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Võ Đông Anh Tuấn
Áp dụng \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)
a)
\(7=\sqrt{49}\\ 3\sqrt{5}=\sqrt{9}\cdot\sqrt{5}=\sqrt{9\cdot5}=\sqrt{45}\\ \text{Vì }\sqrt{49}>\sqrt{45}\text{ nên }7>3\sqrt{5}\)
Vậy \(7>3\sqrt{5}\)
b)
\(2\sqrt{7}+3=\sqrt{4}\cdot\sqrt{7}+3=\sqrt{4\cdot7}+3=\sqrt{28}+3\\ \sqrt{28}+3>\sqrt{25}+3=5+3=8\)
Vậy \(8< 2\sqrt{7}+3\)
c)
\(3\sqrt{6}=\sqrt{9}\cdot\sqrt{6}=\sqrt{9\cdot6}=\sqrt{54}\\ 2\sqrt{15}=\sqrt{4}\cdot\sqrt{15}=\sqrt{4\cdot15}=\sqrt{60}\\ \text{Vì } \sqrt{54}< \sqrt{60}\text{nên }3\sqrt{6}< 2\sqrt{15}\)
Vậy \(3\sqrt{6}< 2\sqrt{15}\)
căn 24< căn 25 =5 :
căn 45<căn 49 =7
=> căn 24+ căn 45 < căn 25+ căn 49 =5+7=12
a) \(2\sqrt{3}=\sqrt{12}\)
\(3\sqrt{2}=\sqrt{18}\)
Vì 12<18 => \(\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) \(12=5+7=\sqrt{25}+\sqrt{49}>\sqrt{24}+\sqrt{45}\)
a)\(1+\sqrt{3}>1+\sqrt{1}=1+1=2\)
Vậy \(1+\sqrt{3}>2\)
c) \(\sqrt{3}-1< \sqrt{4}-1=2-1=1\)
Vậy \(\sqrt{3}-1< 1\)
e) \(\sqrt{2}+\sqrt{5}< \sqrt{16}+\sqrt{16}=4+4=8\)
Vậy \(\sqrt{2}+\sqrt{5}< 8\)
1) \(2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\)\(6+2\sqrt{2}< 6+3=9\)
2) \(4\sqrt{5}=\sqrt{80}>\sqrt{49}=7\)
\(\Rightarrow\)\(9+4\sqrt{5}>9+7=16\)
3) \(2=\sqrt{4}>\sqrt{3}\)
\(\Rightarrow\)\(2-1>\sqrt{3}-1\)
hay \(1>\sqrt{3}-1\)
4) \(9-4\sqrt{5}< 16\)
5) \(\sqrt{2}>\sqrt{1}=1\)
\(\Rightarrow\)\(\sqrt{2}+1>2\)
* \(4\)và \(1+2\sqrt{2}\)
Ta có \(3=\sqrt{9}\)
\(2\sqrt{2}=\sqrt{2^2.2}=\sqrt{8}\)
Ta lại có \(8< 9\Leftrightarrow\sqrt{8}< \sqrt{9}\)
Hay \(2\sqrt{2}< 3\)\(\Leftrightarrow1+2\sqrt{2}< 1+3\Leftrightarrow1+2\sqrt{2}< 4\)
Ta có : \(a)\)\(6+2\sqrt{2}\) và 9
\(\Rightarrow9-6-2\sqrt{2}=3-2\sqrt{2}\)
\(=2-2\sqrt{2}+1\)
\(=(\sqrt{2}-1)^2>0\)
\(\Rightarrow9-6-2\sqrt{2}>0\Rightarrow9>6+2\sqrt{2}\)
\(b)\sqrt{2}+\sqrt{3}\)và 3
\(\Rightarrow\sqrt{[(\sqrt{2}+\sqrt{3})}^2]\)
\(=\sqrt{(5+2\sqrt{6}})\)
\(=\sqrt{(5+\sqrt{24}})=3=\sqrt{9}=\sqrt{(5+\sqrt{16})}\)
\(=\sqrt{(5+24)}>\sqrt{(5+16)}\Rightarrow\sqrt{2+\sqrt{3}}>3\)
\(c)\sqrt{11}-\sqrt{3}\)và 2
\(=\sqrt{11}-\sqrt{3}=\sqrt{[(\sqrt{11}-\sqrt{3}})^2=\sqrt{(14-2\sqrt{33})}\); \(2=\sqrt{4}=\sqrt{(14-10)}=\sqrt{(14-2\sqrt{25})}\Rightarrow\sqrt{(14-2\sqrt{33})}< \sqrt{(14-2\sqrt{25})}\)
\(\Rightarrow\sqrt{11}-\sqrt{3}< 2\)
Chúc bạn học tốt~
a) \(6+2\sqrt{2}=6+\sqrt{2^2.2}=6+\sqrt{8}\)
\(9=6+3=6+\sqrt{9}\)
Ta có: \(\sqrt{9}>\sqrt{8}\)
\(\Rightarrow6+\sqrt{3}>6+\sqrt{8}\)
\(\Rightarrow9>6+2\sqrt{2}\)
b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=2+2.\sqrt{2}.\sqrt{3}+3=5+2.\sqrt{6}=5+\sqrt{2^2.6}=5+\sqrt{24}\)
\(3^2=9=5+4=5+\sqrt{16}\)
Ta có: \(\sqrt{24}>\sqrt{16}\)
\(\Rightarrow5+\sqrt{24}>5+\sqrt{16}\)
\(\Rightarrow\left(\sqrt{2}+\sqrt{3}\right)^2>3^2\)
\(\Rightarrow\sqrt{2}+\sqrt{3}>3\)
c) làm tương tự như câu c
mk ms học lớp 7 nên có gì sai sót thì bỏ qua nha
a) \(2\sqrt{3}=\sqrt{4}.\sqrt{3}=\sqrt{12}< \sqrt{18}=\sqrt{9}.\sqrt{2}=3\sqrt{2}\)
b) \(6\sqrt{5}=\sqrt{36}.\sqrt{5}=\sqrt{36.5}=\sqrt{180}>\sqrt{150}=\sqrt{25}.\sqrt{6}=5\sqrt{6}\)
a) 2√3=√4.√3=√12<√18=√9.√2=3√2
b) 6√5=√36.√5=√36.5=√180>√150=√25.√6=5√6
a: \(2\sqrt[3]{3}=\sqrt[3]{3\cdot2^3}=\sqrt[3]{24}< \sqrt[3]{54}=3\sqrt[3]{2}\)
b: \(24=2\cdot12=2\cdot\sqrt[3]{1728}< 2\cdot\sqrt[3]{1730}\)