Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{3a}{3b}=\dfrac{2c}{2d}=\dfrac{3a-2c}{3b-2d}\)
a/ \(\dfrac{a.c}{b.d}=\dfrac{\left(a+c\right).\left(a-c\right)}{\left(b+d\right).\left(b-d\right)}=\dfrac{a^2-c^2}{b^2-d^2}\)
b/ \(\dfrac{a^2}{b^2}=\dfrac{a}{b}.\dfrac{3a-2c}{3b-2d}=\dfrac{3a^2-2ac}{3b^2-2bd}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh :
a) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a+2c}{3b+2d}\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{-2a+7c}{-2b+7d}\)
cho a, b, c thỏa mãn : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
tính C = \(\dfrac{a^2}{a^2+2bc}+\dfrac{b^2}{b^2+2ac}+\dfrac{c^2}{c^2+2ab}\)
Cho \(a,b,c,d>0\). Chứng minh \(1< \dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< 2\)
Chứng minh bất đẳng thức :
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\) với a,b,c > 0
Cho a,b,c,d >0. C/m 1 <\(\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}\) > 2
Cho a, b, c > 0 thỏa mãn a + b +c =3
Chứng minh : \(\dfrac{a}{b^2+1}\)+\(\dfrac{b}{c^2+1}\)+\(\dfrac{c}{a^2+1}\) \(\ge\) \(\dfrac{3}{2}\)
Cho a,b,c là 3 số dương và \(\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}\)
Chứng minh rằng:\(\dfrac{a+b}{2a-b}+\dfrac{c+b}{2c-b}\ge4\)
Tìm x: a, \(\dfrac{\left|x\right|}{6-a-a^2}=\dfrac{1}{a+3}+\dfrac{1}{2-a}\)
b, \(\dfrac{\left|x\right|-3}{a^2-3a+2}=\dfrac{1}{a-1}-\dfrac{1}{a-2}\)
Chứng minh rằng nếu :
\(\dfrac{bz+cy}{x\left(-ax+by+cz\right)}\) = \(\dfrac{cx+az}{y\left(ax-by+cz\right)}\) = \(\dfrac{ay+bx}{z\left(ax+by-cz\right)}\)
thì : \(\dfrac{x}{a\left(b^2+c^2-a^2\right)}\) = \(\dfrac{y}{b\left(a^2+c^2-b^2\right)}\) = \(\dfrac{z}{c\left(a^2+b^2-c^2\right)}\)
Help me 
