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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{20+\sqrt{20+\sqrt{20+...+\sqrt{20}}}}\sqrt{20}>\sqrt{16}=4\)
\(\Rightarrow4
![](https://rs.olm.vn/images/avt/0.png?1311)
Số này lớn hơn 4 và nhỏ hơn 5 thôi, (rất gần 5)
Tính thế nào được A.
![](https://rs.olm.vn/images/avt/0.png?1311)
D=\(\sqrt{20+\sqrt{20+....+\sqrt{20+\sqrt{25}}}}\)= \(\sqrt{20+\sqrt{20+....+\sqrt{20+5}}}\)=\(\sqrt{20+\sqrt{20+....+\sqrt{25}}}\)
=............=\(\sqrt{20+\sqrt{25}}\)=\(\sqrt{20+5}=5\)
Vậy D=5
![](https://rs.olm.vn/images/avt/0.png?1311)
- Ta có:\(T=\sqrt{20+\sqrt{20+...+\sqrt{20}}}+\sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24}}}>\sqrt{20}+\sqrt[3]{24}>7\)(1)
- Mặt khác:
\(\sqrt{20+\sqrt{20+...+\sqrt{20}}}< \sqrt{20+\sqrt{20+...+\sqrt{20+\sqrt{25}}}}=5\)
- Và:
\(\sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24}}}< \sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24+\sqrt[3]{27}}}}=3\)
- Nên \(T=\sqrt{20+\sqrt{20+...+\sqrt{20}}}+\sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24}}}< 8\)(2).
- Từ (1) và (2), ta có: \(7< T< 8\)đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(\sqrt{49-20\sqrt{6}}-\sqrt{106+20\sqrt{6}}=\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(10+\sqrt{6}\right)^2}=5-2\sqrt{6}-10-\sqrt{6}=-5-3\sqrt{6}\)
b. \(\sqrt{83-20\sqrt{6}}+\sqrt{62-20\sqrt{6}}=\sqrt{\left(5\sqrt{3}-2\sqrt{2}\right)^2}+\sqrt{\left(5\sqrt{2}-2\sqrt{3}\right)^2}=5\sqrt{3}-2\sqrt{2}+5\sqrt{2}-2\sqrt{3}=3\sqrt{3}+3\sqrt{2}\)
c. \(\sqrt{302-20\sqrt{6}}+\sqrt{203-20\sqrt{6}}=\sqrt{\left(10\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(10\sqrt{2}-\sqrt{3}\right)^2}=10\sqrt{3}-\sqrt{2}+10\sqrt{2}-\sqrt{3}=9\sqrt{3}+9\sqrt{2}\)
d. \(\sqrt{601-20\sqrt{6}}-\sqrt{154-20\sqrt{6}}=\sqrt{\left(10\sqrt{6}-1\right)^2}-\sqrt{\left(5\sqrt{6}-2\right)^2}=10\sqrt{6}-1-5\sqrt{6}+2=1+5\sqrt{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(x=\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}\)
\(\Rightarrow x^3=40+3\sqrt[3]{\left(20+14\sqrt[]{2}\right)\left(20-14\sqrt[]{2}\right)}.\left(\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}\right)\)
\(\Rightarrow x^3=40+6x\)
\(\Rightarrow x^3-6x-40=0\)
\(\Rightarrow\left(x-4\right)\left(x^2+4x+10\right)=0\)
\(\Rightarrow x=4\)
Vậy \(\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\sqrt{20+\sqrt{20+\sqrt{20+...+\sqrt{20}}}}< \sqrt{20+\sqrt{20+\sqrt{20+...+\sqrt{25}}}}\)
\(=\sqrt{20+\sqrt{20+\sqrt{20+...+5}}}=\sqrt{20+\sqrt{20+\sqrt{25}}}=\sqrt{20+5}=5\)
\(\Rightarrow\)\(A< 5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
lụi đê ( lụi nhg đúng :D )
\(\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20+....+\sqrt{20}}}}}=A\)
\(20+\sqrt{20+\sqrt{20+\sqrt{20+...+\sqrt{20}}}}=A^2\)
20 + A = A2
GIẢI RA TÌM A