Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất DTSBN :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)
\(\Rightarrow\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\) (ĐPCM)
Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất DTSBN :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)
\(\Rightarrow\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\) (ĐPCM)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). CMR : \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) và \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
\(Cho\) \(\dfrac{a}{b}=\dfrac{c}{d}\). \(CMR:\) \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
CMR : \(\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}\right)^2\) = \(\dfrac{a}{d}\)
cho\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)chứng minh
a).\(\dfrac{a^2-b^2}{c^2-d^2}\)=\(\dfrac{ab}{cd}\)
b)\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)=\(\dfrac{ab}{cd}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) chứng minh rằng \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\).
Chứng minh :
a) \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{ab}{cd}\)
b) \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\)
cho \(\dfrac{a}{b}=\dfrac{c}{d}\) khác \(\pm\)1 và c khác 0 , CMR
a) \(\left(\dfrac{a-b}{c-d}\right)^2=\dfrac{ab}{cd}\)
b)\(\dfrac{5a+3b}{5c+3d}\) =\(\dfrac{5a+3b}{5c+3d}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). Hãy chứng minh rằng :
\(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
\(\dfrac{a+2c}{b+2d}=\dfrac{a-2c}{b-2d}\)
\(\dfrac{a^2+2b^2}{c^2+2d^2}=\dfrac{a^2-2b^2}{c^2-2d^2}\)
\(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) CMR :\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)