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bài này dùng Py-ta-go khá nhìu nhé, a tự hiểu -,-
\(1=\sin^2\alpha+\cos^2\alpha=BN^2+CM^2=AB^2+AC^2+AN^2+AM^2=BC^2+AN^2+AM^2\)
\(=BC^2+\frac{1}{9}\left(AB^2+AC^2\right)=BC^2+\frac{1}{9}BC^2=\frac{10}{9}BC^2\)\(\Rightarrow\)\(BC=\sqrt{\frac{9}{10}}=\frac{3\sqrt{10}}{10}\)
\(\frac{1}{2}\cdot2\sqrt{2}+\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{\sqrt{2}}{2}-7\sqrt{2}=\sqrt{2}+\frac{\sqrt{2}}{12}-7\sqrt{2}=-\frac{71}{12}\sqrt{2}\)
Vì \(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)=3 ==> \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)=9= \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)
ta có \(\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)= \(\frac{2\left(a+b+c\right)}{abc}\)=2
==> đpcm
+) Ta có: \(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\) \(\left(ĐK:x\ge0\right)\)
\(\Leftrightarrow4\sqrt{3x}+2\sqrt{3x}=3\sqrt{3x}+6\)
\(\Leftrightarrow3\sqrt{3x}=6\)
\(\Leftrightarrow\sqrt{3x}=2\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\left(TM\right)\)
Vậy \(S=\left\{\frac{4}{3}\right\}\)
+) Ta có:\(\sqrt{x^2-1}-4\sqrt{x-1}=0\) \(\left(ĐK:x\ge1\right)\)
\(\Leftrightarrow\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{x-1}.\left(\sqrt{x+1}-4\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\\sqrt{x+1}=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\x+1=16\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\left(TM\right)\\x=15\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{1,15\right\}\)
+) Ta có: \(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\) \(\left(ĐK:x\ge0\right)\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)
\(\Leftrightarrow\frac{2.\left(\sqrt{x}-2\right)-\sqrt{x}}{4\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)
Để \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)mà \(4\sqrt{x}\ge0\forall x\)
\(\Rightarrow\)\(\sqrt{x}-4< 0\)
\(\Leftrightarrow\)\(\sqrt{x}< 4\)
\(\Leftrightarrow\)\(x< 16\)
Kết hợp ĐKXĐ \(\Rightarrow\)\(0\le x< 16\)
Vậy \(S=\left\{\forall x\inℝ/0\le x< 16\right\}\)
\(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\) (Đk: x \(\ge\)0)
<=> \(4\sqrt{3x}+2\sqrt{3x}-3\sqrt{3x}=6\)
<=> \(3\sqrt{3x}=6\)
<=> \(\sqrt{3x}=2\)
<=> \(3x=4\)
<=> \(x=\frac{4}{3}\)
\(\sqrt{x^2-1}-4\sqrt{x-1}=0\) (đk: x \(\ge\)1)
<=> \(\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)
<=> \(\sqrt{x-1}\left(\sqrt{x+1}-4\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x-1=0\\x+1=16\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=15\end{cases}}\)(tm)
\(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\) (Đk: x > 0)
<=> \(\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)
<=>\(\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)
<=> \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)
Do \(4\sqrt{x}>0\) => \(\sqrt{x}-4< 0\)
<=> \(\sqrt{x}< 4\) <=> \(x< 16\)
Kết hợp với đk => S = {x|0 < x < 16}