Question

Nếu: 

1/c = 1/2 (1/a + 1/b)

CMR:  (a - c)/(c - b) = a/b

help me (T - T)

Answer 1

\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{2}\left(\frac{a+b}{ab}\right)=\frac{a+b}{2ab}=\frac{1}{\frac{2ab}{a+b}}\)

Từ đây ta có: \(\frac{1}{c}=\frac{1}{\frac{2ab}{a+b}}\Rightarrow c=\frac{2ab}{a+b}\) (hai phân số cùng tử bằng nhau khi cái mẫu của chúng bằng nhau)

Thay vào,ta có: \(\frac{a-c}{c-b}=\frac{a-\frac{2ab}{a+b}}{\frac{2ab}{a+b}-b}=\frac{\frac{a\left(a+b\right)-2ab}{a+b}}{\frac{2ab-b\left(a+b\right)}{a+b}}\)

\(=\frac{\frac{a^2-ab}{a+b}}{\frac{ab-b^2}{a+b}}=\left(\frac{a^2-ab}{a+b}\right):\left(\frac{ab-b^2}{a+b}\right)\)

\(=\frac{a^2-ab}{a+b}.\frac{a+b}{ab-b^2}=\frac{a^2-ab}{ab-b^2}=\frac{a\left(a-b\right)}{b\left(a-b\right)}=\frac{a}{b}^{\left(đpcm\right)}\)

Answer 2

\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

=> \(\frac{1}{c}:\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)

=>\(\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)

=>\(\frac{1}{c}+\frac{1}{c}=\frac{1}{a}+\frac{1}{b}\)

=> \(\frac{1}{c}-\frac{1}{a}=\frac{1}{b}-\frac{1}{c}\)

=>\(\frac{a}{ac}-\frac{c}{ac}=\frac{c}{bc}-\frac{b}{bc}\)(quy đồng mẫu)

=> \(\frac{a-c}{ac}=\frac{c-b}{bc}\)

=> \(\frac{a-c}{c-b}=\frac{ac}{bc}\)(tính chất dãy tỉ số bằng nhau)

hay: \(\frac{a-c}{c-b}=\frac{a}{b}\)(đpcm)

# Kiseki no enzeru #

hok tốt