\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\\ \Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\\ \Leftrightarrow12a=10b\\ \Leftrightarrow6a=5b\Leftrightarrow\dfrac{a}{b}=\dfrac{5}{6}\)

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(b+6\right)\left(a-5\right)\)

nhân ra ik ròi suy ra đpcm :D

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)

\(\Rightarrow ab+5b-6a-30=ab-5b+6a-30\)

\(\Rightarrow5b-6a=-5b+6a\)

\(\Rightarrow10b=12a\)

\(\Rightarrow5b=6a\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{5}{6}\left(đpcm\right)\)

Vậy \(\dfrac{a}{b}=\dfrac{5}{6}\)

\(\dfrac{a+5}{a-5}=\dfrac{a+6}{a-6}\)suy ra \(\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(a+6\right)\)

suy ra: \(6a=5b\)

suy ra: \(\dfrac{a}{b}=\dfrac{5}{6}\)

Một cách giải khác:

TH1: b = -6

VP = 0 => VT = 0 => a = -5

=> \(\dfrac{a}{b}=\dfrac{-5}{-6}=\dfrac{5}{6}\)

TH2: b \(\ne\) -6 nên:

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}=\dfrac{a+5+a-5}{b+6+b-6}=\dfrac{a}{b}=\dfrac{a+5-a+5}{b+6-b+6}=\dfrac{10}{12}=\dfrac{5}{6}\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy \(x=-2004\)

1/ Ta có :

\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+....+\dfrac{1}{49\times50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+.....+\dfrac{1}{50}\right)\)

\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)\times2\)

\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)\)

\(\Rightarrow\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}\)

Hay \(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...+\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)

~ Học tốt nha ~

1: Thay x=16 vào A, ta được:

\(A=\dfrac{6-2\cdot4}{4-5}=\dfrac{-2}{-1}=2\)