1.\(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{6}}=?\)
2.\(\sqrt{11-6\sqrt{2}}+\sqrt{3-2\sqrt{2}}=?\)
Những câu hỏi liên quan
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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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) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=-2\)
frac{3+2sqrt{3}}{sqrt{3}}+frac{2+sqrt{2}}{sqrt{2}+1}-left(sqrt{2}+3right)
0.1cdotsqrt{left(-3right)^2}cdotleft[6sqrt{left(frac{1}{3}right)^2}-sqrt{left(sqrt{3}-2right)^2}right]^2
left(frac{3sqrt{2}+sqrt{6}}{sqrt{12}+2}-frac{sqrt{54}}{3}right)cdotfrac{2}{sqrt{6}}
left(frac{3+2sqrt{3}}{sqrt{3}+2}+frac{2+sqrt{2}}{sqrt{2}+1}right)divleft(1divfrac{1}{sqrt{2}+sqrt{3}}right)
sqrt{frac{5+2sqrt{6}}{5-2sqrt{6}}}+sqrt{frac{5-2sqrt{6}}{5+2sqrt{6}}}
Đọc tiếp
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(0.1\cdot\sqrt{\left(-3\right)^2}\cdot\left[6\sqrt{\left(\frac{1}{3}\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\right]^2\)
\(\left(\frac{3\sqrt{2}+\sqrt{6}}{\sqrt{12}+2}-\frac{\sqrt{54}}{3}\right)\cdot\frac{2}{\sqrt{6}}\)
\(\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right)\div\left(1\div\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
rút gọnAleft(sqrt{28}-2sqrt{14}+sqrt{7}right)cdotsqrt{7}+7sqrt{8}Bsqrt{6+2sqrt{5}}-sqrt{6-2sqrt{5}}Cleft(sqrt{7}-sqrt{10}right)^2+sqrt{280}Ddfrac{sqrt{99}}{sqrt{11}}+sqrt{7}cdotsqrt{63}-sqrt{sqrt{81}}Esqrt{27}left(s-sqrt{5}right)^2cdotleft(3sqrt{48}right)giải chi tiết ra giúp mik nha,cảm ơn nhiều
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rút gọn
A=\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
B=\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
C=\(\left(\sqrt{7}-\sqrt{10}\right)^2+\sqrt{280}\)
D=\(\dfrac{\sqrt{99}}{\sqrt{11}}+\sqrt{7}\cdot\sqrt{63}-\sqrt{\sqrt{81}}\)
E=\(\sqrt{27}\left(s-\sqrt{5}\right)^2\cdot\left(3\sqrt{48}\right)\)
giải chi tiết ra giúp mik nha,cảm ơn nhiều
Thực hiện phép tính:a,left(sqrt{28}-2sqrt{14}+sqrt{7}right)cdotsqrt{7}+7sqrt{8}b,left(sqrt{8}-3sqrt{2}+sqrt{10}right)cdotleft(sqrt{2}-3sqrt{0.4}right)c,left(15sqrt{50}+5sqrt{200}-3sqrt{450}right):sqrt{10}d,sqrt{6+2sqrt{5}}+sqrt{6-2sqrt{5}}e,sqrt{11+6sqrt{2}}-sqrt{11-6sqrt{2}}f,sqrt[3]{5sqrt{2}+7}-sqrt[3]{5sqrt{2}-7}g,sqrt[3]{20+14sqrt{2}}+sqrt[3]{20-14sqrt{2}}h,sqrt[3]{26+15sqrt{3}}-sqrt[3]{26-15sqrt{3}}
Đọc tiếp
Thực hiện phép tính:
\(a,\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(b,\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\cdot\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(c,\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(d,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(e,\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(f,\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(g,\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
\(h,\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}\)
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
\(\sqrt{2\sqrt{11-6\sqrt{2}}-\sqrt{\left(\sqrt{13-2}\right)\cdot\sqrt{17+4\sqrt{3}}}}\)
1) frac{3+2sqrt{3}}{sqrt{3}}+frac{2+sqrt{2}}{sqrt{2}+1}-left(sqrt{2}+3right)2) 0.1sqrt{left(-3right)^2}cdotleft[6sqrt{left(-frac{1}{3}right)^2}-sqrt{left(sqrt{3}-2right)^2}right]^23) left(frac{3+2sqrt{3}}{sqrt{3}+2}+frac{2+sqrt{2}}{sqrt{2}+1}right)divleft(1divfrac{1}{sqrt{2}+sqrt{3}}right)4) left(frac{3sqrt{2}+sqrt{6}}{sqrt{12}+2}-frac{sqrt{54}}{3}right)cdotfrac{2}{sqrt{6}}5) sqrt{frac{5+2sqrt{6}}{5-2sqrt{6}}}+sqrt{frac{5-2sqrt{6}}{5+2sqrt{6}}}
Đọc tiếp
1) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
2) \(0.1\sqrt{\left(-3\right)^2}\cdot\left[6\sqrt{\left(-\frac{1}{3}\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\right]^2\)
3) \(\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right)\div\left(1\div\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
4) \(\left(\frac{3\sqrt{2}+\sqrt{6}}{\sqrt{12}+2}-\frac{\sqrt{54}}{3}\right)\cdot\frac{2}{\sqrt{6}}\)
5) \(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)
\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=10\)
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\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)
\(=\sqrt{3}-1\)
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30 tháng 9 2021 lúc 22:08
* Tính
a. A=\(\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
b. B=\(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
a: Ta có: \(A=\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
\(=\left(2+\sqrt{2}\right):\left(2+\sqrt{2}\right)\)
=1
b: Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+1\)
=1
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TínhAleft(4+sqrt{15}right)left(sqrt{10}-sqrt{6}right)cdotsqrt{4-sqrt{15}}Bleft(3-sqrt{5}right)cdotsqrt{3+sqrt{5}}+left(3+sqrt{5}right)cdotsqrt{3-sqrt{5}}Csqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{3}}}cdotsqrt{2+sqrt{2+sqrt{2+sqrt{3}}}}cdotsqrt{2-sqrt{2+sqrt{2+sqrt{ }}3}}Dsqrt{4+sqrt{15}}+sqrt{4-sqrt{15}}-2sqrt{3-sqrt{5}}Efrac{sqrt{15-10sqrt{2}}+sqrt{13+4sqrt{5}}-sqrt{11+2sqrt{10}}}{2sqrt{3+2sqrt{2}}+sqrt{9-4sqrt{2}}+sqrt{12+8sqrt{2}}}
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Tính
A=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
B=\(\left(3-\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\cdot\sqrt{3-\sqrt{5}}\)
C=\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{ }}3}}\)
D=\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
E=\(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{5}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
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