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a) \(A=\sqrt{10+\sqrt{99}}=\sqrt{10+3\sqrt{11}}=\frac{1}{\sqrt{2}}.\sqrt{20+6\sqrt{11}}\)
\(=\frac{1}{\sqrt{2}}.\sqrt{\left(3+\sqrt{11}\right)^2}=\frac{3+\sqrt{11}}{2}\)
b) \(B=\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
c) bn ktra lại đề
d) ĐK: \(x\ge0\)
\(\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}+1\right)^2}=\sqrt{x}+1\)
e) đk: \(x\ge-1\)
\(\sqrt{2x+3+2\sqrt{x^2+3x+2}}=\sqrt{x+1+2\sqrt{\left(x+1\right)\left(x+2\right)}+x+2}\)
\(=\sqrt{\left(\sqrt{x+1}+\sqrt{x+2}\right)^2}=\sqrt{x+1}+\sqrt{x+2}\)
a,
\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)
=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)
=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
=\(\sqrt{2}+\sqrt{2}-1\)
=\(2\sqrt{2}-1\)
còn tiếp
b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
2) \(A=\sqrt{15a^2-8a\sqrt{15}+16}\\ =\sqrt{\left(a\sqrt{15}-4\right)^2}\)
b) Khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\) thì
\(A=\sqrt{\left[\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)\sqrt{15}-4\right]^2}\)
\(=\sqrt{\left[\left(3+5\right)-4\right]^2}\)
\(=\sqrt{4^2}\)
\(=4\)
a) Ta có: \(\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{2+\sqrt{4+2\sqrt{3}}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2+\sqrt{\left(\sqrt{3}+1\right)^2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2+\left|\sqrt{3}+1\right|}{2-\left|\sqrt{3}-1\right|}\)
\(=\frac{2+\sqrt{3}+1}{2-\sqrt{3}+1}\)(Vì \(\sqrt{3}>1>0\))
\(=\frac{3+\sqrt{3}}{3-\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(a=\frac{2+\sqrt{4+2\sqrt{3}}}{2-\sqrt{4-2\sqrt{3}}}=\frac{2+\sqrt{\left(\sqrt{3}+1\right)^2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\frac{2+\sqrt{3}+1}{2-\sqrt{3}+1}=\frac{3+\sqrt{3}}{3-\sqrt{3}}=\frac{\left(3+\sqrt{3}\right)^2}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)
\(=\frac{12+6\sqrt{3}}{6}=2+\sqrt{3}\)
Xét \(A=\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}}>0\)
\(A^2=6+2\sqrt{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}=6+2\sqrt{2}\)
\(\Rightarrow A=\sqrt{6+2\sqrt{2}}\)
\(\Rightarrow\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}}-\sqrt{6+2\sqrt{2}}=\sqrt{6+2\sqrt{2}}-\sqrt{6+2\sqrt{2}}=0\)