\(\sqrt{3}-\sqrt{27}-\sqrt{144}:\sqrt{36}\)
\(\sqrt{3}×\sqrt{27}-\sqrt{144}:\sqrt{36}\)
\(\left(2\sqrt{9}+3\sqrt{36}\right):4\)
\(\sqrt{7}-\sqrt{8-2\sqrt{7}}\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
\(\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)
\(\sqrt{27}+5\sqrt{12}-2\sqrt{3}=11\sqrt{3}\)
\(\sqrt{3\cdot27}-\sqrt{\dfrac{144}{36}}\)=\(\sqrt{81}-\sqrt{4}\)=9-2=7
\(\dfrac{2\cdot3+3\cdot6}{4}\)=6
\(\sqrt{7}-\sqrt{7-2\cdot\sqrt{7}+1}\)=\(\sqrt{7}-\left(\sqrt{7}-1\right)\)=1
\(\dfrac{\sqrt{3-2\cdot\sqrt{3}+1}}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}\)=\(\dfrac{\sqrt{3}-1}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}\)=\(\dfrac{1}{\sqrt{2}}\)
\(\dfrac{\sqrt{5}\cdot\left(\sqrt{5}+3\right)}{\sqrt{5}}\)+\(\dfrac{\sqrt{3}\cdot\left(1+\sqrt{3}\right)}{\sqrt{3}+1}\)-(\(\sqrt{5}+3\))
=(\(\sqrt{5}+3\))+\(\sqrt{3}\)-(\(\sqrt{5}+3\))=\(\sqrt{3}\)
\(\sqrt{3}\cdot\sqrt{9}+5\cdot\sqrt{4}\cdot3-2\sqrt{3}\)
=\(\sqrt{3}\cdot\left(3+10-2\right)\)
=\(11\sqrt{3}\)
rút gọn và tính biểu thức sau
a, \(3\sqrt{144}-5\sqrt{49}+\dfrac{1}{2}\sqrt{36}\)
\(3\sqrt{144}-5\sqrt{49}+\dfrac{1}{2}\sqrt{36}\)
\(=3.12-5.7+\dfrac{1}{2}.6\)
\(=36-35+3=4\)
D = \(\frac{\sqrt{3}+\sqrt{4}+\sqrt{5}}{\sqrt{6}+\sqrt{8}+\sqrt{10}+\sqrt{27}+\sqrt{36}+\sqrt{45}}\)
=\(\frac{\sqrt{3}+\sqrt{4}+\sqrt{5}}{\sqrt{2}\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)+3\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)}\)
=\(\frac{\sqrt{3}+\sqrt{4}+\sqrt{5}}{\left(\sqrt{2}+3\right)\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)}=\frac{1}{\sqrt{2}+3}\)=\(\frac{3-\sqrt{2}}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}=\frac{3-\sqrt{2}}{7}\)
Tính
\(\dfrac{\sqrt{3}+\sqrt{4}+\sqrt{5}}{\sqrt{6}+\sqrt{8}+\sqrt{10}+\sqrt{27}+\sqrt{36}+\sqrt{45}}\)
\(\dfrac{\sqrt{3}+\sqrt{4}+\sqrt{5}}{\sqrt{6}+\sqrt{8}+\sqrt{10}+\sqrt{27}+\sqrt{36}+\sqrt{45}}\)
\(=\dfrac{\sqrt{3}+\sqrt{4}+\sqrt{5}}{\sqrt{3}.\sqrt{2}+\sqrt{4}.\sqrt{2}+\sqrt{5}.\sqrt{2}+3\sqrt{3}+3\sqrt{4}+3\sqrt{5}}\)
\(=\dfrac{\sqrt{3}+\sqrt{4}+\sqrt{5}}{\sqrt{2}\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)+3\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{4}+\sqrt{5}}{\left(\sqrt{2}+3\right)\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)}\)
\(=\dfrac{1}{\sqrt{2}+3}\)
rút gọn biểu thức
A=2015+\(\sqrt{36}\)-\(\sqrt{25}\)
B=5\(\sqrt{8}\)+\(\sqrt{50}\)-2\(\sqrt{18}\)
C=\(\sqrt{27}\)-2\(\sqrt{12}\)-\(\sqrt{75}\)
D=\(\sqrt{12}\)+\(\sqrt{27}\)-\(\sqrt{48}\)
a: =2015+6-5=2016
b: =10căn 2+5căn 2-6căn 2=9căn 2
c: =3căn 3-4căn 3-5căn 3=-6căn 3
d: =2căn 3+3căn 3-4căn 3=căn 3
\(A=2015+6-5==2015+1=2016\)
\(B=5\sqrt{2^3}+\sqrt{5^2.2}-2\sqrt{3^2.2}\\ =10\sqrt{2}+5\sqrt{2}-6\sqrt{2}\\ =\left(10+5-6\right)\sqrt{2}=9\sqrt{2}\)
\(C=\sqrt{3^3}-2\sqrt{2^2.3}-\sqrt{5^2.3}\\ =3\sqrt{3}-4\sqrt{3}-5\sqrt{3}\\ =\left(3-4-5\right)\sqrt{3}=-6\sqrt{3}\)
\(D=\sqrt{2^2.3}+\sqrt{3^3}-\sqrt{4^2.3}\\ =2\sqrt{3}+3\sqrt{3}-4\sqrt{3}\\ =\left(2+3-4\right)\sqrt{3}=\sqrt{3}\)
Rút gọn:
a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}\)
b)\(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}\)
c)\(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}\)
d)\(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}\)
e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}\)
f)\(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}\dfrac{2}{\sqrt{2}+1}\)
a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}=14\sqrt{2}-9\sqrt{2}+2\sqrt{2}=7\sqrt{2}\)
b) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)
c) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=5\sqrt{6}\)
d) \(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}=12\sqrt{3}+6\sqrt{3}-3\sqrt{3}=15\sqrt{3}\)
e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=2\sqrt{3}+3\sqrt{3}=\left(\sqrt{3}+1\right)=4\sqrt{3}-1\)
f) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)
tinh ket qua \(\sqrt{36}-\sqrt{144}+\sqrt{64}\)
Tính:
1) \(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\)
2) \(\frac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\)
3) \(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}\)
4) \(\sqrt{3}\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right)\)
5) \(\sqrt{144}-\sqrt{25}.\sqrt{4}\)
6) \(\frac{1}{\sqrt{3}-1}-\sqrt{3}+1\)
\(1,\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\)
\(=\left(\sqrt{9.5}\sqrt{4.5}+\sqrt{5}\right).\frac{1}{\sqrt{6}}\)
\(=\frac{2\sqrt{5}}{\sqrt{6}}\)
\(=\frac{\sqrt{30}}{3}\)
1) \(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\)
\(=\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}\)
\(=\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\)
\(=\frac{2\sqrt{5}}{\sqrt{6}}\)
\(=\frac{2\sqrt{5}\sqrt{6}}{\sqrt{6}.\sqrt{6}}\)
\(=\frac{2\sqrt{30}}{6}\)
\(=\frac{\sqrt{30}}{3}\)
1. Áp dụng quy tắc khai phương một thương, hãy tính:
a, \(\sqrt{\dfrac{36}{121}}\) b, \(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}\) c, \(\sqrt{0,0169}\)
d,\(\dfrac{\sqrt{15}}{\sqrt{735}}\) e, \(\sqrt{\dfrac{81}{8}:\sqrt{3\dfrac{1}{8}}}\) g, \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
2. Tính:
a,\(\sqrt{\dfrac{25}{144}}\) b,\(\sqrt{2\dfrac{7}{81}}\) c,\(\sqrt{\dfrac{2,25}{16}}\) d, \(\sqrt{\dfrac{1,21}{0,49}}\)
3. Áp dụng quy tắc chia hai căn bậc hai, hãy tính:
a, \(\sqrt{18}:\sqrt{2}\) b, \(\sqrt{45}:\sqrt{80}\)
c, (\(\sqrt{20}-\sqrt{45}+\sqrt{5}\) ) : \(\sqrt{5}\) d, \(\dfrac{\sqrt{8^2}}{\sqrt{4^5.2^3}}\)
4. Khẳng định nào sau đây là đúng?
A. \(\sqrt{\dfrac{3}{\left(-5\right)^2}}=-\dfrac{\sqrt{3}}{5}\) B. \(\left(\sqrt{\dfrac{-3}{-5}}\right)^2=\dfrac{3}{5}\)
5. Tính.
a, \(\sqrt{2\dfrac{7}{81}}:\dfrac{\sqrt{6}}{\sqrt{150}}\) b, \(\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right):\sqrt{3}\)
c, \(\left(\sqrt{\dfrac{1}{5}-\sqrt{\dfrac{9}{5}}+\sqrt{5}}\right):\sqrt{5}\) d, \(\sqrt{\dfrac{2+\sqrt{3}}{\sqrt{2}}}\)
6. So sánh
a, So sánh \(\sqrt{144-49}\) và \(\sqrt{144}-\sqrt{49}\);
b, Chứng minh rằng , với hai số a,b thỏa mãn a> b> 0 thì \(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
1
a,\(\sqrt{\dfrac{36}{121}}=\sqrt{\dfrac{6^2}{11^2}}=\dfrac{6}{11}\)
\(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}=\sqrt{\dfrac{81}{100}}=\sqrt{\dfrac{9^2}{10^2}}=\dfrac{9}{10}\)
Bài 2:
a: \(\sqrt{\dfrac{25}{144}}=\dfrac{5}{12}\)
b: \(\sqrt{2+\dfrac{7}{81}}=\sqrt{\dfrac{169}{81}}=\dfrac{13}{9}\)
c: \(\sqrt{\dfrac{2.25}{16}}=\dfrac{1.5}{4}=\dfrac{3}{8}\)
d: \(\sqrt{\dfrac{1.21}{0.49}}=\sqrt{\dfrac{121}{49}}=\dfrac{11}{7}\)
Bài3:
a: \(=\sqrt{\dfrac{18}{2}}=\sqrt{9}=3\)
b: \(=\sqrt{\dfrac{45}{80}}=\sqrt{\dfrac{9}{16}}=\dfrac{3}{4}\)
c: \(=\dfrac{2\sqrt{5}-3\sqrt{5}+\sqrt{5}}{\sqrt{5}}=0\)
d: \(=\sqrt{\dfrac{2^6}{2^{10}\cdot2^3}}=\sqrt{\dfrac{1}{2^7}}=\dfrac{1}{8\sqrt{2}}=\dfrac{\sqrt{2}}{16}\)