cho a >=6. chung minh: a^2+18/\(\sqrt{a}>=36+3\sqrt{6}\)
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Lời giải:
\(\frac{18}{2\sqrt{3}-\sqrt{6}}=\frac{18(2\sqrt{3}+\sqrt{6})}{(2\sqrt{3}-\sqrt{6})(2\sqrt{3}+\sqrt{6})}=\frac{36\sqrt{3}+18\sqrt{6}}{6}\)
\(=6\sqrt{3}+3\sqrt{6}\)
$\Rightarrow a=6; b=-3$
$\Rightarrow a+b=6+(-3)=3$
a)
\(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{3}+\sqrt{2}-\sqrt{2}+1=\sqrt{3}+1\)
b)
\(\sqrt{\left(\sqrt{9}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{16}+\sqrt{2}\right)^2}=\sqrt{9}+\sqrt{2}-\sqrt{16}-\sqrt{2}=3-4=-1\)
c)
\(=\sqrt{2\left(2-\sqrt{3}\right)}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
Cho \(A=\sqrt{6+\sqrt{6...+\sqrt{6}}+\sqrt[3]{6+\sqrt[3]{6...+\sqrt[3]{6}}}}\) Chứng minh rằng 4<A<5
\(a,=4\sqrt{6}-15\sqrt{6}+\sqrt{\left(2+\sqrt{6}\right)^2}=-11\sqrt{6}+2+\sqrt{6}=2-10\sqrt{6}\\ b,=\dfrac{\sqrt{3}\left(\sqrt{6}-2\right)}{\sqrt{6}-2}+\dfrac{4\left(\sqrt{3}-1\right)}{2}+\left|3\sqrt{3}-12\right|=\sqrt{3}+2\sqrt{3}-2+12-3\sqrt{3}=10\)
Cho \(A=\sqrt{6+\sqrt{6...+\sqrt{6}}+\sqrt[3]{6+\sqrt[3]{6...+\sqrt[3]{6}}}}\) Chứng minh rằng 4<A<5
\(A=\frac{3-\sqrt{3+\sqrt{3+...+\sqrt{3}}}}{6-\sqrt{3+\sqrt{3+...\sqrt{3}}}}>\frac{3-\sqrt{3+\sqrt{3+...\sqrt{1}}}}{6-\sqrt{3+\sqrt{3+..+\sqrt{1}}}}=\frac{3-2}{6-2}=\frac{1}{4}>\frac{1}{5}\)
<=> A > 1/5