Toán 8 khó.

D

dien0709

Cho x,y,z tỉ lệ a,b,c. CMR:
$(x^2+2y^2+3z^2)(a^2+2b^2+3c^2)=(ax+2by+3cz)^2$ (1)

bỏ ngoặc,lấy VT-VP

$(1)\leftrightarrow 2(ay-bx)^2+3(az-cx)^2+6(bz-cy)^2=0$

$\leftrightarrow ay=bx\to \dfrac{x}{a}=\dfrac{y}{b}$

$az=cx , bz=cy \to ...$

<=>x,y,z tỉ lệ với a,b,c
 
P

phamhuy20011801

Đặt $\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}=k$
$\rightarrow a=xk, b=yk, c=zk$

$(x^2+2b^2+3c^2)(a^2+2b^2+3c^2)=(x^2+2y^2+3z^2)(x^2k^2+2y^2k^2+3z^2k^2)=(x^2k+2y^2k+3z^2k)^2=(ax+2by+3cz)^2$
 
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