Ta có: \(\overrightarrow{B'B}=\overrightarrow{B'A}+\overrightarrow{AB}=\overrightarrow{C'D'}+\overrightarrow{DC}\)
\(\Rightarrow\overrightarrow{B'B}+\overrightarrow{CC'}+\overrightarrow{D'D}=\overrightarrow{C'D'}+\overrightarrow{DC}+\overrightarrow{CC'}+\overrightarrow{D'D}\)
\(=\left(\overrightarrow{CC'}+\overrightarrow{C'D'}\right)+\left(\overrightarrow{D'D}+\overrightarrow{DC}\right)=\overrightarrow{CD'}+\overrightarrow{D'C}=\overrightarrow{CD'}-\overrightarrow{CD'=}\overrightarrow{0}\)
liên hệ làm zì tốn thời gian chả có tác dụng zì !!!!!!!!!!!!!!!!!!!
Ta dễ thấy : \(\hept{\begin{cases}D\left(0;6\right)\\A\left(0;-6\right)\end{cases}}\)
Ta có OG = OH = 3
\(OM=ON=\sqrt{6^2-3^2}=3\sqrt{3}\)
\(\hept{\begin{cases}B\left(-3;3\sqrt{3}\right)\\G\left(-3;-3\sqrt{3}\right)\end{cases}}\)
\(\hept{\begin{cases}C\left(3;3\sqrt{3}\right)\\E\left(3;-3\sqrt{3}\right)\end{cases}}\)
Ta có:
\(xy+yz+zx=0\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+3\left(\frac{1}{xy}\right)\left(\frac{1}{x}+\frac{1}{y}\right)=-\frac{1}{z^3}\)
\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
Ta lại có:
\(M=\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{zx}{y^2}=\frac{xyz}{z^3}+\frac{xyz}{x^3}+\frac{xyz}{y^3}\)
\(=xyz.\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)
Trước tiên ta chứng minh
\(\left(b-c\right)^2\le6\left(\sqrt{b}-\sqrt{c}\right)^2\)
\(\Leftrightarrow\left(\sqrt{b}-\sqrt{c}\right)^2\left(6-\left(\sqrt{b}+\sqrt{c}\right)^2\right)\ge0\)
Cái này đúng.
Từ đây ta có:
\(D\le\sqrt{a+\frac{\left(\sqrt{b}-\sqrt{c}\right)^2}{2}+\sqrt{b}+\sqrt{c}}\)
\(=\sqrt{a+b+c-\frac{\left(\sqrt{b}+\sqrt{c}\right)^2}{2}+\sqrt{b}+\sqrt{c}}\)
\(=\sqrt{3-\left(\left(\frac{\sqrt{b}+\sqrt{c}}{\sqrt{2}}\right)^2-2.\frac{\sqrt{b}+\sqrt{c}}{\sqrt{2}.\sqrt{2}}+\frac{1}{\sqrt{4}}\right)+\frac{1}{2}}\)
\(=\sqrt{3+\frac{1}{2}-\left(\frac{\sqrt{b}+\sqrt{c}}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)^2}\)
\(\le\sqrt{3+\frac{1}{2}}=\sqrt{\frac{7}{2}}\)
Dấu = xảy ra khi \(\hept{\begin{cases}a=2,5\\b=c=0,25\end{cases}}\)
\(\hept{\begin{cases}x^{2000}+y^{2000}=2016,2017=a\left(1\right)\\x^{4000}+y^{4000}=2017,2018=b\left(2\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow\left(x^{2000}+y^{2000}\right)^2=a^2\)
\(\Leftrightarrow x^{4000}+y^{4000}+2x^{2000}y^{2000}=a^2\)
\(\Leftrightarrow x^{2000}y^{2000}=\frac{a^2-b}{2}\)
Ta lại có:
\(\left(x^{2000}+y^{2000}\right)\left(x^{4000}+y^{4000}\right)=ab\)
\(\Leftrightarrow x^{6000}+y^{6000}+x^{2000}y^{2000}\left(x^{2000}+y^{2000}\right)=ab\)
\(\Leftrightarrow x^{6000}+y^{6000}=ab-x^{2000}y^{2000}\left(x^{2000}+y^{2000}\right)\)
\(\Leftrightarrow x^{6000}+y^{6000}=ab-\frac{a^2-b}{2}.a\)
Thay số vô tính là xong
Ta có:
\(y^2+yz+z^2=2-\frac{3x^2}{2}\)
\(\Leftrightarrow2y^2+2yz+2z^2=4-3x^2\)
\(\Leftrightarrow\left(x+y+z\right)^2=4-2x^2-y^2-z^2+2xy+2xz\)
\(=4-\left(x-y\right)^2-\left(x-z\right)^2\le4\)
\(\Rightarrow-2\le x+y+z\le2\)
