Cho dãy tỉ số bằng nhau $\frac{2bz-3cy}{a}$ = $\frac{3cx-az}{2b}$ = $\frac{ay-2bx}{3c}$ . CMR:
$\frac{x}{a}$ = $\frac{y}{2b}$ = $\frac{z}{3c}$
Cho dãy tỉ số bằng nhau $\frac{2bz-3cy}{a}$ = $\frac{3cx-az}{2b}$ = $\frac{ay-2bx}{3c}$ . CMR: $\frac{x}{a}$ = $\frac{y}{2b}$ = $\frac{z}{3c}$
By Melanie
Ta có: `(2bz – 3cy)/a = (3cx – az)/(2b) =(ay-2bx)/(3c)`
`=> (2abz – 3acy)/a^2 = (6bcx – 2baz)/(2b)^2 = (3cay – 6cbx)/(3c^2)`
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
`(2abz – 3acy)/a^2 = (6bcx – 2baz)/(2b)^2 = (3cay – 6cbx)/(3c^2) = (2abz – 3acy + 6bcx – 2baz + 3cay – 6cbx) = 0`
+) `(2abz – 3acy)/a^2 =0`
`=> 2abz – 3acy = 0`
`=> 2abz = 3acy`
`=> 2bz = 3cy`
`=> y/(2b) = z/(3c) (1)`
+) ` (3cay – 6cbx)/(3c^2) =0`
`=> 3cay – 6cbx =0`
`=> 3cay = 6 cbx`
`=> 3ay = 6bx`
`=> ay= 2bx`
`=> y/(2b) = x/a (2)`
Từ `(1)` và `(2)`
`=> x/a = y/(2b) = z/(3c)`
Vậy `x/a = y/(2b) = z/(3c)`
Đáp án:
$\rm \dfrac{2bz-3cy}{a} = \dfrac{3cx-az}{2b} = \dfrac{ay-2bx}{3c} \\ \to \dfrac{2abz-3acy}{a^2} = \dfrac{6bcx-2abz}{4b^2} = \dfrac{3acy-6bcx}{9c^2} \\ Áp \ dụng \ t/c \ dãy \ tỉ \ số \ bằng \ nhau \ ta \ có \ : \\ \dfrac{2abz-3acy}{a^2} = \dfrac{6bcx-2abz}{4b^2} = \dfrac{3acy-6bcx}{9c^2} \\ = \dfrac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2} \\ = \dfrac{0}{a^2+4b^2+9c^2} = 0 \\ \to \begin{cases} \rm \dfrac{2bz-3cy}{a}=0\\ \rm \dfrac{3cx-az}{2b}=0\\ \rm \dfrac{ay-2bx}{3c}=0\end{cases} \to \begin{cases} \rm 2bz-3cy=0\\ \rm 3cx-az=0\\ \rm ay-2bx=0\end{cases} \\ \to \begin{cases} \rm 2bz=3cy\\ \rm 3cx=az\\ \rm ay=2bx \end{cases} \to \begin{cases} \rm \dfrac{y}{2b}=\dfrac{z}{3c} \\ \rm \dfrac{z}{3c} = \dfrac{x}{a} \\ \rm \dfrac{x}{a} = \dfrac{y}{2b} \end{cases} \\ \to \dfrac{x}{a}=\dfrac{y}{2b}=\dfrac{z}{3c} \ \ (đpcm)$