Thực hiện phép tính:\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
Những câu hỏi liên quan
thực hiện phép tính
A=\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
B=\(\sqrt{\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{2}}}\cdot\left(3+\sqrt{5}\right)\)
a) \(A=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\dfrac{2-\sqrt{3}}{1}+\dfrac{2+\sqrt{3}}{1}\)
=4
Đúng 0
Bình luận (0)
Thực hiện các phép tính sau:
a, \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{\sqrt{3}+2}\)
b, \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
c, \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
d, \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
Thực hiện phép tính:\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
= \(\sqrt{2+\sqrt{3}}+\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}\)
= \(\sqrt{2+\sqrt{3}}+\sqrt{4-\left(2+\sqrt{3}\right)}\)
= \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
= \(\sqrt{4-3}\)
= \(\sqrt{1}\)
= \(1\)
Đúng 0
Bình luận (0)
Thực hiện các phép tính sau:a, sqrt{2+sqrt{3}}-sqrt{2-sqrt{3}}b, sqrt{21-12sqrt{3}}-sqrt{3}c, left(sqrt{6}+sqrt{2}right)cdotleft(sqrt{3}-2right)sqrt{sqrt{3}+2}d, left(4+sqrt{15}right)cdotleft(sqrt{10}-sqrt{6}right)cdotsqrt{4-sqrt{15}}e, sqrt{13-sqrt{160}}-sqrt{53+4sqrt{90}}f, sqrt{6-2sqrt{sqrt{2}+sqrt{12}+sqrt{18}-sqrt{128}}}
Đọc tiếp
Thực hiện các phép tính sau:
a, \(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
b, \(\sqrt{21-12\sqrt{3}}-\sqrt{3}\)
c, \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)
d, \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
e, \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)\
f, \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18}-\sqrt{128}}}\)
15 tháng 12 2021 lúc 21:58
Thực hiện phép tính (tính nhanh nếu có thể):
4) \(4\cdot\left(\dfrac{-1}{2}\right)^3+\left|-1\dfrac{1}{2}+\sqrt{\dfrac{9}{4}}\right|:\sqrt{25}\)
5) \(\left[6-3\cdot\left(\dfrac{-1}{3}\right)^2+\sqrt{\dfrac{1}{4}}\right]:\sqrt{0,\left(9\right)}\)
TÍNH :Asqrt{3+sqrt{5+2sqrt{3}}}cdotsqrt{3-sqrt{5+2sqrt{3}}}Bsqrt{4+sqrt{8}}cdotsqrt{2+sqrt{2+sqrt{2}}}cdotsqrt{2-sqrt{2+sqrt{2}}}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}}}}Dleft[4+sqrt{15}right]left[sqrt{10}-sqrt{6}right]cdotsqrt{4-sqrt{15}}Eleft[3-sqrt{5}right]cdotsqrt{3+sqrt{5}}text{ }+left[3+sqrt{5}right]cdotsqrt{3-sqrt{5}}
Đọc tiếp
TÍNH :
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}}\cdot\sqrt{3-\sqrt{5+2\sqrt{3}}}\)
\(B=\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(C=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(D=\left[4+\sqrt{15}\right]\left[\sqrt{10}-\sqrt{6}\right]\cdot\sqrt{4-\sqrt{15}}\)
\(E=\left[3-\sqrt{5}\right]\cdot\sqrt{3+\sqrt{5}}\text{ }+\left[3+\sqrt{5}\right]\cdot\sqrt{3-\sqrt{5}}\)
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}}=\sqrt{\left(3^2\right)-\left(\sqrt{5+2\sqrt{3}}\right)^2}\)
\(=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-2-\sqrt{2}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}.\sqrt{4-2}=\sqrt{2}.\sqrt{2}=2\)
\(C=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(2+\sqrt{2+\sqrt{3}}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}=\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(2+\sqrt{3}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{4-3}=1\)
Đúng 0
Bình luận (0)
\(D=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{4+\sqrt{15}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4^2-15}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
\(E=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}\)
\(=2\sqrt{3-\sqrt{5}}+2\sqrt{3+\sqrt{5}}=\sqrt{2}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)
\(=\sqrt{2}.\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)
Đúng 0
Bình luận (0)
Thực hiện phép tính
a, \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
b,\(\left(1+\sqrt{3}-\sqrt{2}\right)\cdot\left(1+\sqrt{3}+\sqrt{2}\right)\)
a) \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-2\sqrt{40}}-\sqrt{53+12\sqrt{10}}\)
\(=\sqrt{\left(\sqrt{8}\right)^2-2.\sqrt{8}.\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2.3\sqrt{5}.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{8}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=\left|\sqrt{8}-\sqrt{5}\right|-\left|3\sqrt{5}+2\sqrt{2}\right|\)
= √8 - √5 - 3√5 - 2√2 = -4√5
b) (1+√3-√2).(1+√3+√2)= [(1+√3)^2-(√2)^2] = 4+2√3-2=2+2√3
Đúng 0
Bình luận (0)
a) =sprt{13-=sprt{160}} - =sprt{53+4=sprt{90}}
= =sprt{(=sprt{8} - =sprt{5})2 } - =sprt{(=sprt{45} + =sprt{8})2 }
= =sprt{8} - =sprt{5} - =sprt{45} - =sprt{8}
= -3=sprt{5}
b) ( 1 + =sprt{3} - =sprt{2} )( 1+ =sprt{3} + =sprt{2} )
= ( 1 + =sprt{3} )2 - (=sprt{2})2
= 4 + 2=sprt{3} - 2
=2 + 2=sprt{3}
Đúng 0
Bình luận (0)
\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(2+\sqrt{2+\sqrt{3}}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{4-\left(2+\sqrt{3}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\)
\(=\sqrt{4-3}=1\)
ai làm giúp với chịu rồi
\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(\sqrt{2+\sqrt{2+\sqrt{3}}}\right)^2}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{4-\left(\sqrt{2+\sqrt{3}}\right)^2}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{2^2-\left(\sqrt{3}\right)^2}=\sqrt{4-3}=1\)