đề bài là gì ạ
so sánh hay gì ạ
....
a) Ta có:
\(2\sqrt{3}=\sqrt{2^2.3}=\sqrt{12}.\)
Mà \(\sqrt{12}< \sqrt{13}\)
Nên \(2\sqrt{3}< \sqrt{13}\)
b) Ta có:
\(7=\sqrt{49}\)
\(3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}\)
Mà \(\sqrt{45}< \sqrt{49}\)
Nên \(7>3\sqrt{5}\)
c) Ta có:
\(\frac{1}{3}\sqrt{51}=\sqrt{\frac{51}{3^2}}=\sqrt{\frac{17}{3}}.\)
\(\frac{1}{5}\sqrt{150}=\sqrt{\frac{150}{5^2}}=\sqrt{6}\)
Mà \(\sqrt{6}>\sqrt{\frac{17}{3}}\)
Nên \(\Rightarrow\frac{1}{5}\sqrt{150}>\frac{1}{3}\sqrt{51}\)
d) Ta có:
\(\frac{1}{2}\sqrt{6}=\sqrt{\frac{6}{2^2}}\sqrt{\frac{3}{2}}.\)
\(6\sqrt{\frac{1}{2}}=\sqrt{\frac{1.36}{2}}=\sqrt{18}.\)
Mà \(\sqrt{\frac{3}{2}}< \sqrt{18}\)
Nên \(\Rightarrow\frac{1}{2}\sqrt{6}< 6\sqrt{\frac{1}{2}}\)
a) \(2\sqrt{3}=\sqrt{4}.\sqrt{3}=\sqrt{4.3}=\sqrt{12}\) mà \(\sqrt{12}< \sqrt{13}\)\(\Rightarrow2\sqrt{3}< \sqrt{13}\) b) \(3\sqrt{5}=\sqrt{9}.\sqrt{5}=\sqrt{9.5}=\sqrt{45}=6,7082\) mà 6,7082 < 7 nên \(7>3\sqrt{5}\) c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{1}{9}}.\sqrt{51}=\sqrt{\dfrac{1}{9}.51}=\dfrac{\sqrt{51}}{3}\), \(\dfrac{1}{5}\sqrt{150}=\sqrt{\dfrac{1}{25}}.\sqrt{150}=\sqrt{\dfrac{1}{25}.150}=\sqrt{6}\) mà \(\dfrac{\sqrt{51}}{3}< \sqrt{6}\) \(\Rightarrow\)\(\dfrac{1}{3}\sqrt{51}< \dfrac{1}{5}\sqrt{150}\) d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{1}{4}}.\sqrt{6}=\sqrt{\dfrac{1}{4}.6}=\dfrac{\sqrt{6}}{2}=1,2247\),\(6\sqrt{\dfrac{1}{2}}=\sqrt{36}.\sqrt{\dfrac{1}{2}}=\sqrt{36.\dfrac{1}{2}}=3\sqrt{2}=4,2426\) mà 1,2247 < 4,2426 nên\(\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
a,2 căn 3 < căn 13
b, 7> 3 căn 5
c,1/3 căn 51 < 1/5 căn 150
d, 1/2 căn 6 < 6 căn 1/2
a) 2√3<√13
b) 7>3√5
c) 1/3√51<1/5√150
d) 1/2 √6 <6√1/2
a) √12<√13
b) 3√5<7
c) 1/5*√150>1/3*√51
d) 1/2*√6<6*√(1/2)
a) 2căn 3<căn 13
b) 7>3căn 5
c) 1/3căn 51<1/5căn 150
d) 1/2căn 6<6căn 1/2
a) Ta có:
Mà
Nên
b) Ta có:
Mà
Nên
c) Ta có:
Mà
Nên
d) Ta có:
Mà
Nên
a , Ta có : \(2\sqrt{3}=\sqrt{2^2.3}=\sqrt{12}< \sqrt{13}\)
b, Vì \(7=\sqrt{49}\) và \(3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}< \sqrt{49}\)
Vậy \(3\sqrt{5}< 7\)
c, Vì \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{3^2}=\sqrt{\dfrac{17}{3}}}\) và \(\dfrac{1}{5}\sqrt{150}=\sqrt{\dfrac{150}{5^2}=\sqrt{6}}\) nên \(\sqrt{6}>\sqrt{\dfrac{17}{3}}\)
=> \(\dfrac{1}{5}\sqrt{150}>\dfrac{1}{3}\sqrt{51}\)
d, Vì \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{2^2}}=\sqrt{\dfrac{3}{2}}\) và \(6\sqrt{\dfrac{1}{2}}=\sqrt{\dfrac{1.36}{2}}=\sqrt{18}nên\sqrt{18}>\sqrt{\dfrac{3}{2}}\)
=>\(\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
A-2V3 <V13
B- 7> 3V5
C- 1/3V51<1/5 V150
D- 1/2 V6 < 6V1/2
a. \(2\sqrt{3}< \sqrt{13}\)
b. \(7>3\sqrt{5}\)
c. \(\dfrac{1}{3}\sqrt{51}< \dfrac{1}{3}\sqrt{150}\)
d. \(\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
a) \(2\sqrt{3}\) và \(\sqrt{13}\)
\(2\sqrt{3}\) = \(\sqrt{2^2.3}\) = \(\sqrt{4.3}\) = \(\sqrt{12}\)
Vì \(\sqrt{12}\) < \(\sqrt{13}\) nên \(2\sqrt{3}\) < \(\sqrt{13}\)
b) 7 và \(3\sqrt{5}\)
7 = \(\sqrt{49}\)
\(3\sqrt{5}\) = \(\sqrt{3^2.5}\) = \(\sqrt{45}\)
Vì \(\sqrt{49}\) > \(\sqrt{45}\) nên 7 > \(3\sqrt{5}\)
c) \(\dfrac{1}{3}\sqrt{51}\) và \(\dfrac{1}{5}\sqrt{150}\)
\(\dfrac{1}{3}\sqrt{51}\) = \(\sqrt{\dfrac{51}{3^2}}\) = \(\sqrt{\dfrac{17}{3}}\)
\(\dfrac{1}{5}\sqrt{150}\) = \(\sqrt{\dfrac{150}{5^2}}\) = \(\sqrt{6}\)
Vì \(\sqrt{6}\) > \(\sqrt{\dfrac{17}{3}}\) nên \(\dfrac{1}{3}\sqrt{51}\) < \(\dfrac{1}{3}\sqrt{150}\)
d) \(\dfrac{1}{2}\sqrt{6}\) và \(6\sqrt{\dfrac{1}{2}}\)
\(\dfrac{1}{2}\sqrt{6}\) = \(\sqrt{\dfrac{6}{2^2}}\) = \(\sqrt{\dfrac{3}{2}}\)
\(6\sqrt{\dfrac{1}{2}}\) = \(\sqrt{\dfrac{1.36}{2}}\) = \(\sqrt{18}\)
Vì \(\sqrt{18}\) > \(\sqrt{\dfrac{3}{2}}\) nên \(\dfrac{1}{2}\sqrt{6}\) > \(6\sqrt{\dfrac{1}{2}}\)
a) ta có \(2\sqrt{3}=\sqrt{2^2.3}=\sqrt{12}< \sqrt{13}\)
b) vì \(7=\sqrt{49}\) và \(3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}< \sqrt{49}\) vậy \(3\sqrt{5}< 7\)
c) vì \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{3^2}}=\sqrt{\dfrac{17}{3}}\) và \(\dfrac{1}{5}\sqrt{150}=\sqrt{\dfrac{150}{5^2}}=\sqrt{6}\)
nên \(\sqrt{6}>\sqrt{\dfrac{17}{3}}\Rightarrow\dfrac{1}{5}\sqrt{51}\)
d) vì \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{2^2}}=\sqrt{\dfrac{3}{2}}\) và \(6\sqrt{\dfrac{1}{2}}=\sqrt{\dfrac{1.36}{2}}=\sqrt{18}\)
nên \(\sqrt{18}>\sqrt{\dfrac{3}{2}}\Rightarrow\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
a) Ta có:\(2\sqrt{3}=\sqrt{2^2.3}=\sqrt{12}\)
Mà:\(\sqrt{12}< \sqrt{13}\Rightarrow2\sqrt{3}< \sqrt{13}\)
b)Ta có:\(3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}\)
\(7=\sqrt{49}\)
Mà:\(\sqrt{49}>\sqrt{45}\)
\(\Rightarrow7>3\sqrt{5}\)
c,Ta có:
\(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{3^2}}=\sqrt{\dfrac{17}{3}}\)
\(\dfrac{1}{5}\sqrt{150}=\sqrt{\dfrac{150}{5^2}}=\sqrt{6}\)
Mà:\(\sqrt{6}>\sqrt{\dfrac{17}{3}}\)
Nên:\(\dfrac{1}{3}\sqrt{51}< \dfrac{1}{5}\sqrt{150}\)
d)Ta có:
\(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{2^2}}=\sqrt{\dfrac{3}{2}}\)
\(6\sqrt{\dfrac{1}{2}}=\sqrt{\dfrac{36}{2}}=\sqrt{18}\)
Mà:\(\sqrt{\dfrac{3}{2}}< \sqrt{18}\)
Nên \(\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
a,2√3 và √13
2√3 < √13
b,7 và 3√5
7 > 3√5
c, (1/3)√51 và (1/5)√150
(1/3)√51 < (1/5)√150
d,(1/2)√6 và 6(√(1/2)
(1/2)√6 < 6(√(1/2)
a, Ta có 2 căn 3 = căn 2^2.3= căn 12<căn 13
b, Vì 7 =căn 49 và 3 căn 5 = căn 3^2.5 =căn 45< căn 49
Vậy 3 căn 5<7
c, Vì 1/3 căn 51= căn 51/3^2= căn 17/3 và 1/5 căn 150
= căn 150/5^2 = căn 6 nên căn 6 >căn 17/3
=> 1/5căn 150>1/3 căn 51
d, Vì 1/2 căn 6 = căn 6/2^2= căn 3/2 và 6 căn 1/2= căn 1.36/2=căn 18
nên căn 18 > căn 3/2=> 1/2 căn 6< 6 căn 1/2
a) 2\(\sqrt{3}\) < \(\sqrt{13}\)
b) 7 > 3\(\sqrt{5}\)
c) \(\dfrac{1}{3}\sqrt{51}\) < \(\dfrac{1}{5}\sqrt{150}\)
d) \(\dfrac{1}{2}\sqrt{6}\) < 6\(\sqrt{\dfrac{1}{2}}\)