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 \(x_1=x_2=x_3=x_4=x_5=6\)

a)

\(\Rightarrow A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)

\(\Rightarrow A=\frac{1}{5}+\frac{2}{7}\)

\(\Rightarrow A=\frac{17}{35}\)

b)

\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{56}-\frac{1}{61}\right)\)

\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{61}\right)\)

\(\Rightarrow B=5.\frac{50}{671}=\frac{250}{671}\)

c)

\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)

\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)

\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)

\(\Rightarrow C=1-1-\frac{1}{25}\)

\(\Rightarrow C=\frac{1}{25}\)

yutyugubhujyikiu

c.\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)

\(\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:-\frac{41}{21}}\)

\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}\)

\(\frac{100}{-\frac{100}{41}}=-41\)

a. \(\frac{4}{9}:-\frac{1}{7}+6\frac{5}{9}:-\frac{1}{7}\)

\(\left(\frac{4}{9}+6\frac{5}{9}\right):-\frac{1}{7}\)

\(7:-\frac{1}{7}=-49\)

b. \(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)

\(\left(\frac{7}{3}+\frac{7}{2}\right):\left(-\frac{25}{6}+\frac{22}{7}\right)+\frac{15}{2}\)

\(\frac{35}{6}:-\frac{43}{42}+\frac{15}{2}\)

\(-\frac{245}{43}+\frac{15}{2}=\frac{155}{86}\)

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

cho x₁, x₂ , x₃ là 3 số thực khác 0 thỏa mãn x₁ + x₂ + x₃ = a ; x₁x₂ + x₂x₃ + x₁x₃ = 0 ; x₁x₂x₃ = b

CMR: a/b < 0

lấy máy tính bấm nha bn 

thi ai cho 

giúp mình cách tính nhanh ik

a) \(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)

\(=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)

\(=\frac{1}{5}+\frac{2}{7}\)

\(=\frac{7}{35}+\frac{10}{35}\)

\(=\frac{17}{35}\)

Vậy \(A=\frac{17}{35}\)

b) \(B=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.61}\)

\(=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{56.61}\right)\)

\(=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}\right)\)

\(=5.\left(\frac{1}{11}-\frac{1}{61}\right)\)

\(=5.\left(\frac{61}{671}-\frac{11}{671}\right)\)

\(=5.\frac{50}{671}\)

\(=\frac{250}{671}\)

Vậy \(B=\frac{250}{671}\)