Cho a ; b ; c ; x ; y ; z \(\ne\) 0 tm : \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
CM: \(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}=\dfrac{\left(a+b+c\right)^2}{x+y+z}\)
cho \(\dfrac{1}{c}\)=\(\dfrac{1}{2}\)\((\dfrac{1}{a}\)+\(\dfrac{1}{b}\))(với a,b,c \(\ne\)0;b\(\ne\)c)
chứng minh rằng \(\dfrac{a}{b}\)=\(\dfrac{a-c}{c-b}\)
Cho : \(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\) Tính M biết :
M =\(\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
Cho :\(\dfrac{a}{b}=\dfrac{c}{d}CMR:\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}v\text{à}\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Cho a ; b;c tm : abc =1 . Tính
M = \(\dfrac{1}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{abc+bc+b}\)
Cho a ; b ; c > 0 và : \(\dfrac{2b+c-a}{a}=\dfrac{2c+a-b}{b}=\dfrac{2a+b-c}{c}\) . Tính :
\(\dfrac{\left(3a-2b\right).\left(3b-2c\right).\left(3c-2a\right)}{abc}\)
Cho a, b, c > 0 và \(\dfrac{a+b}{3}\) = \(\dfrac{b+c}{4}\) = \(\dfrac{c+a}{5}\). Tính M = 10a + b - 7c + 2017
Biết :
\(\dfrac{bz-cy}{a}=\dfrac{cx-ay}{b}=\dfrac{ay-bx}{c}\) CM : \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Cm ta có tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) nếu có đẳng thức:
( a + b + c + d ).(a-b-c+d ) = ( a-b+c-d).(a+b-c-d)
