Ta có:\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ab}{cd}\)
\(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}=\dfrac{a}{c}=\dfrac{b}{d}\)
Vậy \(\dfrac{a}{c}=\dfrac{b}{d}\left(\text{đ}pcm\right)\)
Biết \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với a,b,c,d khác 0. Chứng minh rằng: \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng
a) \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)
CMR \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) và \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh :
a) \(\dfrac{3a+5b}{2a-7b}=\dfrac{3c+5d}{2c-7d}\)
b) \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{ab}{cd}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}.CMR\)
a, \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b, \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
c, \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
cho \(\dfrac{a}{b}\) =\(\dfrac{c}{d}\) cm rằng
a) \(\dfrac{a}{a-b}\) =\(\dfrac{c}{c-d}\) b)\(\dfrac{a}{b}\) =\(\dfrac{a+c}{b+d}\) c) \(\dfrac{a}{3a+d}\) =\(\dfrac{c}{3c+d}\) d)\(\dfrac{a.c}{b.d}\) =\(\dfrac{a^2+c^2}{b^2+c^2}\) e)\(\dfrac{a.b}{c.d}\) =\(\dfrac{a^2-b^2}{c^2-d^2}\) f)\(\dfrac{a.b}{c.d}\) =\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
mn giúp mk vs ạ! thanks
Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a.b}{c.d}\) với a;b;c;d khác 0 và c khác +- d
CMR: \(\dfrac{a}{b}=\dfrac{c}{d} \) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)
Chứng minh rằng : Nếu \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) thì \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
mọi người ơi giúp mik với ai làm đc mik tick cho
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). CMR: \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
Các bạn giúp mình với ạ!
