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b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)
\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)
mà 80>75
nên \(4\sqrt{5}>5\sqrt{3}\)
a) \(1=\sqrt{1}< \sqrt{2}\)
b) \(2=\sqrt{4}>\sqrt{3}\)
c) \(6=\sqrt{36}< \sqrt{41}\)
d) \(7=\sqrt{49}>\sqrt{47}\)
e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)
f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)
g) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\)
h) \(\sqrt{3}>0>-\sqrt{12}\)
i) \(5=\sqrt{25}< \sqrt{29}\)
\(\Rightarrow-5>-\sqrt{29}\)
a: \(1< \sqrt{2}\)
nên \(2< \sqrt{2}+1\)
b: \(2\sqrt{31}=\sqrt{124}\)
\(10=\sqrt{100}\)
mà 124>100
nên \(2\sqrt{31}>10\)
c: \(-3\sqrt{11}=-\sqrt{99}\)
\(-\sqrt{12}=-\sqrt{12}\)
mà 99>12
nên \(-3\sqrt{11}< -\sqrt{12}\)
Bài 1:
Để M có nghĩa thì \(\left\{{}\begin{matrix}x+4\ge0\\2-x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\le2\end{matrix}\right.\Leftrightarrow-4\le x\le2\)
Số giá trị nguyên thỏa mãn điều kiện là:
\(\left(2+4\right)+1=7\)
Đặt:
\(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)\)
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\right)\)
\(A=\dfrac{1}{\sqrt{2}}\left(\left|1+\sqrt{5}\right|+\left|\sqrt{5}-1\right|\right)\)
\(A=\dfrac{1}{\sqrt{2}}\left(1+\sqrt{5}+\sqrt{5}-1\right)\)
\(A=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
Ta có: \(A^2=\left(\sqrt{10}\right)^2=10\)
\(B=\left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)
Mà: \(4\sqrt{5}>1\)
Nên: \(A^2< B^2\)
\(\Rightarrow A< B\)
Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5}+1+\sqrt{5}-1\right)=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
=>A^2=(căn 10)^2=10=9+1
Đặt B=2+căn 5
=>B^2=(2+căn 5)^2=9+4căn 5
1<4căn 5
=>9+1<9+4căn 5
=>A^2<B^2
=>A<B
a) Ta có:
\(2=1+1=1+\sqrt{1}\)
Mà: \(1< 2\Rightarrow\sqrt{1}< \sqrt{2}\)
\(\Rightarrow1+\sqrt{1}< \sqrt{2}+1\)
\(\Rightarrow2< \sqrt{2}+1\)
b) Ta có:
\(1=2-1=\sqrt{4}-1\)
Mà: \(4>3\Rightarrow\sqrt{4}>\sqrt{3}\)
\(\Rightarrow\sqrt{4}-1>\sqrt{3}-1\)
\(\Rightarrow1>\sqrt{3}-1\)
c) Ta có:
\(10=2\cdot5=2\sqrt{25}\)
Mà: \(25< 31\Rightarrow\sqrt{25}< \sqrt{31}\)
\(\Rightarrow2\sqrt{25}< 2\sqrt{31}\)
\(\Rightarrow10< 2\sqrt{31}\)
d) Ta có:
\(-12=-3\cdot4=-3\sqrt{16}\)
Mà: \(16>11\Rightarrow\sqrt{16}>\sqrt{11}\)
\(\Rightarrow-3\sqrt{16}< -3\sqrt{11}\)
\(\Rightarrow-12< -3\sqrt{11}\)
\(A=\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{5}+1-\sqrt{5}=1\)
\(B=\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)
Do đó: A=B
\(\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}=\left|\sqrt{5}+1\right|-\sqrt{5}=1\)
\(\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}\right)^3+1^3+3.2+3\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)
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