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b: \(\sqrt{3}-1=\sqrt{4-2\sqrt{3}}\)
mà \(4-3\sqrt{3}< 4-2\sqrt{3}\)
nên \(\sqrt{4-3\sqrt{3}}< \sqrt{3}-1\)
Đề này sai rồi bạn vì \(4-3\sqrt{3}< 0\)
a) \(\left(-\dfrac{1}{3}\sqrt{63}\right)^2=\dfrac{1}{9}\cdot63=7\)
\(\left(-2\sqrt{2}\right)^2=8\)
mà 7<8
nên \(-\dfrac{1}{3}\sqrt{63}>-2\sqrt{2}\)
b) Ta có: \(\left(2\sqrt{55}\right)^2=4\cdot55=220\)
\(\left(\dfrac{3}{5}\sqrt{750}\right)=\dfrac{9}{25}\cdot750=270\)
mà 220<270
nên \(2\sqrt{55}< \dfrac{3}{5}\sqrt{750}\)
hay \(-2\sqrt{55}< -\dfrac{3}{5}\sqrt{750}\)
a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)
mà 112<117
nên \(4\sqrt{7}< 3\sqrt{13}\)
b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
mà \(\dfrac{21}{4}>\dfrac{36}{7}\)
nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)
d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)
\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)
a)
\(7\sqrt{2}=\sqrt{49.2}=\sqrt{98}\\ 2\sqrt{8}=\sqrt{4.8}=\sqrt{32}\\ 5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)
Do 98 > 50 > 32 > 28 nên \(\sqrt{98}>\sqrt{50}>\sqrt{32}>\sqrt{28}\)
=> \(7\sqrt{2}>5\sqrt{2}>2\sqrt{8}>\sqrt{28}\)
b)
\(3\sqrt{10}=\sqrt{9.10}=\sqrt{90}\\ 5\sqrt{3}=\sqrt{25.3}=\sqrt{75}\)
\(\dfrac{20}{\sqrt{5}}=\dfrac{20\sqrt{5}}{5}=4\sqrt{5}=\sqrt{16.5}=\sqrt{80}\)
\(12\sqrt{\dfrac{2}{3}}=\sqrt{144.\dfrac{2}{3}}=\sqrt{96}\)
Do 96 > 90 > 80 > 75 => \(\sqrt{96}>\sqrt{90}>\sqrt{80}>\sqrt{75}\)
=> \(12\sqrt{\dfrac{2}{3}}>3\sqrt{10}>\dfrac{20}{\sqrt{5}}>5\sqrt{3}\)
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) \(=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) \(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}=2+\sqrt{3}-1-\sqrt{3}=1\)
c) \(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}+1+\sqrt{7}-1=2\sqrt{7}\)
d) \(=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{2}+1=\sqrt{5}+1\)
c.
(\sqrt{5}-\sqrt{3})-(\sqrt{10}-\sqrt{7})=(\sqrt{5}+\sqrt{7})-(\sqrt{3}+\sqrt{10})
Mà:
\((\sqrt{5}+\sqrt{7})^2=12+\sqrt{35}< 12+\sqrt{36}=18\)
\((\sqrt{3}+\sqrt{10})^2=13+\sqrt{30}>13+\sqrt{25}=18\)
\(\Rightarrow \sqrt{3}+\sqrt{10}> \sqrt{5}+\sqrt{7}\Rightarrow \sqrt{5}-\sqrt{3}< \sqrt{10}-\sqrt{7}\)
Lời giải:
a.
$5+\sqrt{2}>5+\sqrt{1}=6$
$4+\sqrt{3}< 4+\sqrt{4}=6$
$\Rightarrow 5+\sqrt{2}>4+\sqrt{3}$
b.
$\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$
$\sqrt{5}-\sqrt{3}=\frac{5-3}{\sqrt{5}+\sqrt{3}}=\frac{2}{\sqrt{5}+\sqrt{3}}< \frac{2}{\sqrt{2}}=\sqrt{2}$
Vậy $\sqrt{8}-\sqrt{2}>\sqrt{5}-\sqrt{2}$