Cho: \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)và x, y, z khác 0
CMR: \(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)
Cho x,y,z,a,b,c khác 0 và \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\).Chứng minh rằng
a) \(\dfrac{a^2}{x}=\dfrac{b^2}{y}=\dfrac{c^2}{z}=\dfrac{\left(a+b+c\right)^2}{x+y+z}\)
b) \(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)}=\dfrac{1}{a^2+b^2+c^2}\)
Cho \(\dfrac{bz+cy}{x\left(-ax+by+cz\right)}=\dfrac{cx+az}{y\left(ax-by+cz\right)}=\dfrac{ay+bx}{z\left(ax+by-cz\right)}\)
CMR : \(\dfrac{ay+bx}{c}=\dfrac{bz+cy}{a}=\dfrac{cx+az}{b}\)
b) \(\dfrac{x}{a\left(b^2+c^2-a^2\right)}=\dfrac{y}{b\left(a^2+c^2-b^2\right)}=\dfrac{z}{c\left(a^2+b^2-c^2\right)}\)
Cho: \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)và x, y, z khác 0
CMR: \(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)
Giả sử điều cần c/m là đúng . Khi đó , ta có :
\(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)
\(\Leftrightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)
\(\Leftrightarrow x^2a^2+y^2a^2+z^2a^2+x^2b^2+y^2b^2+z^2b^2+x^2c^2+y^2c^2+z^2c^2\)
\(=x^2a^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)
\(\Leftrightarrow y^2a^2+z^2a^2+x^2b^2+z^2b^2+x^2c^2+y^2c^2=2axby+2bycz+2axcz\)
\(\Leftrightarrow y^2a^2+z^2a^2+x^2b^2+z^2b^2+x^2c^2+y^2c^2-2axby-2bycz-2axcz=0\) \(\Leftrightarrow\left(y^2a^2-2axby+b^2x^2\right)+\left(b^2z^2-2bycz+c^2y^2\right)+\left(x^2c^2-2axcz+a^2z^2\right)=0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(cx-az\right)^2=0\left(1\right)\)
Do \(\left\{{}\begin{matrix}\left(ay-bx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\\\left(cx-az\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(cx-az\right)^2\ge0\left(2\right)\)
Từ ( 1 ) ; ( 2 ) \(\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\bz-cy=0\\cx-az=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\bz=cy\\cx=az\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{b}{y}=\dfrac{c}{z}\\\dfrac{c}{z}=\dfrac{a}{x}\end{matrix}\right.\) \(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Điều này đúng với GT đề bài cho
\(\Rightarrow\) Điều cần c/m là đúng
\(\Rightarrow\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\left(đpcm\right)\)
cho \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) rut gon\(\dfrac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{\left(ax+by+cz\right)^22}\)
Cho a,b,c,x,y,z là các số thực khác 0,thỏa:
\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\).CMR:\(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)}=\dfrac{1}{a^2+b^2+c^2}\)
Đặt x/a=y/b=z/c=k
=>x=ak; y=bk; z=ck
\(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{a^2k^2+b^2k^2+c^2k^2}{a^4k^2+b^4k^2+c^4k^2}=\dfrac{1}{a^2+b^2+c^2}\)
Cho \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\ne0\). Chứng minh:
\(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)
Đặt x/a=y/b=z/c=k
=>x=ak; y=bk; z=ck
\(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{a^2k^2+b^2k^2+c^2k^2}{\left(a\cdot ak+b\cdot bk+c\cdot ck\right)^2}\)
\(=\dfrac{k^2\left(a^2+b^2+c^2\right)}{k^2\left(a^2+b^2+c^2\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)
C/m rằng nếu \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\) với x,y,z khác 0 thì \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)
Trừ cả 2 vế cho \(a^2x^2+b^2y^2+c^2z^2\), ta có:
\(a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2bycz+2axcz\)
\(\Rightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2axby-2bycz-2axcz=0\)
\(\left(a^2y^2+b^2x^2-2axby\right)+\left(a^2z^2+c^2z^2-2axcz\right)+\left(b^2z^2+c^2y^2-2bycz\right)=0\)
\(\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
Mà \(\left\{{}\begin{matrix}\left(ay-bx\right)^2\ge0\\\left(az-cx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\az-cx=0\\bz-cy=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
=> đpcm
Cho biết x, y, z khác 0 và \(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
Chứng minh rằng \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
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.
Cho \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\ne0\) . Rút gọn biểu thức :
\(A=\dfrac{\left(x^2+y^2+z^2\right)\cdot\left(a^2+b^2+c^2\right)}{\left(ax+by+cz\right)^2}\)
Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\) \(\left(k\ne0\right)\) \(\Rightarrow\left\{{}\begin{matrix}x=a.k\\y=b.k\\z=c.k\end{matrix}\right.\)
Ta có :
\(A=\dfrac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{\left(ax+by+cz\right)^2}\)
\(A=\dfrac{\left[\left(a.k\right)^2+\left(b.k\right)^2+\left(c.k\right)^2\right]\cdot\left(a^2+b^2+c^2\right)}{\left(a.a.k+b.b.k+c.c.k\right)^2}\)
\(A=\dfrac{\left(a^2k^2+b^2k^2+c^2k^2\right)\cdot\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}\)
\(A=1\)