\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
Những câu hỏi liên quan
chứng minh :a) 11+6sqrt{2} (3+sqrt{2})^2 b) sqrt{11+6sqrt{2}}+sqrt{11-6sqrt{2}}6 c) sqrt{8-2sqrt{7}}-sqrt{8+2sqrt{7}} -2 d) sqrt{49-12sqrt{5}}-sqrt{49+12sqrt{5}}-4
Đọc tiếp
chứng minh :a) 11+6\(\sqrt{2}\)= (3+\(\sqrt{2}\))\(^2\)
b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)=6
c) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)= -2
d) \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)=-4
a: \(\left(3+\sqrt{2}\right)^2=3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2\)
\(=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b: \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c: \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d: \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-2\cdot3\sqrt{5}\cdot2+4}-\sqrt{45+2\cdot3\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
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a) \(\left(3+\sqrt{2}\right)^2=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d) \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
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a) 11+6\(\sqrt{2}\) = \(\left(3+\sqrt{2}\right)^2\)
b) 8-2\(\sqrt{7}\)=\(\left(\sqrt{7}-1\right)^2\)
c)\(\sqrt{11+6\sqrt{2}}=\sqrt{11-6\sqrt{2}}=6\)
d) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=-2\)
a)\(\sqrt{\left(2\sqrt{2}-3\right)^2+\sqrt{15}}\)
b)\(\sqrt{\left(\sqrt{10}-3\right)}^2+\sqrt{\left(\sqrt{10}-4\right)^2}\)
c)\(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)
d)\(\sqrt{11}+6\sqrt{2}+\sqrt{11-6\sqrt{2}=6}\)
b: =căn 10-3+4-căn 10=1
a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)
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\(\sqrt{\left(\sqrt{7}-5\right)^2}+\sqrt{\left(2-\sqrt{7}\right)^2}\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
Lời giải:
a. \(=|\sqrt{7}-5|+|2-\sqrt{7}|=5-\sqrt{7}+(\sqrt{7}-2)=3\)
b. \(=\sqrt{(3+\sqrt{2})^2}-\sqrt{(3-\sqrt{2})^2}=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=(3+\sqrt{2})-(3-\sqrt{2})=2\sqrt{2}\)
c.
\(=\sqrt{(3+2\sqrt{2})^2}+\sqrt{(3-2\sqrt{2})^2}=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
$=(3+2\sqrt{2})+(3-2\sqrt{2})=6$
d.
$=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}$
$=|\sqrt{5}+1|-|\sqrt{5}-1|=\sqrt{5}+1-(\sqrt{5}-1)=2$
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C/m: \(\sqrt{11+6\sqrt{2}}\) + \(\sqrt{11-6\sqrt{2}}\) = 6
Ta có VT:
\(VT=\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\)
\(=3+\sqrt{2}+3-\sqrt{2}\)
\(=6=VP\left(dpcm\right)\)
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\(VT=\sqrt{9+2\cdot3\cdot\sqrt{2}+2}+\sqrt{9-2\cdot3\cdot\sqrt{2}+2}\)
\(=3+\sqrt{2}+3-\sqrt{2}\)
=6=VP
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tính:a) sqrt{dfrac{1}{8}}.sqrt{2}.sqrt{125}.sqrt{dfrac{1}{5}}b)sqrt{sqrt{2}-1}.sqrt{sqrt{2}+1}c) sqrt{11-6sqrt{2}}.sqrt{11+6sqrt{2}}d) sqrt{12-6sqrt{3}}.sqrt{dfrac{1}{3-sqrt{3}}}e) dfrac{sqrt{15}-sqrt{6}}{sqrt{35}-sqrt{14}}f) dfrac{2sqrt{15}-2sqrt{10}+sqrt{6}-3}{2sqrt{5}-2sqrt{10}-sqrt{3}+sqrt{6}}g) left(dfrac{1}{5-2sqrt{6}}+dfrac{2}{5+2sqrt{6}}right)left(15+2sqrt{6}right)
Đọc tiếp
tính:
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}\)
c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}\)
d) \(\sqrt{12-6\sqrt{3}}.\sqrt{\dfrac{1}{3-\sqrt{3}}}\)
e) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
f) \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
g) \(\left(\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)
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b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2-1}=1\)
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c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}=\sqrt{\left(11-6\sqrt{2}\right)\left(11+6\sqrt{2}\right)}=\sqrt{121-72}=\sqrt{49}=7\)
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Xem thêm câu trả lời
\(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2^6\right)}\)
rút gọn:giải chi tiết hộ mình nha
a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
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Rút gọn biểu thức
1) \(\sqrt{6\sqrt{2}+11}\) - \(\sqrt{11-6\sqrt{2}}\)
2) (\(\sqrt{3}\) - 2)\(\sqrt{7+4\sqrt{3}}\)
1: =3+căn 2-3+căn 2
=2căn 2
2: =(căn 3-2)(căn 3+2)
=3-4=-1
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1. Tính
a) \(\sqrt[3]{(\sqrt{2}+3)(11+6\sqrt{2})}\sqrt[3]{(\sqrt{2}+-3)(11-6\sqrt{2})}\)
b) (\((\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}-\sqrt[3]{2})\)
c)\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
