Câu hỏi của erza

Question

Câu 1: Cho tỉ lệ thức: $\frac{a}{b}=\frac{c}{d}$CMR:a)$\frac{a^n+b^n}{c^n+d^n}=\frac{a^n-b^n}{c^n-d^n}$       b) $\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}$Câu 2: CMR: nếu \(\frac{a...

ST · Accepted Answer

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}$$x_5^2=x_4x_6\Rightarrow\frac{x4}{x5}=\frac{x5}{x6}$Đến đây thfi làm giống câu 2Câu trả lời: Câu 1:a, $\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^n}{c^n}=\frac{b^n}{d^n}=\frac{a^n+b^n}{c^n+d^n}=\frac{a^n-b^n}{c^n-d^n}$b,Ta có: $\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{b}{d}\cdot\frac{a}{c}\Rightarrow\frac{a^2}{b^2}=\frac{ab}{cd}$$\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

Trần Tân · Answer

cho x1, x2 , x3 là 3 số thực khác 0 thỏa mãn x1 + x2 + x3 = a ; x1x2 + x2x3 + x1x3 = 0 ; x1x2x3 = bCMR: a/b < 0Câu trả lời: cho x1, x2 , x3 là 3 số thực khác 0 thỏa mãn x1 + x2 + x3 = a ; x1x2 + x2x3 + x1x3 = 0 ; x1x2x3 = bCMR: a/b < 0

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