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Đặt \(\frac{x_1-1}{5}=\frac{x_2-2}{4}=\frac{x_3-3}{3}=\frac{x_4-4}{2}=\frac{x_5-5}{1}=k\)
Áp dụng TC DTSBN ta có :
\(k=\frac{\left(x_1-1\right)+\left(x_2-2\right)+\left(x_3-3\right)+\left(x_4-4\right)+\left(x_5-5\right)}{5+4+3+2+1}\)
\(=\frac{x_1+x_2+x_3+x_4+x_5-15}{15}=\frac{30-15}{15}=1\)
\(\frac{x_1-1}{5}=1\Rightarrow x_1=6;\frac{x_2-2}{4}=1\Rightarrow x_2=6;\frac{x_3-3}{3}=1\Rightarrow x_3=6;\frac{x_4-4}{2}=1\Rightarrow x_4=6;\frac{x^5-5}{2}=1\Rightarrow x_5=6\)
Vậy \(x_1=x_2=x_3=x_4=x_5=6\)
Theo TCDTSBN ta có:
\(\frac{x1}{x2}=\frac{x2}{x3}=....=\frac{x2008}{x2009}=\frac{x1+x2+...+x2008}{x2+x3+...+x2009}\)
Ta có: \(\frac{x1}{x2}=\frac{x1+x2+...+x2008}{x2+x3+....+x2009}\left(1\right)\)
\(\frac{x2}{x3}=\frac{x1+x2+...+x2008}{x2+x3+...+x2009}\left(2\right)\)
............
\(\frac{x2008}{x2009}=\frac{x1+x2+...+x2008}{x2+x3+...+x2009}\left(2008\right)\)
Nhân (1),(2),....(2008) vế với vế:
\(\frac{x1}{x2}\cdot\frac{x2}{x3}\cdot\cdot\cdot\cdot\frac{x2008}{x2009}=\frac{x1}{x2009}=\left(\frac{x1+x2+...+x2008}{x2+x3+...+x2009}\right)^{2008}\)
Vậy...
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x_1}{x_2}=\frac{x_2}{x_3}=\frac{x_3}{x_4}=...=\frac{x_{2008}}{x_{2009}}=\frac{x_1+x_2+x_3+...+x_{2008}}{x_2+x_3+x_4+...+x_{2009}}\)
=> \(\frac{x_1}{x_2}=\frac{x_1+x_2+x_3+...+x_{2008}}{x_2+x_3+x_4+...+x_{2009}}\)
\(\frac{x_2}{x_3}=\frac{x_1+x_2+x_3+...+x_{2008}}{x_2+x_3+x_4+...+x_{2009}}\)
\(\frac{x_3}{x_4}=\frac{x_1+x_2+x_3+...+x_{2008}}{x_2+x_3+x_4+...+x_{2009}}\)
..........
\(\frac{x_{2008}}{x_{2009}}=\frac{x_1+x_2+x_3+...+x_{2008}}{x_2+x_3+x_4+...+x_{2009}}\)
Như vậy nhân các vế lại ta có \(\frac{x_1}{x_2}.\frac{x_2}{x_3}.\frac{x_3}{x_4}.....\frac{x_{2008}}{x_{2009}}=\frac{x_1.x_2.x_3...x_{2008}}{x_2.x_3.x_4....x_{2009}}=\frac{x_1}{x_{2009}}\) (đpcm)
Câu 1:
a, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^n}{c^n}=\frac{b^n}{d^n}=\frac{a^n+b^n}{c^n+d^n}=\frac{a^n-b^n}{c^n-d^n}\)
b,Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{b}{d}\cdot\frac{a}{c}\Rightarrow\frac{a^2}{b^2}=\frac{ab}{cd}\)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ac}{cd}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{ac}{bd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)
Ta lại có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a+b}{c+d}\cdot\frac{a+b}{c+d}\Rightarrow\frac{ab}{cd}=\left(\frac{a+b}{c+d}\right)^2\left(2\right)\)
Từ (1) và (2) => \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
Câu 2:
\(\frac{a1}{a2}=\frac{a2}{a3}=....=\frac{a2017}{a2018}=\frac{a1+a2+...+a2017}{a2+a3+....+a2018}\)
\(\Rightarrow\frac{a1}{a2}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(1\right)\)
\(\frac{a2}{a3}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(2\right)\)
..............
\(\frac{a2017}{a2018}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(2017\right)\)
Nhân các vế (1),(2)....(2017) ta được:
\(\frac{a1}{a2}\cdot\frac{a2}{a3}\cdot\cdot\cdot\cdot\cdot\frac{a2017}{a2018}=\frac{a1}{a2018}=\left(\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\right)^{2017}\)
Vậy...
Câu 3:
\(x_2^2=x_1x_3\Rightarrow\frac{x1}{x2}=\frac{x2}{x3}\)
\(x_3^2=x_2x_4\Rightarrow\frac{x2}{x3}=\frac{x3}{x4}\)
\(x_4^2=x_3x_5\Rightarrow\frac{x3}{x4}=\frac{x4}{x5}\)
\(x_5^2=x_4x_6\Rightarrow\frac{x4}{x5}=\frac{x5}{x6}\)
Đến đây thfi làm giống câu 2
Bỏ x4 đi nhé bn
Theo t/c dãy tỉ số=nhau:
\(\frac{x_1-1}{3}=\frac{x_2-2}{2}=\frac{x_3-3}{1}=\frac{x_1-1+x_2-2+x_3-3}{3+2+1}\)\(=\frac{\left(x_1+x_2+x_3\right)-\left(1+2+3\right)}{6}=\frac{30-6}{6}=\frac{24}{6}=4\)
=>x1-1=4.3=12=>x1=13
x2-2=4.2=8=>x2=10
x3-3=4=>x3=7