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\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c},c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(1\right)\)
\(\frac{a^3}{b^3}=\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\left(2\right)\)
=> đpcm
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\left(1\right)\)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{abc}{bcd}=\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(đpcm\right)\)
b, Tỉ số = nhau + tất vào là xông
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
a)Với x1 = x2 = 1
\( \implies\) \(f\left(1\right)=f\left(1.1\right)\)
\( \implies\) \(f\left(1\right)=f\left(1\right).f\left(1\right)\)
\( \implies\)\(f\left(1\right).f\left(1\right)-f\left(1\right)=0\)
\( \implies\) \(f\left(1\right).\left[f\left(1\right)-1\right]=0\)
\( \implies\) \(\orbr{\begin{cases}f\left(1\right)=0\\f\left(1\right)-1=0\end{cases}}\)
Mà \(f\left(x\right)\) khác \(0\) ( với mọi \(x\) \(\in\) \(R\) ; \(x\) khác \(0\) )
\( \implies\) \(f\left(1\right)\) khác \(0\)
\( \implies\) \(f\left(1\right)-1=0\)
\( \implies\) \(f\left(1\right)=1\)
b)Ta có : \(f\left(\frac{1}{x}\right).f\left(x\right)=f\left(\frac{1}{x}.x\right)\)
\( \implies\) \(f\left(\frac{1}{x}\right).f\left(x\right)=f\left(1\right)=1\)
\( \implies\) \(f\left(\frac{1}{x}\right).f\left(x\right)=1\)
\( \implies\) \(f\left(\frac{1}{x}\right)=\frac{1}{f\left(x\right)}\)
\( \implies\) \(f\left(x^{-1}\right)=\left[f\left(x\right)\right]^{-1}\)