ND
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NN
28 tháng 10 2020
a) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}\)
\(=\frac{\sqrt{2\left(4-\sqrt{7}\right)}-\sqrt{2\left(4+\sqrt{7}\right)}+2}{\sqrt{2}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+2}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}+2}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+2}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|+2}{\sqrt{2}}=\frac{\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)+2}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1+2}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0\)
b) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)
\(=\frac{\sqrt{2\left(6+\sqrt{11}\right)}-\sqrt{2\left(6-\sqrt{11}\right)}+3.2}{\sqrt{2}}\)
\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}+6}{\sqrt{2}}\)
\(=\frac{\sqrt{11+2\sqrt{11}+1}-\sqrt{11-2\sqrt{11}+1}+6}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-1\right)^2}+6}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{11}+1\right|-\left|\sqrt{11}-1\right|+6}{\sqrt{2}}\)
\(=\frac{\left(\sqrt{11}+1\right)-\left(\sqrt{11}-1\right)+6}{\sqrt{2}}\)
\(=\frac{\sqrt{11}+1-\sqrt{11}+1+6}{\sqrt{2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}\)
KT
2 tháng 7 2018
a) \(A=\sqrt{4-\sqrt{15}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\)\(\sqrt{2}A=\sqrt{8-2\sqrt{15}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{5}-\sqrt{3}-\left(\sqrt{3}+1\right)=\sqrt{5}-1\)
\(\Rightarrow\)\(A=\frac{\sqrt{5}-1}{\sqrt{2}}\)
b) tương tự câu a
c) \(\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}-\sqrt{6-2\sqrt{5+\sqrt{13-4\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}-\sqrt{6-2\sqrt{5+\sqrt{\left(\sqrt{12}-1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{5-\left(\sqrt{12}+1\right)}}-\sqrt{6-2\sqrt{5+\left(\sqrt{12}-1\right)}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}-\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}-\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}-\sqrt{6-2\left(\sqrt{3}+1\right)}\)
\(=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)=2\)
7 tháng 8 2016
\(\sqrt{13+4\sqrt{3}}\cdot\sqrt{28+6\sqrt{3}}-5\sqrt{3}\)
\(=\sqrt{\left(2\sqrt{3}+1\right)^2}\cdot\sqrt{\left(3\sqrt{3}+1\right)^2}-5\sqrt{3}\)
\(=\left(2\sqrt{3}+1\right)\left(3\sqrt{3}+1\right)-5\sqrt{3}\)
\(=6\cdot3+2\sqrt{3}+3\sqrt{3}+1-5\sqrt{3}\)
\(=18+1=19\)
5 tháng 10 2020
b) \(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)
19 tháng 5 2021
a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)
Thay x = 4 => \(\sqrt{x}=2\) vào B ta được :
\(B=\frac{2+5}{2-3}=-7\)
19 tháng 5 2021
b, Ta có : Với \(x\ge0;x\ne9\)
\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)
Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
2 tháng 9 2017
\(13-4\sqrt{3}=\left(2\sqrt{3}\right)^2-2.2\sqrt{2}.1+1^2=\left(2\sqrt{3}-1\right)^2\)
a) \(\left(\sqrt{5}+\sqrt{3}\right)\sqrt{8-2\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
câu này \(\sqrt{15}\)đúng hơn \(\sqrt{5}\)
b) \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)c) \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}=\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}=-2\sqrt{2}\)
B
12 tháng 7 2019
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
12 tháng 7 2019
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
A = 2 + 2
A = 4
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(B=\sqrt{2}+3+3-\sqrt{2}\)
B = 3 + 3
B = 6
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(C=3+2\sqrt{2}+3-2\sqrt{2}\)
C = 3 + 3
C = 6
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(D=\sqrt{5}+2-\sqrt{5}+2\)
D = 2 + 2
D = 4
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(E=\sqrt{5}+1-\sqrt{5}+1\)
E = 1 + 1
E = 2