Thực hiện phép tính sau
\(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
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11 tháng 8 2020
Đặt: \(A=\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\)
=> \(A^2=\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
=> \(A^2=2\sqrt{5}+2\sqrt{5-4}\)
=> \(A^2=2\sqrt{5}+2\)
=> \(A^2=2\left(\sqrt{5}+1\right)\)
=> \(A=\sqrt{2\left(\sqrt{5}+1\right)}\)
=> \(\frac{A}{\sqrt{\sqrt{5}+1}}=\frac{\sqrt{2\left(\sqrt{5}+1\right)}}{\sqrt{\sqrt{5}+1}}=\sqrt{2}\)
Đặt: \(B=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)
=> \(VT=\frac{A}{\sqrt{\sqrt{5}+1}}-B=\sqrt{2}-\left(\sqrt{2}-1\right)=\sqrt{2}-\sqrt{2}+1=1\)
VẬY KẾT QUẢ CỦA PHÉP TÍNH = 1.
16 tháng 10 2021
a: Ta có: \(A=\sqrt{8}-2\sqrt{18}+3\sqrt{50}\)
\(=2\sqrt{2}-6\sqrt{2}+15\sqrt{2}\)
\(=11\sqrt{2}\)
b: Ta có: \(B=\sqrt{125}-10\sqrt{\dfrac{1}{20}}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
\(=5\sqrt{5}-\sqrt{5}+\sqrt{5}-1\)
\(=5\sqrt{5}-1\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 9 2023
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
TA
3 tháng 8 2021
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\\ =\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-2-\sqrt{3}\\ =\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\\ =\sqrt{2}\)
TN
25 tháng 7 2017
\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2-\sqrt{3}}\)
\(=\sqrt{2^2-\sqrt{3}^2}=\sqrt{4-3}=1\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
21 tháng 9 2023
\(a,\left(2\sqrt{3}-\sqrt{2}\right)^2+2\sqrt{24}=\left[\left(2\sqrt{3}\right)^2-2.2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2\right]+2\sqrt{24}\\ =\left[12-4\sqrt{6}+2\right]+2\sqrt{24}=14-4\sqrt{6}+4\sqrt{6}=14\\ b,\left(3\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+2\sqrt{3}\right)-\sqrt{60}=3\sqrt{5}.\sqrt{5}-2\sqrt{3}.\sqrt{3}+3\sqrt{5}.2\sqrt{3}-\sqrt{3}.\sqrt{5}-\sqrt{60}\\ =15-6+6\sqrt{15}-\sqrt{15}-\sqrt{2^2.15}\\ =9+3\sqrt{15}\)
Dat A=\(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
=> A2=\(2+\sqrt{3}+2-\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=4-2\sqrt{4-3}=2\)
Suy ra A=\(\sqrt{2}\)
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