A = \(10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)

A : B = 10 

\(\frac{x_1-x_2}{k_1}=\frac{x_2-x_3}{k_2}=\frac{x_1-x_3}{k_3}=\frac{x_1-x_2+x_2-x_3-x_1-x_3}{k_1+k_2-k_3}=\frac{0}{k_1+k_2-k_3}=0\)

=> x₁ - x₂ = x₂ - x₃ = x₁ - x₃= 0

=> x₁ = x₂ = x₃ (đpcm)

\(\sqrt{6}+\sqrt{20}\) và \(7\)

\(\sqrt{6}^2+\sqrt{20}^2\) và \(7^2\)

\(\sqrt{6}^2+\sqrt{20}^2=6+20=26\)

\(7^2=49\) 

Mà 26<49 nên \(\sqrt{6}^2+\sqrt{20}^2

\(\frac{y^2-x^2}{3}=\frac{x^2+y^2}{5}=\frac{y^2-x^2-x^2-y^2}{3-5}=\frac{0}{3-5}=0\)

=> y² - x² = 0

=> x² + y² = 0

=> y² - x² = x² + y²

=> x² = y² = 0

=> x = y = 0

Bài này mik ko chắc

\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=....=\frac{a_{2008}}{a_{2009}}=\frac{a_1+a_2+....+a_{2008}}{a_2+a_3+....+a_{2009}}\)

=> \(\left(\frac{a_1}{a_2}\right)^{2008}=\left(\frac{a_2}{a_3}\right)^{2008}=....=\left(\frac{a_{2008}}{a_{2009}}\right)^{2008}=\left(\frac{a_1+a_2+....+a_{2008}}{a_2+a_3+....+a_{2009}}\right)^{2008}\)

\(=\frac{a_1.a_2....a_{2008}}{a_2.a_3....a_{2009}}=\frac{a_1}{a_{2009}}\)

=> \(\left(\frac{a_1+a_2+....+a_{2008}}{a_2+a_3+....+a_{2009}}\right)^{2008}=\frac{a_1}{a_{2009}}\)

=> Đpcm

\(\frac{x}{-8}=\frac{y}{7}=\frac{-3x}{24}=\frac{4y}{28}=\frac{-3x-4y}{24-28}=\frac{16}{-4}=-4\)

=> x = -4.(-8) = 32

     y = -4.7 = -28

Trời như quanh năm suốt tháng mới thấy Huân già trả lời câu hỏi,tui ủng hộ 1 tick nha!

b² = ac

=> \(\frac{a}{b}=\frac{b}{c}\)

=> \(\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{ab}{bc}=\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)

=> \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)

=> đpcm

\(\frac{a^2+b^2}{b^2+c^2}=\frac{a.a+a.c}{a.c+c.c}=\frac{a.\left(a+c\right)}{c.\left(a+c\right)}=\frac{a}{c}\)