Question

4. CMR : Nếu a² = bc thì \(\dfrac{a+b}{a-b}\)=\(\dfrac{c+a}{c-a}\)

5. Tìm x, y , z biết :

\(\dfrac{y+z+1}{x}\)=\(\dfrac{x+z+2}{y}\)=\(\dfrac{x+y-3}{z}\)=\(\dfrac{1}{x+y+z}\)

Answer 1

4) Ta có: a²=bc => aa=bc =>\(\dfrac{a}{b}=\dfrac{c}{a}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{a}=k\left(k\ne0\right)\)

=> a=bk ; c=ak

+)\(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\left(1\right)\)

+) \(\dfrac{c+a}{c-a}=\dfrac{ak+a}{ak-a}=\dfrac{a\left(k+1\right)}{a\left(k-1\right)}=\dfrac{k+1}{k-1}\left(2\right)\)

Từ (1) và (2) => \(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

5) phải xét 2 trường họp dài lắm nên mình chả muốn làm ~~