So sánh:
1-2căn3 và căn2 - căn6
3- 2 căn 3 và 2 căn 6 - 5
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\(\sqrt{11+4\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{8+2.2\sqrt{2}.\sqrt{3}+3}-\sqrt{2-2\sqrt{2}.\sqrt{3}+3}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=2\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=3\sqrt{2}\)
\(\sqrt{11+4\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{8+4\sqrt{6}+3}-\sqrt{3-2\sqrt{6}+2}\)
\(=\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|2\sqrt{2}+\sqrt{3}\right|-\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\left(2\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{2}\right)\)
\(=2\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=3\sqrt{2}\)
\(3\sin^2\alpha+2\sin\alpha=1\Leftrightarrow3\sin^2\alpha+3\sin\alpha-\sin\alpha-1=0\Leftrightarrow3\sin\alpha\left(\sin\alpha+1\right)-\left(\sin\alpha+1\right)=0\Leftrightarrow\left(3\sin\alpha-1\right)\left(\sin\alpha+1\right)=0\)Do \(\sin\alpha>0\)nên \(\sin\alpha+1>0\)suy ra \(3\sin\alpha-1=0\Leftrightarrow\sin\alpha=\frac{1}{3}\)
Ta có: \(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos\alpha=\sqrt{1-\sin^2\alpha}=\frac{2\sqrt{2}}{3}\)
\(\Rightarrow\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{1}{3}}{\frac{2\sqrt{2}}{3}}=\frac{\sqrt{2}}{4}\)
ta có \(x^2-5x+7=x^2-2.\frac{5}{2}.x+\frac{25}{4}+\frac{3}{4}\)
\(=\left(x-\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)\(\Rightarrow\sqrt{x^2-5x+7}\ge\frac{\sqrt{3}}{2}\)
\(\Rightarrow M=\frac{1}{\sqrt{x^2-5x+7}}\le\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\)
\(A=\sqrt{3-\sqrt{6}-\sqrt{3}+\sqrt{2}}+\sqrt{2-\sqrt{3}}\)
\(\Rightarrow\sqrt{2}A=\sqrt{6-2\sqrt{6}-2\sqrt{3}+2\sqrt{2}}+\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2+1-2\sqrt{6}-2\sqrt{3}+2\sqrt{2}}+\sqrt{3-2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+\left(\sqrt{2}\right)^2+1^2-2\sqrt{3}.\sqrt{2}-2\sqrt{3}.1+2\sqrt{2}.1}+\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1^2}\)
\(=\sqrt{\left(1+\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=|1+\sqrt{2}-\sqrt{3}|+|\sqrt{3}-1|\)
\(=1+\sqrt{2}-\sqrt{3}+\sqrt{3}-1=\sqrt{2}\)
PT \(\Leftrightarrow\sqrt{2x+1}=7-x\)\(\Leftrightarrow\hept{\begin{cases}7-x\ge0\\2x+1=\left(7-x\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le7\\2x+1-49+14x-x^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\le7\\-x^2+16x-48=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le7\\-\left(x-12\right)\left(x-4\right)=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\le7\\x=12\left(ktm\right);x=4\left(tm\right)\end{cases}}\)
Vậy pt có 1 nghiệm là x = 4