1. \(\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}\) .
2. \(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}.}\)
3. \(\sqrt{11}-\sqrt{20-6\sqrt{11}}=3\)
4.\(\sqrt{41+12\sqrt{5}}-\sqrt{41-12\sqrt{5}}=2\sqrt{5.}\)
Những câu hỏi liên quan
Rút gọn biểu thức:
1) sqrt{12}+5sqrt{3}-sqrt{48}
2) 5sqrt{5}+sqrt{20}-3sqrt{45}
3) 2sqrt{32}+4sqrt{8}-5sqrt{18}
4) 3sqrt{12}-4sqrt{27}+5sqrt{48}
5) sqrt{12}+sqrt{75}-sqrt{27}
6) 2sqrt{18}-7sqrt{2}+sqrt{162}
7) 3sqrt{20}-2sqrt{45}+4sqrt{5}
8) left(sqrt{2}+2right)sqrt{2}-2sqrt{2}
9) dfrac{1}{sqrt{5}-1}-dfrac{1}{sqrt{5}+}
10) dfrac{1}{sqrt{5}-2}+dfrac{1}{sqrt{5}+2}
11) dfrac{2}{4-3sqrt{2}}-dfrac{2}{4+3sqrt{2}}
12) dfrac{2+sqrt{2}}{1+sqrt{2}}
13) left(sqrt{28}-2sqrt{14}+sqrt{7}right)sqr...
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Rút gọn biểu thức:
1) \(\sqrt{12}+5\sqrt{3}-\sqrt{48}\)
2) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
3) \(2\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
4) \(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
5) \(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
7) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
8) \(\left(\sqrt{2}+2\right)\sqrt{2}-2\sqrt{2}\)
9) \(\dfrac{1}{\sqrt{5}-1}-\dfrac{1}{\sqrt{5}+}\)
10) \(\dfrac{1}{\sqrt{5}-2}+\dfrac{1}{\sqrt{5}+2}\)
11) \(\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}\)
12) \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\)
13) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
14) \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)
15) \(\left(\sqrt{6}-\sqrt{5}\right)^2-\sqrt{120}\)
16) \(\left(2\sqrt{3}-3\sqrt{2}\right)^2+2\sqrt{6}+3\sqrt{24}\)
17) \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+3\right)^2}\)
18) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
19) \(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
20) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
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\(\sqrt{\left(2\sqrt{2}-3\right)^2}+2\sqrt{2}\)
\(\sqrt{\left(\sqrt{10}-3\right)^2}+\sqrt{\left(\sqrt{10}-4\right)^2}\)
\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}\)
\(\sqrt{49a^2}+3a\)
\(3\sqrt{9a^6}-6a^3\)
\(\sqrt{\left(2\sqrt{2}-3\right)^2}+2\sqrt{2}=\left|2\sqrt{2}-3\right|+2\sqrt{2}=3-2\sqrt{2}+2\sqrt{2}=3\)
\(\sqrt{\left(\sqrt{10}-3\right)^2}+\sqrt{\left(\sqrt{10}-4\right)^2}=\left|\sqrt{10}-3\right|+\left|\sqrt{10}-4\right|\)
\(=\sqrt{10}-3+4-\sqrt{10}=1\)
\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|2-\sqrt{3}\right|=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\)
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\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}=\sqrt{\left(6-\sqrt{5}\right)^2}-\sqrt{\left(6+\sqrt{5}\right)^2}\)
\(=6-\sqrt{5}-6-\sqrt{5}=-2\sqrt{5}\)
\(A=\sqrt{49a^2}+3a=7\left|a\right|+3a\)
Nếu \(a\ge0\)thì: \(A=7a+3a=10a\)
Nếu \(a< 0\)thì: \(A=-7a+3a=-4a\)
\(B=3\sqrt{9a^6}-6a^3=9\left|a^3\right|-6a^3\)
Nếu \(a\ge0\)thì: \(B=9a^3-6a^3=3a^3\)
Nếu \(a< 0\)thì: \(B=-9a^3-6a^3=-15a^3\)
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Bài 1: Rút gọn biểu thức1) sqrt{12}-sqrt{27}+sqrt{48} 2) left(sqrt{25}+sqrt{20}-sqrt{80}right):sqrt{5}3) 2sqrt{27}-sqrt{frac{16}{3}}-sqrt{48}-sqrt{8frac{1}{3}} 4) frac{1}{sqrt{5}-sqrt{3}}-frac{1}{sqrt{5}+sqrt{3}}5) left(sqrt{125}-sqrt{12}-2sqrt{5}right)left(3sqrt{5}-sqrt{3}+sqrt{27}right) 6) left(3sqrt{20}-sqrt{125}-15sqrt{frac{1}{5}}right).sqrt{5}7) left(6sqrt{128}-frac{3}{5}sqrt{50}+7sqrt{8}right):3sqrt{2} 8) left(2sqrt{48}-frac{3}{2}sqrt{frac{4}{3}}+sqrt{27}right).2sqrt{3...
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Bài 1: Rút gọn biểu thức
1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\) 2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)
3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\) 4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)
5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\) 6) \(\left(3\sqrt{20}-\sqrt{125}-15\sqrt{\frac{1}{5}}\right).\sqrt{5}\)
7) \(\left(6\sqrt{128}-\frac{3}{5}\sqrt{50}+7\sqrt{8}\right):3\sqrt{2}\) 8) \(\left(2\sqrt{48}-\frac{3}{2}\sqrt{\frac{4}{3}}+\sqrt{27}\right).2\sqrt{3}\)
9) \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}\) 10) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
11) \(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\) 12) \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
13) \(\sqrt{15-6\sqrt{6}}\) 14) \(\sqrt{8-2\sqrt{15}}\) 15) \(\sqrt[3]{-2}.\sqrt[3]{32}+\sqrt{2}.\sqrt{32}\)
Tính giá trị các biểu thức:
a.\(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\sqrt{3}\)
b.\(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
c.\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)3\sqrt{6}\)
d.\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
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a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
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d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)
\(=19\sqrt{3\sqrt{2}}\)
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Rút gọn :
1, sqrt{8}-3sqrt{32}+sqrt{72}
2, 6sqrt{12}-sqrt{20}-2sqrt{27}+sqrt{125}
3 , 3sqrt{112}-7sqrt{216}+4sqrt{54}-2sqrt{252}-3sqrt{96}
4, 3sqrt{3}left(3+2sqrt{6}-sqrt{33}right)
6, sqrt{2}left(sqrt{8}-sqrt{32}+3sqrt{18}right)
7, (5sqrt{6}-4sqrt{10}+7sqrt{30}):sqrt{2}
8, left(2sqrt{28}-3sqrt{7}+5sqrt{63}right)sqrt{112}
9, left(5sqrt{3}+3sqrt{5}right):sqrt{15}
10, left(4sqrt{27}-2sqrt{48}-5sqrt{75}right):2sqrt{3}
11, left(1+sqrt{3}-sqrt{2}right).left(1+sqrt{3}+sqrt{2}right)
các bạn...
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Rút gọn :
1, \(\sqrt{8}-3\sqrt{32}+\sqrt{72}\)
2, \(6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}\)
3 , \(3\sqrt{112}-7\sqrt{216}+4\sqrt{54}-2\sqrt{252}-3\sqrt{96}\)
4, \(3\sqrt{3}\left(3+2\sqrt{6}-\sqrt{33}\right)\)
6, \(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\)
7, \((5\sqrt{6}-4\sqrt{10}+7\sqrt{30}):\sqrt{2}\)
8, \(\left(2\sqrt{28}-3\sqrt{7}+5\sqrt{63}\right)\sqrt{112}\)
9, \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
10, \(\left(4\sqrt{27}-2\sqrt{48}-5\sqrt{75}\right):2\sqrt{3}\)
11, \(\left(1+\sqrt{3}-\sqrt{2}\right).\left(1+\sqrt{3}+\sqrt{2}\right)\)
các bạn ơi ! giúp mik với đi !
a) frac{2sqrt{5}-4sqrt{10}}{3sqrt{10}}b)frac{6sqrt{6}-2sqrt{12}+3-sqrt{2}}{2sqrt{6}+1}c)frac{sqrt{3left(3-sqrt{11}right)^2}}{sqrt{6}left(3-sqrt{11}right)}d)frac{5sqrt{7}-4sqrt{35}+7sqrt{5}}{sqrt{35}}e) left(sqrt{3}-1right)sqrt{2sqrt{19+8sqrt{3}-4}}g) sqrt{47+sqrt{5}}.sqrt{7-sqrt{2+sqrt{5}}.}sqrt{7+sqrt{2+sqrt{5}}}
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a) \(\frac{2\sqrt{5}-4\sqrt{10}}{3\sqrt{10}}\)
b)\(\frac{6\sqrt{6}-2\sqrt{12}+3-\sqrt{2}}{2\sqrt{6}+1}\)
c)\(\frac{\sqrt{3\left(3-\sqrt{11}\right)^2}}{\sqrt{6}\left(3-\sqrt{11}\right)}\)
d)\(\frac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
e) \(\left(\sqrt{3}-1\right)\sqrt{2\sqrt{19+8\sqrt{3}-4}}\)
g) \(\sqrt{47+\sqrt{5}}.\sqrt{7-\sqrt{2+\sqrt{5}}.}\sqrt{7+\sqrt{2+\sqrt{5}}}\)
1) left(sqrt{6}-sqrt{8}right)left(sqrt{6}+sqrt{8}right).2)left(3-sqrt{5}right)left(3+sqrt{5}right)3)sqrt{7-4sqrt{3}}+sqrt{7+4sqrt{3}}4)sqrt{2sqrt{3}-4}+sqrt{2sqrt{3}+4}5)sqrt{4sqrt{6}+11}-sqrt{11-4sqrt{6}}6)sqrt{10+2sqrt{11}}-sqrt{10-2sqrt{11}}7)sqrt{5-2sqrt{7-2sqrt{6}}}AI ĐÓ TỐT BỤNG GIÚP MK ZỚI:((
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1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right).\)
2)\(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
3)\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
4)\(\sqrt{2\sqrt{3}-4}+\sqrt{2\sqrt{3}+4}\)
5)\(\sqrt{4\sqrt{6}+11}-\sqrt{11-4\sqrt{6}}\)
6)\(\sqrt{10+2\sqrt{11}}-\sqrt{10-2\sqrt{11}}\)
7)\(\sqrt{5-2\sqrt{7-2\sqrt{6}}}\)
AI ĐÓ TỐT BỤNG GIÚP MK ZỚI:((
1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)
\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)
\(=6-8=-2\)
2) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=3^2-\left(\sqrt{5}\right)^2\)
\(=9-5=4\)
3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)
=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn
5) Đặt \(A=\sqrt{4\sqrt{6}+11}-\sqrt{11-4\sqrt{6}}\)
\(\Rightarrow A^2=\left(\sqrt{11+4\sqrt{6}}-\sqrt{11-4\sqrt{6}}\right)^2\)
\(=11+4\sqrt{6}-2\sqrt{\left(11+4\sqrt{6}\right)\left(11-4\sqrt{6}\right)}+11-4\sqrt{6}\)
\(=22-2\sqrt{121-96}\)
\(=22-2\sqrt{5}\)
=> \(A=\sqrt{22-2\sqrt{5}}\)
6) Đặt \(B=\sqrt{10+2\sqrt{11}}-\sqrt{10-2\sqrt{11}}\)
\(\Leftrightarrow B^2=\left(\sqrt{10+2\sqrt{11}}-\sqrt{10-2\sqrt{11}}\right)^2\)
\(=10+2\sqrt{11}-2\sqrt{\left(10+2\sqrt{11}\right)\left(10-2\sqrt{11}\right)}+10-2\sqrt{11}\)
\(=20-2\sqrt{100-44}\)
\(=20-4\sqrt{14}\)
=> \(B=\sqrt{20-4\sqrt{14}}\)
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Bài 1: Tính
1, Aleft(1-frac{5+sqrt{5}}{1+sqrt{5}}right).left(frac{5-sqrt{5}}{1-sqrt{5}}-1right)
2, Bleft(frac{3sqrt{125}}{15}-frac{10-4sqrt{6}}{sqrt{5}-2}right).frac{1}{sqrt{5}}
3, Cleft(frac{sqrt{1000}}{100}-frac{5sqrt{2}-2sqrt{5}}{2sqrt{5}-8}right).frac{sqrt{10}}{10}
4, Dfrac{1}{sqrt{49+20sqrt{6}}}-frac{1}{sqrt{49-20sqrt{6}}}+frac{1}{sqrt{7-4sqrt{3}}}
5, Efrac{1}{sqrt{4-2sqrt{3}}}-frac{1}{sqrt{7-sqrt{48}}}+frac{3}{sqrt{14-6sqrt{5}}}
6, Ffrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}...
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Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
20 tháng 8 2021 lúc 21:02
Tính
a,\(\sqrt{4+\sqrt{15}}\)
b,\(\left(3-\sqrt{2}\right)\sqrt{11+6\sqrt{2}}\)
c,\(\left(\sqrt{5}+\sqrt{7}\right)\sqrt{12-2\sqrt{35}}\)
a: Ta có: \(\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
b: Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{2}}\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
=9-2
=7
c: Ta có: \(\left(\sqrt{7}+\sqrt{5}\right)\cdot\sqrt{12-2\sqrt{35}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=2
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