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\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Leftrightarrow x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)+y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)+z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)
Dấu "=" xảy ra tại x=y=z=0
Khi đó T=0
Ta có:
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
<=> \(\left(a^2+b^2+c^2\right)\)\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\left(a^2+b^2+c^2\right)\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)
<=> \(x^2+y^2+z^2=\left(a^2+b^2+c^2\right)\frac{x^2}{a^2}+\left(a^2+b^2+c^2\right)\frac{y^2}{b^2}+\left(a^2+b^2+c^2\right)\frac{z^2}{c^2}\)
<=> \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2=0\)
vì a, b , c khác 0 nên \(\frac{\left(b^2+c^2\right)}{a^2};\frac{\left(c^2+a^2\right)}{b^2};\frac{\left(b^2+a^2\right)}{c^2}\ne0\)
\(\frac{\left(b^2+c^2\right)}{a^2}x^2\ge0;\frac{\left(a^2+c^2\right)}{b^2}y^2\ge0;\frac{\left(a^2+b^2\right)}{c^2}z^2\ge0\)với mọi x, y, z
=> \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2\ge0\)với mọi x; y; z
Do đó: \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2=0\)
=> x = y = z = 0
Vậy T = 0
1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
a, \(9x^2+y^2+2z^2-18x-6y+4z+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Vì \(\left\{{}\begin{matrix}9\left(x-1\right)^2\ge0\\\left(y-3\right)^2\ge0\\2\left(z+1\right)^2\ge0\end{matrix}\right.\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Mà \(9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
Vậy...
1) Đặt \(B=x^2+y^2+z^2\)
\(C=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)
Ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)
\(\Leftrightarrow-2\left(xy+yz+xz\right)=x^2+y^2+z^2\)
Suy ra: \(C=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow A=\dfrac{B}{C}=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
2) \(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\)
\(\Leftrightarrow x^2+xy-2xy-2y^2=0\)
\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)
Do \(x+y\ne0\) nên \(x-2y=0\Leftrightarrow x=2y\)
Do đó: \(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
Ta có : \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\)
\(\Rightarrow\dfrac{bcx}{abc}+\dfrac{acy}{abc}+\dfrac{abz}{abc}=0\)
\(\Rightarrow\dfrac{bcx+acy+abz}{abc}=0\)
\(\Rightarrow bcx+acy+abz=0\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=10\)
\(\Rightarrow\left(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\right)^2=10^2\)
\(\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{ab}{xy}+\dfrac{bc}{yz}+\dfrac{ac}{xz}\right)=100\)
\(\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{abz+bcx+acy}{xyz}\right)=100\)
\(\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{0}{xyz}\right)=100\)
\(\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}=100\)
\(\Rightarrow S=100\)
Ta đặt x/a=m;y/b=n;c/z=k suy ra m+n+k=0 suy ra
a/x=1/m;y/b=1/n và c/z=1/k từ đó ta có bài toán m+n+k=0 và 1/n+1/m+1/k=1
Mk chỉ gợi ý đến đó thôi nha các bạn tự làm tiếp nha nhớ tick mk vs
Lời giải:
Ta có:
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Rightarrow x^2+y^2+z^2=\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)(a^2+b^2+c^2)\)
\(\Leftrightarrow x^2+y^2+z^2=x^2+\frac{x^2b^2}{a^2}+\frac{x^2c^2}{a^2}+y^2+\frac{y^2a^2}{b^2}+\frac{y^2c^2}{b^2}+z^2+\frac{z^2a^2}{c^2}+\frac{z^2b^2}{c^2}\)
\(\Leftrightarrow \frac{x^2b^2}{a^2}+\frac{x^2c^2}{a^2}+\frac{y^2a^2}{b^2}+\frac{y^2c^2}{b^2}+\frac{z^2a^2}{c^2}+\frac{z^2b^2}{c^2}=0(*)\)
Bởi vì mỗi số hạng trong tổng $(*)$ đều là những số không âm, cho nên để tổng các số không âm bằng $0$ thì bản thân mỗi số đó phải bằng $0$
Do đó:
\(\Leftrightarrow \frac{x^2b^2}{a^2}=\frac{x^2c^2}{a^2}=\frac{y^2a^2}{b^2}=\frac{y^2c^2}{b^2}=\frac{z^2a^2}{c^2}=\frac{z^2b^2}{c^2}=0\)
Do $a,b,c\neq 0$ nên \(x^2=y^2=z^2=0\Rightarrow x=y=z=0\)
Khi đó:\(T=x^{2016}+y^{2016}+z^{2016}=0\)