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\(C=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)
\(C^2=\left(\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\right)^2\)
\(C^2=x^2+2\sqrt{x^2-1}-2\sqrt{\left(x^2+2\sqrt{x^2-1}\right)\left(x^2-2\sqrt{x^2-1}\right)}+x^2-2\sqrt{x^2-1}\)
\(C^2=2x^2-2\sqrt{x^4-2x^2\sqrt{x^2-1}+2x^2\sqrt{x^2-1}-\left(2\sqrt{x^2-1}\right)^2}\)
\(C^2=2x^2-2\sqrt{x^4-4\left(x^2-1\right)}\)
\(C^2=2x^2-2\sqrt{x^4-4x^2+4}\)
\(C=\sqrt{2x^2-2\sqrt{x^4-4x^2+4}}\)
Thay: \(x=\sqrt{5}\) vào C, ta có:
\(C=\sqrt{2\sqrt{5}^2-2\sqrt{\sqrt{5}^4-4\sqrt{5}^2+4}}\)
\(C=\sqrt{10-2\sqrt{25-20+4}}\)
\(C=\sqrt{10-2\sqrt{9}}\)
\(C=\sqrt{10-6}\)
\(C=\orbr{\begin{cases}-2\\2\end{cases}}\)
Mà theo bài ra: \(\sqrt{x^2+2\sqrt{x^2-1}}>\sqrt{x^2-2\sqrt{x^2-1}}\)
\(\Rightarrow\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}>0\)
\(\Rightarrow C=2\)
2.
A=\(\sqrt{\sqrt{\left(\sqrt{16}-\sqrt{12}\right)^2}}-\sqrt{\sqrt{\left(\sqrt{16}+\sqrt{12}\right)^2}}\)
\(=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{1}\right)^2}\)
\(=\sqrt{3}-1-\left(\sqrt{3}+1\right)\)
\(=\sqrt{3}-1-\sqrt{3}-1\)
\(=-2\)
B= \(\sqrt{5-2\sqrt{2+\sqrt{\left(\sqrt{8}+\sqrt{1}\right)^2}}}\)
\(=\sqrt{5-2\sqrt{2+\sqrt{8}+1}}\)
\(=\sqrt{5-2\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{5-2\sqrt{\left(\sqrt{2}+\sqrt{1}\right)^2}}\)
\(=\sqrt{5-2\sqrt{2}-2}\)
\(=\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}-\sqrt{1}\right)^2}\)
\(=\sqrt{2}-1\)
Bài 1 bạn nhóm , trục như thường nhé :D
Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)
\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)
\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)
\(D=-\sqrt{6}\left(do:D< 0\right)\)
a,\(\sqrt{\left(\sqrt{3}-1\right)^2}\) \(+\sqrt{\left(\sqrt{3}+1\right)^2}=2\sqrt{3}\)
b. \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=3\sqrt{5}\)
c,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=4\)
d.\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2\sqrt{2}\)
\(A=\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{\left(1+\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{3}+1\)
\(=1+\sqrt{2}+\sqrt{3}+\sqrt{3}+1=\sqrt{2}+2\sqrt{3}+2\)
a ) Để ý thấy \(16\sqrt{3}=2.2\sqrt{3}.4=2.\sqrt{12}.4\) , như vậy , ta sẽ tách :
\(28=12+16\) \(\Rightarrow\sqrt{\sqrt{28+16\sqrt{3}}=\sqrt{\sqrt{12+16+16\sqrt{3}}}}=\sqrt{\sqrt{\left(\sqrt{12}+4\right)^2}}=\sqrt{\sqrt{12}+4}\)
\(=\sqrt{3+2.\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b ) \(4\sqrt{3}=2.2\sqrt{3}\), tách \(7=4+3\)
c ) \(24\sqrt{5}=2.\sqrt{5}.12=2.\sqrt{5}.2.6=2.\sqrt{20}.6\) , tách : \(56=20+36\)
d ) \(2\sqrt{11}=2.11.1\) , tách : \(12=11+1\)
e ) \(4\sqrt{2}=2.\sqrt{2}.2.1=2.\sqrt{8}.1\) , tách : \(9=8+1\)
a) \(\sqrt{\sqrt{28+16\sqrt{3}}}\)
\(=\sqrt{\sqrt{\left(2\sqrt{3}\right)^2+2\cdot2\sqrt{3}\cdot4+16}}\)
\(=\sqrt{\sqrt{\left(2\sqrt{3}+4\right)^2}}\) \(=\sqrt{2\sqrt{3}+4}\)
\(=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b)\(\sqrt{7+4\sqrt{3}}=\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
c) \(\sqrt{\sqrt{56-24\sqrt{5}}}=\sqrt{\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot6}+36}\)
\(=\sqrt{\sqrt{\left(2\sqrt{5}-6\right)^2}}=\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
d) \(\sqrt{12-2\sqrt{11}}=\sqrt{11-2\sqrt{11}+1}\)
\(=\sqrt{\left(\sqrt{11}-1\right)^2}=\sqrt{11}-1\)
e) \(\sqrt{9+4\sqrt{2}}=\sqrt{\left(2\sqrt{2}\right)^2+2\cdot2\sqrt{2}+1}\)
\(=\sqrt{\left(2\sqrt{2}+1\right)^2}=2\sqrt{2}+1\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
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
a/ \(\sqrt[4]{17+12\sqrt{2}}-\sqrt{2}\)
= \(\sqrt[4]{9+2×3×2\sqrt{2}+8}-\sqrt{2}\)
= \(\sqrt{3+2\sqrt{2}}-\sqrt{2}\)
= \(\sqrt{2}+1-\sqrt{2}\)= 1
Mấy câu còn lại giải tương tự