Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Suy ra: \(VT=\dfrac{bk^2\left(b+d\right)}{dk^2\left(d-b\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\)
\(VP=\dfrac{b^2+bd}{d^2-bd}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\)
\(\Rightarrow VT=VP\rightarrowđpcm.\)
cho \(\dfrac{a}{b}=\dfrac{c}{d}\) , cmr: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}.CMR\)
a, \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b, \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)
c, \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}\)
cho \(\dfrac{b}{a}=\dfrac{c}{d}\)cmr:
a,\(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
b,\(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
c,\(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
d,\(\dfrac{ac}{bd}=\dfrac{a^2+b^2}{b^2+d^2}\)
e,\(\dfrac{a.b}{c.d}=\dfrac{a^2-b^2}{c^2-d^2}\)
f,\(\dfrac{a.b}{c.d}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Cho \(a^2=ac=bd\) CMR
\(\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=\dfrac{a+c-c}{b+c-d}^3\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng \(\dfrac{ac}{bd}=\dfrac{a^2-c^2}{b^2-d^2}\) ( với giả thiết các tỉ số đều có nghĩa )
Cho \(\dfrac{a}{b}=\dfrac{c}{d} \) . Chứng minh :
a, \(\dfrac{a^2+c^2}{b^2+d^2} =\dfrac{ac}{bd}\)
b, \(\dfrac{a^2+c^2}{b^2+d^2} = \dfrac{a^2-c^2}{b^2-d^2}\)
c, \(\dfrac{(a+c)^2}{(b+d)^2} = \dfrac{(a-c)^2}{b-d)^2}\)
d, \(\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{c^2+d^2}{c^2-d^2}\)
e, \(\dfrac{(a-b )^2}{(c-d)^2} = \dfrac{a^2+b^2}{c^2+d^2}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) chứng minh rằng:
a) \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
b)\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
c)\(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Cho a.b.c.d khác 0 và \(b^2=ac\) ; \(c^2=bd\)
CMR : \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
Cho a,b,c,d thỏa mã \(b^2=ac;c^2=bd\)
CM \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
