HOC24
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Ta có:
\(\left\{{}\begin{matrix}x^2+xy+\dfrac{y^2}{3}=2019\\z^2+\dfrac{y^2}{3}=1011\\x^2+xz+z^2=1008\end{matrix}\right.\Leftrightarrow x^2+xy+\dfrac{y^2}{3}=z^2+\dfrac{y^2}{3}+x^2+xz+z^2\)
\(\Rightarrow xy=2z^2+xz\Leftrightarrow xy+xz=2z^2+2xz\)
\(\Rightarrow x\left(y+z\right)=2z\left(x+z\right)\Leftrightarrow\dfrac{2z}{x}=\dfrac{y+z}{x+z}\left(đpcm\right)\)
\(\dfrac{\sqrt{a}}{\dfrac{\sqrt{a}}{\sqrt{b}}-1}\ge\dfrac{\sqrt{b}}{1-\dfrac{\sqrt{b}}{\sqrt{a}}}\Leftrightarrow\dfrac{\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}}{\left(\dfrac{\sqrt{a}}{\sqrt{b}}-1\right)\left(1-\dfrac{\sqrt{b}}{\sqrt{a}}\right)}\ge0\)
\(\Leftrightarrow\left(\dfrac{\sqrt{a}}{\sqrt{b}}-1\right)\left(1-\dfrac{\sqrt{b}}{\sqrt{a}}\right)>0\Leftrightarrow\dfrac{\sqrt{a}}{\sqrt{b}}-1-1+\dfrac{\sqrt{b}}{\sqrt{a}}\ge0\)
\(\Leftrightarrow\dfrac{\sqrt{a}}{\sqrt{b}}+\dfrac{\sqrt{b}}{\sqrt{a}}-2\ge0\Leftrightarrow\dfrac{\sqrt{a}}{\sqrt{b}}+\dfrac{\sqrt{b}}{\sqrt{a}}\ge2\) (Đúng theo AM-GM)
Vậy bđt cần cm đúng
\("="\Leftrightarrow a=b\)
\(\dfrac{yc-bz}{x}=\dfrac{za-xc}{y}=\dfrac{xb-ya}{z}\)
\(\Rightarrow\dfrac{xyc-xbz}{x^2}=\dfrac{yza-xyc}{y^2}=\dfrac{xbz-yza}{z^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{xyc-xbz}{x^2}=\dfrac{yza-xyc}{y^2}=\dfrac{xbz-yza}{z^2}\)
\(=\dfrac{xyc-xbz+yza-xyc+xbz-yza}{x^2+y^2+z^2}=0\)
\(\Rightarrow\left\{{}\begin{matrix}yc=bz\\za=xc\\xb=ya\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y}{b}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{y}{b}\end{matrix}\right.\Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\left(đpcm\right)\)
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\ge1-\dfrac{3}{2}=-\dfrac{1}{2}\)
\("="\Leftrightarrow x=0\)
Lớp 8 học Véc-tơ làm gì cho đau đầu
Gian lận :v
Áp dụng liên tiếp bđt Cauchy-Schwarz và AM-GM
\(\dfrac{x}{1+y^2}+\dfrac{y}{1+x^2}=\dfrac{x^2}{x+y^2x}+\dfrac{y^2}{y+x^2y}\)
\(\ge\dfrac{\left(x+y\right)^2}{x+y+y^2x+x^2y}=\dfrac{4}{x+y+xy\left(x+y\right)}\)
\(=\dfrac{4}{2+2xy}\ge\dfrac{4}{2+\dfrac{\left(x+y\right)^2}{2}}=\dfrac{4}{4}=1\)
\("="\Leftrightarrow x=y=1\)