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\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)
Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)
Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)
\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)
Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)
Nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
Do \(\sqrt{1}=1;\sqrt{2}+\sqrt{3}+\sqrt{4}< 3.\sqrt{4}=6\)\(;\sqrt{5}+\sqrt{6}+...+\sqrt{9}< 5.\sqrt{9}=15\)
\(\Rightarrow\sqrt{1}+\sqrt{2}+...+\sqrt{9}< 1+6+15=22\)(1)
Cung co:\(5.\sqrt{5}>5.\sqrt{4}=10\)\(\Rightarrow5.\sqrt{5}+12>10+12=22\)(2)
Tu (1) va (2) =>....
a: \(\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\)
\(\left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)
mà \(-2\sqrt{105}>-2\sqrt{120}\)
nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
b: \(\left(\sqrt{2}+\sqrt{8}\right)^2=10+2\cdot4=16=12+4\)
\(\left(3+\sqrt{3}\right)^2=12+6\sqrt{3}\)
mà \(4< 6\sqrt{3}\)
nên \(\sqrt{2}+\sqrt{8}< 3+\sqrt{3}\)
Cách 1:
a, Ta có: \(2\sqrt{3}=\sqrt{4.3}=\sqrt{12}\); \(3\sqrt{2}=\sqrt{9.2}=\sqrt{18}\)
Vì 18 > 12 \(\Rightarrow\sqrt{18}>\sqrt{12}\)\(\Rightarrow3\sqrt{2}>2\sqrt{3}\)
b, Ta có: \(4\sqrt{3}=\sqrt{16.3}=\sqrt{48}\); \(3\sqrt{4}=\sqrt{9.4}=\sqrt{36}\)
Vì 48 > 36 \(\Rightarrow\sqrt{48}>\sqrt{36}\)\(\Rightarrow4\sqrt{3}>3\sqrt{4}\)
Cách 2:
Đặt \(A=2\sqrt{3}\)\(\Rightarrow A^2=\left(2\sqrt{3}\right)^2=4.3=12\)
\(B=3\sqrt{2}\)\(\Rightarrow B^2=\left(3\sqrt{2}\right)^2=9.2=18\)
Vì 12 < 18 => A2 < B2 => A < B
b, Đặt \(A=4\sqrt{3}\)\(\Rightarrow A^2=\left(4\sqrt{3}\right)^2=16.3=48\)
\(B=3\sqrt{4}\)\(\Rightarrow B^2=\left(3\sqrt{4}\right)^2=9.4=36\)
Vì 48 > 36 => A2 > B2 => A > B
Bình Tất cả lên
a) 2sqrt(3) < 3sqrt(2)