\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk; c=dk
\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: a/a+b=c/c+d
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk; c=dk
\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: a/a+b=c/c+d
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) Chứng minh rằng \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)chứng minh rằng : \(\dfrac{a^3}{b^3}=\dfrac{a}{d}\)
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) Chứng minh rằng \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
cho \(\dfrac{a^2+b^2}{c^2+d^2}\)= \(\dfrac{ab}{cd}\).Chứng minh rằng: hoặc \(\dfrac{a}{b}\)= \(\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}\)= \(\dfrac{d}{c}\)
Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?
Chứng minh rằng \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\) biết \(\dfrac{a}{b}=\dfrac{c}{d}\)
cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{e}{f}\) chứng minh rằng \(\dfrac{a^4}{b^4}=\dfrac{a}{f}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\left(b,d\ne0\right)\).Chứng minh rằng
\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)
\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
Cho tỉ lệ thức \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\). Chứng minh rằng \(\dfrac{a.d}{c.d}=\dfrac{a^2-b^2}{b^2-d^2}\)và \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)