\(\sqrt{x^2-12}=2\)
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\(\frac{\sqrt{2}+\sqrt{3}}{2+\sqrt{6}}=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(2-\sqrt{6}\right)}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}\)
\(=\frac{2\sqrt{2}-2\sqrt{3}+2\sqrt{3}-3\sqrt{2}}{4-6}\)
\(=\frac{-\sqrt{2}}{-2}\)
\(=\frac{\sqrt{2}}{2}\)
cộng hai vế ta được: 2tan\(\alpha\)=\(\frac{31}{12}\)\(\Rightarrow\)tan\(\alpha\)=\(\frac{31}{24}\)
=> cot\(\alpha\)=\(\frac{17}{24}\)
mik nham r . hai cau nay rieng biet nha , ko lien quan j toi nhau
Đặt \(\hept{\begin{cases}a=\frac{x}{y}\\b=\frac{y}{z}\\c=\frac{z}{x}\end{cases}}\) Ta có: \(A=\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}=\frac{1}{\frac{x}{y}+2}+\frac{1}{\frac{y}{z}+2}+\frac{1}{\frac{z}{x}+2}\)
\(=\frac{y}{x+2y}+\frac{z}{y+2z}+\frac{x}{z+2x}\)
Cần cm \(A\le1\Leftrightarrow2A\le2\)
\(\Leftrightarrow\frac{2y}{x+2y}+\frac{2z}{y+2z}+\frac{2x}{z+2x}\le2\)
\(\Leftrightarrow\left(1-\frac{2y}{x+2y}\right)+\left(1-\frac{2z}{y+2z}\right)+\left(1-\frac{2x}{z+2x}\right)\ge1\)
\(\Leftrightarrow\frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x}\ge1\)
\(\Leftrightarrow\frac{x^2}{x^2+2xy}+\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}\ge1\)
bđt này đúng theo cauchy-schwarz. dấu bằng xảy ra khi a=b=c=1
\(\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\3\sqrt{2x}-\sqrt{3y}=\frac{9}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\6\sqrt{2x}-2\sqrt{3y}=9\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\7\sqrt{2x}=14\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\\sqrt{2x}=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\2x=2^2=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{2.2}+2\sqrt{3y}=5\\x=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2+2\sqrt{3y}=5\\x=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2\sqrt{3y}=3\\x=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{3y}=\frac{3}{2}\\x=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3y=\left(\frac{3}{2}\right)^2=\frac{9}{4}\\x=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=\frac{3}{4}\\x=2\end{cases}}\)
\(ĐKXĐ:x^2-12\ge0\Rightarrow x^2\ge12\Rightarrow x\ge-2\sqrt{3}\)
\(\sqrt{x^2-12}=2\)
\(\Leftrightarrow x^2-12=4\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
cảm ơn bạn