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) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\Rightarrow x=15\)
\(\frac{x-1}{21}=\frac{3}{\left(x+1\right)}\)
=> (x + 1)(x - 1) = 3.21
=> x2 + x - x - 1 = 63
=> x2 - 1 = 63
=> x2 = 64
=> x = \(\pm\)8
b) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\left(đk:x\ne0\right)\)
=> \(\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)
=> \(\frac{7}{x}+\frac{1}{5}-\frac{1}{45}=\frac{29}{45}\)
=> \(\frac{7}{x}=\frac{29}{45}+\frac{1}{45}-\frac{1}{5}\)
=> \(\frac{7}{x}=\frac{7}{15}\)
=> x = 15
Vậy x = 15
a)
ĐK \(x+1\ne0\)
\(x\ne-1\)
\(\left(x-1\right)\left(x+1\right)=3\cdot21\)
\(x^2-1^2=63\)
\(x^2=64\)
\(x=\pm\sqrt{64}=\pm8\)
b)
\(\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)
\(\frac{7}{x}+\frac{1}{5}-\frac{1}{45}=\frac{29}{45}\)
\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
