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a) \(\frac{x-1}{21}=\frac{3}{x+1}\)( ĐKXĐ : x khác -1 )
<=> ( x - 1 )( x + 1 ) = 21.3
<=> x2 - 1 = 63
<=> x2 = 64
<=> x2 = ( ±8 )2
<=> x = ±8 ( tmđk )
b) \(\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)( ĐKXĐ : x khác 0 )
<=> \(\frac{7}{x}+\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
<=> \(\frac{7}{x}=\frac{7}{15}\)
<=> x = 15 ( tmđk )
a) \(\frac{x-1}{21}=\frac{3}{x+1}\Leftrightarrow\left(x-1\right)\left(x+1\right)=3.21\)
\(\Leftrightarrow x^2-1=63\Rightarrow x^2=63+1=64\Rightarrow x=\pm8\)
b) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\Rightarrow x=15\)
\(a,\frac{x-1}{21}=\frac{3}{x+1}\)
\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x^2=8^2\)
\(\Leftrightarrow x=\pm8\)
\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Leftrightarrow x=15\)
Vậy x = 15
Bài cuối tương tự
\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{9}{45}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\Rightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)
\(\Rightarrow x=\frac{7}{\frac{21}{45}}=15\)
Vậy \(x=15\).
Đặt B = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{8.9.10}\)
=> 2B = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{8.9.10}\)
=> 2B = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\)
=> 2B = \(\frac{1}{1.2}-\frac{1}{9.10}\)
2B = \(\frac{22}{45}\)
B = \(\frac{22}{45}:2\)
=> B = \(\frac{11}{45}\)
Ta có : \(\frac{11}{45}.x=\frac{22}{45}\)
=> x = \(\frac{22}{45}:\frac{11}{45}\)
=> x = \(\frac{2}{1}\)
\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\frac{22}{45}.x=\frac{44}{45}\)
x =2
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)
\(\Leftrightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+..+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\Leftrightarrow\frac{22}{45}.x=\frac{44}{45}\Leftrightarrow x=2\)
Vậy x=2
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)
\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+......+\frac{2}{8.9.10}\right).x=\frac{22}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{22}{45}\)
\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{22}{45}\)
\(\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)
\(\frac{11}{45}.x=\frac{22}{45}\)
\(x=\frac{22}{45}:\frac{11}{45}\)
\(x=2\)
\(\frac{x+1}{5}=\frac{-10}{16}\Rightarrow x+1=\frac{5.\left(-10\right)}{16}=\frac{-25}{8}\Rightarrow x=\frac{-25}{8}-1=-\frac{33}{8}\)
\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{-37}{45}\)
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\frac{8}{45}=\frac{-37}{45}\)
\(x=\frac{-37}{45}-\frac{8}{45}\)
\(x=-1\)
e. \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Rightarrow x=15\)
f. \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{22}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{22}{45}\)
\(\Rightarrow x=2\)