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Ta có :
+) \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\)
+) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{zc}\right)=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)
Lời giải:
Đặt $\frac{x}{a}=m; \frac{y}{b}=n; \frac{z}{c}=p$ với $m,n,p>0$.
BĐT cần chứng minh tương đương với:
(m^2a+n^2b+p^2c)(a+b+c)\geq (am+bn+cp)^2$
$\Leftrightarrow m^2(ab+ac)+n^2(ba+bc)+p^2(ca+cb)\geq 2abmn+2amcp+2bncp$
$\Leftrightarrow ab(m^2-2mn+n^2)+bc(n^2-2np+p^2)+ca(m^2-2mp+p^2)\geq 0$
$\Leftrightarrow ab(m-n)^2+bc(n-p)^2+ca(m-p)^2\geq 0$
(luôn đúng với $a,b,c>0$)
Ta có đpcm.
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{abz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)
\(\Rightarrow\dfrac{abz+bxz+cxy}{xyz}=0\)
\(\Rightarrow abz+bxz+cxy=0\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=0\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\left(\dfrac{0}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\) ( đpcm )
Lời giải:
\(\frac{(ax+by+cz)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
\(\Rightarrow (ax+by+cz)^2=(a^2+b^2+c^2)(x^2+y^2+z^2)\)
\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz=a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
\(\Leftrightarrow 2axby+2bycz+2axcz=a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2\)
\(\Leftrightarrow (a^2y^2+b^2x^2-2axby)+(a^2z^2+c^2x^2-2axcz)+(b^2z^2+c^2y^2-2bycz)=0\)
\(\Leftrightarrow (ay-bx)^2+(az-cx)^2+(bz-cy)^2=0\)
Vì bản thân mỗi số hạng đều không âm nên để tổng của chúng bằng $0$ thì:
\((ay-bx)^2=(az-cx)^2=(bz-cy)^2=0\Rightarrow ay=bx; az=cx; bz=cy\)
\(\Rightarrow \frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Ta có đpcm.
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=\dfrac{xbc+yac+zab}{abc}=1\\ \Rightarrow xbc+yac+zab=abc\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=\dfrac{ayz+bxz+cxy}{xyz}=0\\ \Rightarrow ayz+bxz+cxy=0\)
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(abc\right)^2}\)
\(\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc+yac+zab\right)^2}\\ =\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2abc\left(ayz+bxz+cxy\right)}\)
\(\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2abc.0}\\ =\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}=1\)
vậy \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)(đpcm)
\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.\left(\dfrac{c}{z}+\dfrac{b}{y}+\dfrac{a}{x}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.0=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)
Ta có : \(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
\(\Leftrightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2bycz=a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)
\(\Leftrightarrow2axby+2axvz+2bycz=a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2\)
\(\Leftrightarrow a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2-2axby-2azcx-2bycz=0\)
\(\Leftrightarrow\left(a^2y^2-2axby+b^2x^2\right)+\left(a^2z^2-2azcx+c^2x^2\right)+\left(b^2z^2-2bycz+c^2y^2\right)=0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
Do \(\left(ay-bx\right)^2\ge0;\left(az-cx\right)^2\ge0;\left(bz-cy\right)^2\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\az-cx=0\\bz-cy=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{c}{z}=\dfrac{a}{x}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)
:D
\(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+x^2}=a^2+b^2+c^2\)
\(\Leftrightarrow\left(x^2+y^2+x^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)\(\Leftrightarrow a^2x^2+b^2x^2+c^2x^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+x^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2bycz\)\(\Leftrightarrow\left(a^2y^2+2axby+b^2x^2\right)+\left(a^2z^2+2axcz+c^2x^2\right)+\left(b^2z^2+2bycz+c^2y^2\right)=0\)\(\Leftrightarrow\left(ay+bx\right)^2+\left(az+cx\right)^2+\left(bz+cy\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{a}{x}=\dfrac{c}{z}\\\dfrac{b}{y}=\dfrac{c}{z}\end{matrix}\right.\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)
x,y,z ở đâu ra vậy
Ở đề bài chứ ở đâu bạn!