CMR nếu 2(x+y)=5(y+z)=3(z+x) thì $\frac{x-y}{4}$= $\frac{y-z}{5}$ 24/08/2021 Bởi Liliana CMR nếu 2(x+y)=5(y+z)=3(z+x) thì $\frac{x-y}{4}$= $\frac{y-z}{5}$
Đáp án: Ta có : $2(x+y)=5(y+z)=3(z+x)$ => $\frac{2(x + y)}{30} = \frac{5(y+z)}{30} = \frac{3(z+x)$}{30} $ => $\frac{x+y}{15}$ = $\frac{y+z}{6}$ = $\frac{z+x}{10}$ = $\frac{z+x-(y+z)}{10-6}$ = $\frac{x+y-(z+x)}{15-10}$ = $\frac{x-y}{4}$ = $\frac{y-z}{5}$ => $\frac{x-y}{4}$ = $\frac{y-z}{5}$ Giải thích các bước giải: Bình luận
2(x+y)=5(y+z)=3(z+x) => $\frac{2(x+y)}{30}$ = $\frac{5(y+z)}{30}$= $\frac{3(z+x)}{30}$ => $\frac{x+y}{15}$ = $\frac{y+z}{6}$= $\frac{x+z}{10}$ Áp dụng TC dãy tỉ số bằng nhau $\frac{x+y}{15}$ = $\frac{x+z}{10}$= $\frac{x+y- x- z}{15- 10}$= $\frac{y-z}{5}$ $\frac{x+z}{10}$ = $\frac{y+z}{6}$= $\frac{x+z- y-z}{10-6}$= $\frac{x-y}{4}$ => $\frac{y-z}{5}$= $\frac{x-y}{4}$ Bình luận
Đáp án:
Ta có :
$2(x+y)=5(y+z)=3(z+x)$
=> $\frac{2(x + y)}{30} = \frac{5(y+z)}{30} = \frac{3(z+x)$}{30} $
=> $\frac{x+y}{15}$ = $\frac{y+z}{6}$ = $\frac{z+x}{10}$ = $\frac{z+x-(y+z)}{10-6}$ = $\frac{x+y-(z+x)}{15-10}$ = $\frac{x-y}{4}$ = $\frac{y-z}{5}$
=> $\frac{x-y}{4}$ = $\frac{y-z}{5}$
Giải thích các bước giải:
2(x+y)=5(y+z)=3(z+x)
=> $\frac{2(x+y)}{30}$ = $\frac{5(y+z)}{30}$= $\frac{3(z+x)}{30}$
=> $\frac{x+y}{15}$ = $\frac{y+z}{6}$= $\frac{x+z}{10}$
Áp dụng TC dãy tỉ số bằng nhau
$\frac{x+y}{15}$ = $\frac{x+z}{10}$= $\frac{x+y- x- z}{15- 10}$= $\frac{y-z}{5}$
$\frac{x+z}{10}$ = $\frac{y+z}{6}$= $\frac{x+z- y-z}{10-6}$= $\frac{x-y}{4}$
=> $\frac{y-z}{5}$= $\frac{x-y}{4}$