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a) Ta có: \(\frac{1}{5}\sqrt{150}=\frac{1}{5}\cdot5\sqrt{6}=\sqrt{6}=\frac{1}{3}\cdot\sqrt{6\cdot9}=\frac{1}{3}\sqrt{54}>\frac{1}{3}\sqrt{51}\)
b) Ta có: \(\frac{1}{2}\sqrt{6}=\sqrt{\frac{6}{4}}< \sqrt{\frac{36}{2}}=6\sqrt{\frac{1}{2}}\)
a) Vì \(5,\left(6\right)< 6\)\(\Rightarrow\)\(\frac{51}{9}< \frac{150}{25}\)
\(\Rightarrow\)\(\sqrt{\frac{51}{9}}< \sqrt{\frac{150}{25}}\)
\(\Rightarrow\)\(\frac{1}{3}\sqrt{51}< \frac{1}{5}\sqrt{150}\)
b) Vì \(1,5< 18\)\(\Rightarrow\)\(\frac{6}{4}< \frac{36}{2}\)
\(\Rightarrow\)\(\sqrt{\frac{6}{4}}< \sqrt{\frac{36}{2}}\)
\(\Rightarrow\)\(\frac{1}{2}\sqrt{6}< 6\sqrt{\frac{1}{2}}\)
\(A=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}:2\sqrt{2}=\frac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}.\left(\sqrt{3}-1\right)}.\frac{1}{2\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}.\left(\sqrt{3}-1\right)}.\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}.\left(\sqrt{3}-1\right)}.\frac{1}{2\sqrt{2}}\)
\(=\frac{1}{2.2}=\frac{1}{4}\)
Giả sử A > B
<=> 19 >\(5\sqrt{3}+6\sqrt{2}\)
<=> (6 + 3 - \(2\sqrt{3}\sqrt{6}\)
) + (10 - 5\(\sqrt{3}\))>0
<=> (\(\sqrt{6}-\sqrt{3}\))2 + (10 - \(5\sqrt{3}\))>0
Mà 10 - 5\(\sqrt{3}\)> 10 - 5\(\sqrt{4}\) = 0
Vậy A > B
a) Ta thấy:
\(\left(3+\sqrt{5}\right)^2=\left(\sqrt{9}+\sqrt{5}\right)^2=9+5+2\sqrt{45}=14+2\sqrt{45}\)
\(\left(2\sqrt{2}+\sqrt{6}\right)^2=\left(\sqrt{8}+\sqrt{6}\right)^2=8+6+2\sqrt{48}=14+2\sqrt{48}\)
Vì \(45< 48\)
\(\Rightarrow\sqrt{45}< \sqrt{48}\)
\(\Rightarrow2\sqrt{45}< 2\sqrt{48}\)
\(\Rightarrow14+2\sqrt{45}< 14+2\sqrt{48}\)
\(\Rightarrow\left(3+\sqrt{5}\right)^2< \left(2\sqrt{2}+\sqrt{6}\right)^2\)
Do \(3+\sqrt{5}>0;2\sqrt{2}+\sqrt{6}>0\)
\(\Rightarrow3+\sqrt{5}< 2\sqrt{2}+6\)
b) Ta thấy:
Vì \(26>3\)
\(\Rightarrow\sqrt{26}>\sqrt{3}\)
\(\Rightarrow\sqrt{26}+1>\sqrt{3}\)
\(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{27}+\sqrt{3}\)
Mà \(\sqrt{27}+\sqrt{3}=3\sqrt{3}+\sqrt{3}=4\sqrt{3}=\sqrt{48}\)
\(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{48}\)
\(a)\)\(A=\sqrt{3\left(\sqrt{2}\right)^2-2\left(\sqrt{2}\right)-\sqrt{2}.\sqrt{2}-1}=\sqrt{6-2\sqrt{2}-2-1}\)
\(A=\sqrt{3-2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-\sqrt{1}\right|=\sqrt{2}-1\)
\(b)\)\(C=\left(1-\sqrt{2}\right)^2+\sqrt{8}-2=1-2\sqrt{2}+2+2\sqrt{2}-2=1\)
\(c_1)\)\(D=\sqrt{\left(1-\sqrt{x}\right)^2}.\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}\) ( đề sai r mem :3 )
\(D=\left|\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right|=\left|x-1\right|\)
\(c_2)\)\(D=\left|x-1\right|=\left|2020-1\right|=2019\)
\(A=\sqrt{2}+1-\sqrt{2}+1=2\)