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a) \(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)
b) \(3\sqrt{5}=\sqrt{45}>\sqrt{27}\)
c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{9}}< \sqrt{\dfrac{54}{9}}=6=\sqrt{\dfrac{150}{25}}=\dfrac{1}{5}\sqrt{150}\)
d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{4}}=\sqrt{\dfrac{3}{2}}< \sqrt{\dfrac{36}{2}}=6\sqrt{\dfrac{1}{2}}\)
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 \(3\sqrt{12}>2\sqrt{16}\)
a) Ta có:
\(2\sqrt{3}=\sqrt{2^2.3}=\sqrt{12}.\)
Mà \(\sqrt{12}< \sqrt{13}\)
Nên \(2\sqrt{3}< \sqrt{13}\)
a: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+\dfrac{23}{9}\sqrt{7}+\dfrac{16}{9}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
b:\(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}+\dfrac{5}{6}\sqrt{6}\)
\(=2\sqrt{6}+\dfrac{5}{6}\sqrt{6}=\dfrac{17}{6}\sqrt{6}\)
c: \(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\sqrt{\dfrac{5-2\sqrt{6}}{12}}\)
\(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)
a: \(E=1+1=2\)
b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)
\(=6+2\sqrt{6}-\sqrt{2}+2\sqrt{5}\)
d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)
Câu 1:
a: \(\dfrac{2}{5}\sqrt{75}-0,5\cdot\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\cdot\sqrt{12}\)
\(=\dfrac{2}{5}\cdot5\sqrt{3}-0,5\cdot4\sqrt{3}+10\sqrt{3}-\dfrac{2}{3}\cdot2\sqrt{3}\)
\(=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}\)
\(=10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)
b: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
\(=\dfrac{\sqrt{3}\cdot3\sqrt{3}-2\sqrt{3}}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{9-6}\)
\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+3-\sqrt{6}\)
\(=\dfrac{\sqrt{3}}{\sqrt{2}}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)
c: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
=\(\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
Bài 2:
a: 
b: Phương trình hoành độ giao điểm là:
\(3x+2=-x-4\)
=>4x=-6
=>x=-3/2
Thay x=-3/2 vào y=-x-4, ta được:
\(y=-\left(-\dfrac{3}{2}\right)-4=\dfrac{3}{2}-4=-\dfrac{5}{2}\)
Vậy: \(A\left(-\dfrac{3}{2};-\dfrac{5}{2}\right)\)
c: Vì (d2)//(d) nên \(\left\{{}\begin{matrix}a=-1\\b\ne-4\end{matrix}\right.\)
Vậy: (d2): y=-x+b
Thay x=-2 và y=5 vào (d2), ta được:
\(b-\left(-2\right)=5\)
=>b+2=5
=>b=5-2=3
Vậy: (d2): y=-x+3
\(a,=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\left(\sqrt{7}+2\right)}{3}-\dfrac{5\left(4-\sqrt{7}\right)}{9}\\ =\dfrac{\sqrt{7}-5-3+\sqrt{7}}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{2\sqrt{7}-8}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\sqrt{7}-4+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{27\sqrt{7}-20+5\sqrt{7}}{9}=\dfrac{32\sqrt{7}-20}{9}\)
\(b,=\dfrac{2\left(\sqrt{6}+2\right)}{2}+\dfrac{2\left(\sqrt{6}-2\right)}{2}+\dfrac{5\sqrt{6}}{6}\\ =\sqrt{6}+2+\sqrt{6}-2+\dfrac{5\sqrt{6}}{6}\\ =\dfrac{12\sqrt{6}+5\sqrt{6}}{6}=\dfrac{17\sqrt{6}}{6}\)
\(c,=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}-\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}\\ =\dfrac{2\sqrt{5}}{5+2\sqrt{6}-5}=\dfrac{2\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{30}}{6}\)
a: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
b: \(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}+\dfrac{5\sqrt{6}}{6}\)
\(=2\sqrt{6}+\dfrac{5}{6}\sqrt{6}=\dfrac{17}{6}\sqrt{6}\)
a)
Có:
\(2\sqrt{29}=\sqrt{4.29}=\sqrt{116}\\ 3\sqrt{13}=\sqrt{9.13}=\sqrt{117}\)
Vì \(\sqrt{117}>\sqrt{116}\) nên \(3\sqrt{13}>2\sqrt{29}\)
b)
Có:
\(\dfrac{5}{4}\sqrt{2}=\sqrt{\dfrac{25}{16}.2}=\sqrt{\dfrac{25}{8}}\)
\(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{9}{4}.\dfrac{3}{2}}=\sqrt{\dfrac{27}{8}}\)
Do \(\sqrt{\dfrac{27}{8}}>\sqrt{\dfrac{25}{8}}\) nên \(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}>\dfrac{5}{4}\sqrt{2}\)
c)
Có:
\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)
\(4\sqrt{3}=\sqrt{16.3}=\sqrt{48}\)
Vì \(\sqrt{50}>\sqrt{48}\) nên \(5\sqrt{2}>4\sqrt{3}\)
d)
Có:
\(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\dfrac{25}{4}.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}\)
\(6\sqrt{\dfrac{1}{37}}=\sqrt{36.\dfrac{1}{37}}=\sqrt{\dfrac{36}{37}}\)
lại có: \(\dfrac{25}{24}>\dfrac{36}{37}\)
\(\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}\)