 1,6 x 48 + 1,6 x 2 + 50 x 1,6
= 1,6 x [ 48 + 50 + 2 ]
= 1,6 x 100
= 160
2. 1/1x3 + 1/3x5 + 1/5x7 + 1/7x9
= 1/1 + 1/3 - 1/3 + 1/5 - 1/5 + 1/7 - 1/7 + 1/9
= 1/1 - 1/9
= 8/9
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{9}\right)+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\frac{8}{9}+\frac{1}{x}=1\)
\(\Rightarrow\frac{4}{9}+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{x}=1-\frac{4}{9}\)
\(\Rightarrow\frac{1}{x}=\frac{5}{9}\)
\(\Rightarrow x=\frac{1.9}{5}\)
\(\Rightarrow x=\frac{9}{5}\)
Vậy x = \(\frac{9}{5}\)
b) \(\frac{2}{3}-\frac{1}{3}.\left(x-2\right)=\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}.\left(x-2\right)=\frac{2}{3}-\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}.\left(x-2\right)=\frac{5}{12}\)
\(\Rightarrow x-2=\frac{5}{12}:\frac{1}{3}\)
\(\Rightarrow x-2=\frac{5}{4}\)
\(\Rightarrow x=\frac{5}{4}+2\)
\(\Rightarrow x=\frac{13}{4}\)
Vậy x = \(\frac{13}{4}\)
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