CMR : `a(b-c)(b+c-a)^2 +c(a-b)(a+b-c)^2 =b(a-c)(a+c-b)^2`
2 câu trả lời
Gỉa sử : `a(b-c)(b+c-a)^2 +c(a-b)(a+b-c)^2 =b(a-c)(a+c-b)^2`
`<=>a(b-c)(b+c-a)^2 +c(a-b)(a+b-c)^2 -b(a-c)(a+c-b)^2 =0` $(*)$
Đặt $\begin{cases}a+b-c=x\\ b+c-a=y\\a+c-b=z\end{cases}$ `=>` $\begin{cases}a=\dfrac{x+z}{2}\\b=\dfrac{x+y}{2}\\c=\dfrac{y+z}{2}\end{cases}$
Từ $(*)$ ta có :
`VT=(x+y)/2((x+y)/2-(y+z)/2).y^2 +(y+z)/2((x+z)/2-(x+y)/2).x^2 -1/4(x+y)(x-y).z^2`
`=(x+y)/2 . (x-z)/2 .y^2 +(y+z)/2 . (z-y)/2 . x^2 -1/4(x^2 -y^2 )z^2`
`=1/4(x^2 -z^2).y^2 +1/4(z^2 -y^2 ).x^2 -1/4(x^2 -y^2).z^2 =0=VP`
`->` Điều giả sử đã đúng
Vậy `a(b-c)(b+c-a)^2 +c(a-b)(a+b-c)^2 =b(a-c)(a+c-b)^2`
$\textit{Đáp án + Giải thích các bước giải:}$
`a(b-c)(b+c-a)^2+c(a-b)(a+b-c)^2=b(a-c)(a+c-b)^2`
`<=>a(b-c)(b+c-a)^2+c(a-b)(a+b-c)^2-b(a-c)(a+c-b)^2=0`
`<=>a(b-c)(b+c-a)^2+c(a-b)(a+b-c)^2+b(c-a)(a+c-b)^2=0`
`<=>(ab-ac)(b+c-a)^2+(ca-cb)(a+b-c)^2+(bc-ab)(a+c-b)^2=0`
`<=>ab(b+c-a)^2-ca(b+c-a)^2+ca(a+b-c)^2-bc(a+b-c)^2+bc(a+c-b)^2-ab(a+c-b)^2=0`
`<=>ab[(b+c-a)^2-(a+c-b)^2]+bc[(a+c-b)^2-(a+b-c)^2]+ca[(a+b-c)^2-(b+c-a)^2]=0`
`<=>ab(b+c-a-a-c+b)(b+c-a+a+c-b)+bc(a+c-b-a-b+c)(a+c-b+a+b-c)+ca(a+b-c-b-c+a)(a+b-c+b+c-a)=0`
`<=>ab.2c(2b-2a)+bc.2a(2c-2b)+ca.2b(2a-2c)=0`
`<=>4abc(b-a)+4abc(c-b)+4abc(a-c)=0`
`<=>4abc(b-a+c-b+a-c)=0`
`<=>4abc.0=0`
`<=>0=0` (đúng)
Ta có $đpcm$
