1tìm x,y,z
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}.\)
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5},xyz=810\)
2tìm x:
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
3
\(CMRtừ:\frac{a+b}{a-b}=\frac{c+d}{c-d}\ne1\)
\(tacó:\frac{a}{b}=\frac{c}{d}\)
3:
\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
\(\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\dfrac{2b}{2d}=\dfrac{b}{d}\) (1)
\(\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b+a-b}{c+d+c-d}=\dfrac{2a}{2c}=\dfrac{a}{c}\) (2)
Từ (1) và (2) suy ra \(\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\) \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\) ĐPCM
nguyễn họ hoàng ok