Có \(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a.b}{c.d}\) (1)
Có \(\dfrac{a}{c}=\dfrac{b}{d}\) => \(\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2\)
=> \(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)
ADTCDTSBN ta có
\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\) (2)
Từ (1) và (2) =>\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a.b}{c.d}\) (đpcm)
cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)
chứng minh
a. \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
b. \(\dfrac{a\cdot b}{c\cdot d}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
c.\(\dfrac{2008\cdot a-2009\cdot b}{2009\cdot c+2010\cdot d}=\dfrac{2008\cdot c-2009\cdot d}{2009\cdot a+2010\cdot b}\)
Cho bốn số dương a, b, c, d . Biết rằng : b=\(\dfrac{a+c}{2}\)và c=\(\dfrac{2bd}{b+d}\)
CM : 4 số này lập thành tỉ lệ thức
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\)và b, d khác 0. CMR \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
\(Cho\dfrac{a}{b}=\dfrac{c}{d}CMR\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2-b^2}{c^2-d^2}\)
biet \(\dfrac{b\cdot z-x\cdot y}{a}=\dfrac{c\cdot x-a\cdot z}{b}=\dfrac{a\cdot y-b\cdot x}{c}\)
CM , \(\dfrac{a}{b}=\dfrac{b}{y}=\dfrac{c}{z}\)
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 \(\dfrac{a}{b}\) =\(\dfrac{c}{d}\) 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 \(\dfrac{a}{b}=\dfrac{d}{c}\)
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, \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
