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\(\sqrt{6+\sqrt{6+...+\sqrt{6}}}>\sqrt{6}=\sqrt{\frac{150}{25}}>\sqrt{\frac{144}{25}}=\frac{12}{5}\)
\(\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}>\sqrt[3]{6}=\sqrt[3]{\frac{750}{125}}>\sqrt[3]{\frac{729}{125}}=\frac{9}{5}\)
\(\Rightarrow A>\frac{12}{5}+\frac{9}{5}=\frac{21}{5}>4\)
\(\sqrt{6+\sqrt{6+...+\sqrt{6}}}< \sqrt{6+\sqrt{6+...+\sqrt{9}}}=3\)
\(\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}< \sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{8}}}=2\)
\(\Rightarrow A< 3+2=5\)
\(\Rightarrow4< A< 5\Rightarrow\left[A\right]=4\)
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\(DặtA=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}\)
Ta có so sánh A với 3
A^2 = 6 + \(\sqrt{6+\sqrt{6+\sqrt{6}}}\)
3^2 = 9 = 6 + 3
Tiếp tục lại so sánh
\(\sqrt{6+\sqrt{6+\sqrt{6}}với}3\)
\(\sqrt{6+\sqrt{6+\sqrt{6}}}^2=6+\sqrt{6+\sqrt{6}}\)
3^2 =9= 6 + 3
Tiếp tục một lần nữa
BBAy giờ ta chỉ việc so sánh :
\(\sqrt{6}và3\)
6 < 9 => văn 6 < căn 9 => căn 6 < 3
làm tiếp nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(\sqrt{6+2\sqrt{2}\sqrt{3-\left(\sqrt{3}+1\right)^2}}=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
b/ \(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4+\sqrt{15}\right)=2\left(16-15\right)\)
\(M=\sqrt{\frac{\left(3\sqrt{3}-4\right)\left(2\sqrt{3}-1\right)}{\left(2\sqrt{3}+1\right)\left(2\sqrt{3}-1\right)}}+\sqrt{\frac{\left(\sqrt{3}+4\right)\left(5+2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}}\)
\(M=\sqrt{\frac{18-3\sqrt{3}-8\sqrt{3}+4}{11}}+\sqrt{\frac{5\sqrt{3}+6+20+8\sqrt{3}}{13}}\)
\(M=\sqrt{\frac{11\left(2-\sqrt{3}\right)}{11}}+\sqrt{\frac{13\left(2+\sqrt{3}\right)}{13}}=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(M=\frac{1}{\sqrt{2}}\left(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\right)\)
\(M=\frac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\right)\)
\(M=\frac{1}{\sqrt{2}}\left(\sqrt{3}-1+\sqrt{3}+1\right)=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có VT:
\(VT=\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\)
\(=3+\sqrt{2}+3-\sqrt{2}\)
\(=6=VP\left(dpcm\right)\)
\(VT=\sqrt{9+2\cdot3\cdot\sqrt{2}+2}+\sqrt{9-2\cdot3\cdot\sqrt{2}+2}\)
\(=3+\sqrt{2}+3-\sqrt{2}\)
=6=VP
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1. \(2M-N=\dfrac{2}{2-\sqrt{3}}-\sqrt{6}.\sqrt{2}=\dfrac{2-2\sqrt{3}\left(2-\sqrt{3}\right)}{2-\sqrt{3}}=\)\(\dfrac{2-4\sqrt{3}+6}{2-\sqrt{3}}=\dfrac{8-4\sqrt{3}}{2-\sqrt{3}}=4\)
Đáp án C
2. Ta có: A= \(-x+\sqrt{\left(6-x\right)^2}=-x+\left|6-x\right|\)
Mà x>6 \(\Rightarrow6-x< 0\)A=-x-6+x=-6
Đáp án C
3. Vẽ đồ thị hàm f(x) ta có:
Ta thấy f(2)<f(3), chọn Đáp án A
4.
Khi đó, bán kính của đường tròn bằng \(\dfrac{2}{3}\)đường cao của tam giác đều ABC
Ta có: \(R=\dfrac{2}{3}.\dfrac{a\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{3}\)
Đáp án A
Đặt \(A=\sqrt{6+\sqrt{6+\sqrt{6+...}}}=3\)
\(A^2=6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}\)
\(A^2=6+A\)
\(A^2-3A+2A-6=0\)
\(A\left(A-3\right)+2\left(A-3\right)=0\)
\(\Rightarrow\left(A-3\right)\left(A+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}A-3=0\\A+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}A=3\left(n\right)\\A=-2\left(l\right)\end{cases}}}\)
Vậy \(\sqrt{6+\sqrt{6+\sqrt{6+...}}}=3\left(đpcm\right)\)