Tìm x; y biết: $\frac{x – a}{m}$ = $\frac{y-b}{n}$ và x + y = k 07/08/2021 Bởi Reese Tìm x; y biết: $\frac{x – a}{m}$ = $\frac{y-b}{n}$ và x + y = k
$\dfrac{x-a}{m}=\dfrac{y-b}{n}=\dfrac{x-a+y-b}{m+n}=\dfrac{k-a-b}{m+n}$ $→\dfrac{x-a}{m}=\dfrac{k-a-b}{m+n}$ $x-a=\dfrac{(k-a-b).m}{m+n}$ $x-a=\dfrac{ka-am-bm}{m+n}$ $x=\dfrac{ka-am-bm}{m+n}+a$ $x=\dfrac{ka-am-bm}{m+n}+\dfrac{a(m+n)}{m+n}$ $x=\dfrac{ka-am-bm+am+an}{m+n}$ $x=\dfrac{ka-bm+an}{m+n}$ $→\dfrac{y-b}{n}=\dfrac{k-a-b}{m+n}$ $y-b=\dfrac{(k-a-b).n}{m+n}$ $y-b=\dfrac{kn-an-bn}{m+n}$ $y=\dfrac{kn-an-bn}{m+n}+b$ $y=\dfrac{kn-an-bn}{m+n}+\dfrac{b(m+n)}{m+n}$ $y=\dfrac{kn-an-bn+bm+bn}{m+n}$ $y=\dfrac{kn-an+bm}{m+n}$ Vậy $x=\dfrac{ka-bm+an}{m+n}, y=\dfrac{kn-an+bm}{m+n}$ Bình luận
Đáp án: `x=\frac{km-bm+an}{m+n};y=\frac{kn-an+bm}{m+n}` Giải thích các bước giải: Áp dụng tính chất của dãy tỉ số bằng nhau, ta có: `\frac{x-a}{m}=\frac{y-b}{n}=\frac{x-a+y-b}{m+n}=\frac{k-a-b}{m+n}` Ta có: `\frac{x-a}{m}=\frac{k-a-b}{m+n}` `⇒x-a=\frac{(k-a-b).m}{m+n}=\frac{km-am-bm}{m+n}` `⇒x=\frac{km-am-bm}{m+n}+a=\frac{km-am-bm+am+an}{m+n}=\frac{km-bm+an}{m+n}` `\frac{y-b}{n}=\frac{k-a-b}{m+n}` `⇒y-b=\frac{(k-a-b).n}{m+n}=\frac{kn-an-bn}{m+n}` `⇒y=\frac{kn-an-bn}{m+n}+b=\frac{kn-an-bn+bm+bn}{m+n}=\frac{kn-an+bm}{m+n}` Bình luận
$\dfrac{x-a}{m}=\dfrac{y-b}{n}=\dfrac{x-a+y-b}{m+n}=\dfrac{k-a-b}{m+n}$
$→\dfrac{x-a}{m}=\dfrac{k-a-b}{m+n}$
$x-a=\dfrac{(k-a-b).m}{m+n}$
$x-a=\dfrac{ka-am-bm}{m+n}$
$x=\dfrac{ka-am-bm}{m+n}+a$
$x=\dfrac{ka-am-bm}{m+n}+\dfrac{a(m+n)}{m+n}$
$x=\dfrac{ka-am-bm+am+an}{m+n}$
$x=\dfrac{ka-bm+an}{m+n}$
$→\dfrac{y-b}{n}=\dfrac{k-a-b}{m+n}$
$y-b=\dfrac{(k-a-b).n}{m+n}$
$y-b=\dfrac{kn-an-bn}{m+n}$
$y=\dfrac{kn-an-bn}{m+n}+b$
$y=\dfrac{kn-an-bn}{m+n}+\dfrac{b(m+n)}{m+n}$
$y=\dfrac{kn-an-bn+bm+bn}{m+n}$
$y=\dfrac{kn-an+bm}{m+n}$
Vậy $x=\dfrac{ka-bm+an}{m+n}, y=\dfrac{kn-an+bm}{m+n}$
Đáp án: `x=\frac{km-bm+an}{m+n};y=\frac{kn-an+bm}{m+n}`
Giải thích các bước giải:
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
`\frac{x-a}{m}=\frac{y-b}{n}=\frac{x-a+y-b}{m+n}=\frac{k-a-b}{m+n}`
Ta có:
`\frac{x-a}{m}=\frac{k-a-b}{m+n}`
`⇒x-a=\frac{(k-a-b).m}{m+n}=\frac{km-am-bm}{m+n}`
`⇒x=\frac{km-am-bm}{m+n}+a=\frac{km-am-bm+am+an}{m+n}=\frac{km-bm+an}{m+n}`
`\frac{y-b}{n}=\frac{k-a-b}{m+n}`
`⇒y-b=\frac{(k-a-b).n}{m+n}=\frac{kn-an-bn}{m+n}`
`⇒y=\frac{kn-an-bn}{m+n}+b=\frac{kn-an-bn+bm+bn}{m+n}=\frac{kn-an+bm}{m+n}`