tìm x: x + 4/5*9 + 4/9*13 +...+ 4/41*45=29/45
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\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}\)
\(\Rightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Rightarrow x=15\)
Vậy x = 15
_Chúc bạn học tốt_
=>x-(1/5-1/9+1/9-1/13+...+1/41-1/45)=37/45
=>x-8/45=37/45
=>x=1
Gọi A=4/5.9+4/9.13+4/13.17+...+4/41.45
4.A=4.4/5.9+4.4/9.13+...+4.4/41.45
A=4/5-4/9+4/9-4/13+...+4/41-4/45
A=4/5-4/45
A=32/45
Vậy 4/5.9+4/9.13+...+4/41.45=32/45
\(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{29}{45}\)
=>\(x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\)
=>\(x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{29}{45}\)
=>\(x+\dfrac{8}{45}=\dfrac{29}{45}\)
=>\(x=\dfrac{29-8}{45}=\dfrac{21}{45}=\dfrac{7}{15}\)