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Cho \(A=\sqrt{6+\sqrt{6...+\sqrt{6}}+\sqrt[3]{6+\sqrt[3]{6...+\sqrt[3]{6}}}}\) Chứng minh rằng 4<A<5
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho \(A=\sqrt{6+\sqrt{6...+\sqrt{6}}+\sqrt[3]{6+\sqrt[3]{6...+\sqrt[3]{6}}}}\) Chứng minh rằng 4<A<5
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6+\sqrt[3]{6}}}}}\)
\(\Rightarrow D< \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6+\sqrt[3]{8}}}}}\)
\(\Rightarrow D< \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6+2}}}}\)
\(\Rightarrow D< \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{8}}}}\)
\(\Rightarrow D< \sqrt[3]{6+\sqrt[3]{8}}=\sqrt[3]{6+2}=\sqrt[3]{8}\)
\(\Rightarrow D< 2\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Em thử nhá, ko chắc đâu ạ. Em chỉ làm đc một cái thôi
Gọi biểu thức trên là A
*Chứng minh A > 1/6
Đặt \(x=\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6}}}}\left(\text{n dấu căn}\right)\)
Thì \(x=\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6}}}}< \sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{9}}}}=\sqrt{6+3}=3\) (1)
Và \(x^2-6=\sqrt{6+\sqrt{6+...+\sqrt{6}}}\left(\text{n -1 dấu căn}\right)\)
Biểu thức trở thành \(A=\frac{3-x}{9-x^2}=\frac{1}{3+x}\). Từ (1) suy ra \(A>\frac{1}{3+3}=\frac{1}{6}\)(*)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{3}{2}\sqrt{6}+\frac{2.1}{3}\sqrt{2.3}-\frac{4.1}{2}\sqrt{3.2}\)
\(=\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-2\sqrt{6}=\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-2\right)\)
\(=\sqrt{6}\left(\frac{9}{6}+\frac{4}{6}-\frac{12}{6}\right)=\sqrt{6}.\frac{1}{6}=\frac{\sqrt{6}}{6}\)
Vậy \(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn không sửa thì m sửa.
Sửa đề: \(P=\sqrt[3]{\sqrt{\frac{2303}{27}}+6}-\sqrt[3]{\sqrt{\frac{2303}{27}}-6}\)
\(P^3=\sqrt{\frac{2303}{27}}+6-\left(\sqrt{\frac{2303}{27}}-6\right)-\frac{3.11.P}{3}\)
\(\Leftrightarrow P^3=12-11P\)
\(\Leftrightarrow P^3+11P-12=0\)
\(\Leftrightarrow\left(P-1\right)\left(P^2+P+12\right)=0\)
Vì \(P^2+P+12>0\) nên ta có
\(P=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)\(\Leftrightarrow A=\left[\left(\sqrt{57}+6\right)+\left(3\sqrt{6}+\sqrt{38}\right)\right]\left[\left(\sqrt{57}+6\right)-\left(3\sqrt{6}+\sqrt{38}\right)\right]\)\(\Leftrightarrow A=\left(\sqrt{57}+6\right)^2-\left(3\sqrt{6}+\sqrt{38}\right)^2\)
\(\Leftrightarrow A=57+12\sqrt{57}+36-54-12\sqrt{57}-38\)
\(\Leftrightarrow A=1\)
b) \(B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{1+4\sqrt{3}+\left(2\sqrt{3}\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{\sqrt{8+4\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}}{\sqrt{6}+\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}=1\)
c)\(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{3^2-2\times3\times2\sqrt{5}+\left(2\sqrt{5}\right)^2}}}\)
\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(\Leftrightarrow C=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
\(C< \sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{9}}}}}\)
\(\Rightarrow C< \sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{9}}}}\)
\(\Rightarrow C< \sqrt{6+\sqrt{9}}=\sqrt{9}=3\) (đpcm)