Cho a,b,c,x,y,z khác 0 thõa mãn x/a = y/b=z/c . Chứng minh rằng : a^2/x + b^2/y + c^2/z = ( a+ b + c ) ^2 /x+Y+Z 03/08/2021 Bởi Gianna Cho a,b,c,x,y,z khác 0 thõa mãn x/a = y/b=z/c . Chứng minh rằng : a^2/x + b^2/y + c^2/z = ( a+ b + c ) ^2 /x+Y+Z
Ta có: `x/a=y/b=z/c=k` `⇒x=ak,y=bk,z=ck` `⇒a^2/x+b^2/y+c^2/z` `=a^2/ak+b^2/bk+c^2/ck` `=a/k+b/k+c/k` `={a+b+c}/k` Ta lại có: `{(a+b+c)^2}/{x+y+z}` `={(a+b+c)^2}/{ak+bk+ck}` `={(a+b+c)^2}/{k(a+b+c)}` `={a+b+c}/k` `⇒a^2/x+b^2/y+c^2/z` `={(a+b+c)^2}/{x+y+z}` Bình luận
Giải thích các bước giải: Ta có :$\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k$ $\to x=ak, y=bk, z=ck$ $\to \dfrac{a^2}{x}+\dfrac{b^2}y+\dfrac{c^2}z=\dfrac{a^2}{ak}+\dfrac{b^2}{bk}+\dfrac{c^2}{ck}=\dfrac ak+\dfrac bk+\dfrac ck=\dfrac{a+b+c}{k}$ Lại có :$\dfrac{(a+b+c)^2}{x+y+z}=\dfrac{(a+b+c)^2}{ak+bk+ck}=\dfrac{(a+b+c)^2}{k(a+b+c)}=\dfrac{a+b+c}{k}$ $\to \dfrac{a^2}{x}+\dfrac{b^2}y+\dfrac{c^2}z=\dfrac{(a+b+c)^2}{x+y+z}$ Bình luận
Ta có:
`x/a=y/b=z/c=k`
`⇒x=ak,y=bk,z=ck`
`⇒a^2/x+b^2/y+c^2/z`
`=a^2/ak+b^2/bk+c^2/ck`
`=a/k+b/k+c/k`
`={a+b+c}/k`
Ta lại có:
`{(a+b+c)^2}/{x+y+z}`
`={(a+b+c)^2}/{ak+bk+ck}`
`={(a+b+c)^2}/{k(a+b+c)}`
`={a+b+c}/k`
`⇒a^2/x+b^2/y+c^2/z`
`={(a+b+c)^2}/{x+y+z}`
Giải thích các bước giải:
Ta có :
$\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k$
$\to x=ak, y=bk, z=ck$
$\to \dfrac{a^2}{x}+\dfrac{b^2}y+\dfrac{c^2}z=\dfrac{a^2}{ak}+\dfrac{b^2}{bk}+\dfrac{c^2}{ck}=\dfrac ak+\dfrac bk+\dfrac ck=\dfrac{a+b+c}{k}$
Lại có :
$\dfrac{(a+b+c)^2}{x+y+z}=\dfrac{(a+b+c)^2}{ak+bk+ck}=\dfrac{(a+b+c)^2}{k(a+b+c)}=\dfrac{a+b+c}{k}$
$\to \dfrac{a^2}{x}+\dfrac{b^2}y+\dfrac{c^2}z=\dfrac{(a+b+c)^2}{x+y+z}$