a^2=bc
nên a/b=c/a
Đặt a/b=c/a=k
=>a=bk; c=ak
=>c=bk*k=b*k^2
\(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{k+1}{k-1}\)
\(\dfrac{a+c}{a-c}=\dfrac{bk+bk^2}{bk-bk^2}=\dfrac{k+1}{1-k}\)
Do đó: \(\dfrac{a+b}{a-b}=-\dfrac{a+c}{a-c}\)
Cho : \(\dfrac{a}{c}=\dfrac{c}{b}.CMR:\\ a,\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\\ b,\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
Cho \(\dfrac{\overline{ab}+\overline{bc}}{a+c}=\dfrac{\overline{bc}+\overline{ca}}{b+c}=\dfrac{\overline{ca}+\overline{ab}}{c+a}\)
CMR : a = b = c
Cho a2= bc. CMR:
\(\dfrac{a+b}{a-b}\) = \(\dfrac{c+a}{c-d}\)
CMR nếu \(\dfrac{x^2-yz}{a}=\dfrac{y^2-x\text{z}}{b}=\dfrac{z^2-\text{yx}}{c}th\text{ì \dfrac{a^2-bc}{x}=\dfrac{b^2-ca}{y}=\dfrac{c^2-ab}{z}}\)
Cho tam giác ABC có BC=a; AC=b; AB=c và trung tuyến CD=m.
CMR \(\dfrac{a+b-c}{2}< m< \dfrac{a+b}{2}\)
Cho\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\). CMR trong 3 số a,b,c tồn tại 2 số bằng nhau
Cho : \(\dfrac{a}{c}=\dfrac{c}{b}\) CMR :
a.\(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)
b.\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
Bài 1: CMR : Nếu a2 = b.c thì \(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\) Đảo lại có đúng không?
Bài 2: Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) CMR: \(\left(\dfrac{a+b+c}{b+c+d}\right)\)3 =\(\dfrac{a}{d}\)
Bài 3: Cho dãy tỉ số: \(\dfrac{b.z-c.y}{a}=\dfrac{c.x-a.z}{b}=\dfrac{a.y-b.x}{c}\) , CMR: \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
Cho \(\dfrac{a}{c}=\dfrac{c}{b}\) . CMR : \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)
