Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(A=\sqrt{5\sqrt{5\sqrt{5...\sqrt{5}}}}< \sqrt{5\sqrt{5\sqrt{5...\sqrt{25}}}}=5\)
\(B=\sqrt{20+\sqrt{20+...+\sqrt{20}}}< \sqrt{20+\sqrt{20+...+\sqrt{25}}}=5\)
\(\Rightarrow A+B< 5+5=10\)
~ ~ ~
\(A=\sqrt{\dfrac{37}{4}-\sqrt{49+12\sqrt{5}}}\)
\(=\sqrt{\dfrac{37}{4}-\sqrt{\left(3\sqrt{5}+2\right)^2}}\)
\(=\sqrt{\dfrac{29}{4}-3\sqrt{5}}\)
\(=\sqrt{\dfrac{29-12\sqrt{5}}{4}}\)
\(=\sqrt{\dfrac{\left(2\sqrt{5}-3\right)^2}{4}}\)
\(=\dfrac{\sqrt{5}}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}\left(\sqrt{5}-\dfrac{3}{2}\right)\)
\(>\sqrt{5}-\dfrac{3}{2}=B\)
~ ~ ~
\(C=\dfrac{16\sqrt{36}-20\sqrt{48}+10\sqrt{3}}{\sqrt{12}}\)
\(=\dfrac{96-80\sqrt{3}+10\sqrt{3}}{\sqrt{12}}\)
\(=\dfrac{96-70\sqrt{3}}{2\sqrt{3}}\)
\(=16\sqrt{3}-35\)
\(>16\sqrt{3}-36=B\)
~ ~ ~
\(A=\sqrt{47+\sqrt{5}}\cdot\sqrt{47-\sqrt{5}}\)
\(=\sqrt{2204}=2\sqrt{551}\)
\(B=5-2\sqrt{6}+10+\sqrt{6}=15-\sqrt{6}\)
\(A=\sqrt{\left(47+\sqrt{5}\right)\left(47-\sqrt{5}\right)}=2\sqrt{551}\)
\(B=5-2\sqrt{6}+10+\sqrt{6}=15-\sqrt{6}\)
a: \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Leftrightarrow A^3=9+4\sqrt{5}+9-4\sqrt{5}+3\cdot A\)
=>A^3-3A-18=0
=>A=3
b: \(B=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
=>\(B^3=5\sqrt{2}+7-5\sqrt{2}+7+3B\)
=>B^3-3B-14=0
=>B=2,82
c: \(C^3=20+14\sqrt{2}-14\sqrt{2}+20-6C\)
=>C^3+6C-40=0
=>C=2,84
\(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\Leftrightarrow A^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\Leftrightarrow A^3=4+3\sqrt[3]{-1}.A\Leftrightarrow A^3=4-3A\Leftrightarrow A^3+3A-4=0\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)(1)
Ta có \(A^2+A+4>0\)
Vậy (1)\(\Leftrightarrow A-1=0\Leftrightarrow A=1\)
Vậy A=1
\(B=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\Leftrightarrow B^3=5\sqrt{2}+7-5\sqrt{2}+7-3\sqrt[3]{\left(5\sqrt{2}+7\right)\left(5\sqrt{2}-7\right)}\left(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\right)\Leftrightarrow B^3=14-3\sqrt[3]{1}.B\Leftrightarrow B^3=14-3B\Leftrightarrow B^3+3B-14=0\Leftrightarrow\left(B-2\right)\left(B^2+2B+7\right)=0\left(2\right)\)
Ta lại có \(B^2+2B+7>0\)
Vậy (2)\(\Leftrightarrow B-2=0\Leftrightarrow B=2\)
Vậy B=2
\(C=\sqrt[3]{20+14\sqrt{2}}-\sqrt[3]{14\sqrt{2}-20}=\sqrt[3]{\left(\sqrt{2}\right)^3+3.\left(\sqrt{2}\right)^2.2+3.\sqrt{2}.4+8}-\sqrt[3]{\left(\sqrt{2}\right)^3-3.\left(\sqrt{2}\right)^2.2+3.\sqrt{2}.4-8}=\sqrt[3]{\left(\sqrt{2}+2\right)^2}-\sqrt[3]{\left(\sqrt{2}-2\right)}=\sqrt{2}+2-\sqrt{2}+2=4\)
Ta có: \(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5}}}}< \sqrt{5\sqrt{5\sqrt{5...\sqrt{25}}}}=...=5\)
\(\sqrt{20+\sqrt{20+...+\sqrt{20}}}< \sqrt{20+\sqrt{20+...+\sqrt{25}}}=...=5\)
Vậy A+B<5+5=10 (ĐPCM)
vì A nhỏ hơn hoặc bằng 3 và B nhỏ hơn hoặc bằng 5 =>A+B nhỏ hơn hoặc bằng 8 => A+B<10