Có bạn nào giúp mình giải 2 bài này với ạ. mình đg cần gấp. cảm ơn nhiều !!!!!!!!
12.
a.
\(\sqrt{11-6\sqrt2}-\sqrt{11+6\sqrt2}=\sqrt{9-2.3.\sqrt2+2}-\sqrt{9+2.3.\sqrt2+2}\)
\(=\sqrt{\left(3-\sqrt2\right)^2}-\sqrt{\left(3+\sqrt2\right)^2}=\left|3-\sqrt2\right|-\left|3+\sqrt2\right|\)
\(=3-\sqrt2-\left(3+\sqrt2\right)=-2\sqrt2\)
b.
\(\sqrt{\left(2-\sqrt5\right)^2}+\sqrt{14-6\sqrt5}=\sqrt{\left(2-\sqrt5\right)^2}+\sqrt{9-2.3.\sqrt5+5}\)
\(=\sqrt{\left(2-\sqrt5\right)^2}+\sqrt{\left(3-\sqrt5\right)^2}=\left|2-\sqrt5\right|+\left|3-\sqrt5\right|\)
\(=\sqrt5-2+3-\sqrt5=1\)
c.
\(\left(2+\sqrt7\right)\sqrt{11-4\sqrt7}=\left(2+\sqrt7\right)\sqrt{7-2.2.\sqrt7+4}=\left(2+\sqrt7\right)\sqrt{\left(\sqrt7-2\right)^2}\)
\(=\left(2+\sqrt7\right).\left|\sqrt7-2\right|=\left(\sqrt7+2\right)\left(\sqrt7-2\right)\)
\(=7-2^2=3\)
d.
\(\sqrt{\left(3+\sqrt2\right)^2}+\sqrt{6-4\sqrt2}=\left|3+\sqrt2\right|+\sqrt{4-2.2\sqrt2+2}=\left|3+\sqrt2\right|+\sqrt{\left(2-\sqrt2\right)^2}\)
\(=3+\sqrt2+\left|2-\sqrt2\right|=3+\sqrt2+2-\sqrt2=5\)
e.
\(\sqrt{9-3\sqrt8}-\frac{\sqrt3-1}{\sqrt2}+\sqrt{5-2\sqrt6}-\sqrt{2-\sqrt3}\)
\(=\sqrt{6-2.\sqrt{6.3}+3}-\frac{\sqrt3-1}{\sqrt2}+\sqrt{3-2\sqrt{3.2}+2}-\sqrt{\frac{3-2\sqrt3+1}{2}}\)
\(=\sqrt{\left(\sqrt6-\sqrt3\right)^2}-\frac{\sqrt3-1}{\sqrt2}+\sqrt{\left(\sqrt3-\sqrt2\right)^2}-\sqrt{\frac{\left(\sqrt3-1\right)^2}{2}}\)
\(\left|\sqrt6-\sqrt3\right|-\frac{\sqrt3-1}{\sqrt2}+\left|\sqrt3-\sqrt2\right|-\frac{\left|\sqrt3-1\right|}{\sqrt2}\)
\(=\sqrt6-\sqrt3-\frac{\sqrt3-1}{\sqrt2}+\sqrt3-\sqrt2-\frac{\sqrt3-1}{\sqrt2}\)
\(=\sqrt6-\sqrt2-2.\left(\frac{\sqrt3-1}{\sqrt2}\right)=\sqrt6-\sqrt2-\left(\sqrt6-\sqrt2\right)=0\)
12f. (hoc24 bị lỗi phần phân số nên chịu khó dịch :D)
\(\frac{2-\sqrt3}{\sqrt2+\sqrt{2+\sqrt3}}+\frac{2+\sqrt3}{\sqrt2-\sqrt{2-\sqrt3}}=\frac{\sqrt2\left(2-\sqrt3\right)}{2+\sqrt{4+2\sqrt3}}+\frac{\sqrt2\left(2+\sqrt3\right)}{2-\sqrt{4-2\sqrt3}}\)
\(=\frac{\sqrt2\left(2-\sqrt3\right)}{2+\sqrt{\left(\sqrt3+1\right)^2}}+\frac{\sqrt2\left(2+\sqrt3\right)}{2-\sqrt{\left(\sqrt2-1\right)^2}}=\frac{\sqrt2\left(2-\sqrt3\right)}{2+\sqrt3+1}+\frac{\sqrt2\left(2+\sqrt3\right)}{2-\left(\sqrt3-1\right)}\)
\(=\sqrt2\left(\frac{2-\sqrt3}{3+\sqrt3}+\frac{2+\sqrt3}{3-\sqrt3}\right)=\sqrt2\left(\frac{\left(2-\sqrt3\right)\left(3-\sqrt3\right)+\left(2+\sqrt3\right)\left(3+\sqrt3\right)}{\left(3-\sqrt3\right)\left(3+\sqrt3\right)}\right)\)
\(=\sqrt2\cdot\frac{18}{9-3}=3\sqrt2\)
13a.
\(\sqrt{19-8\sqrt3}+\sqrt{4-2\sqrt3}=\sqrt{16-2.4.\sqrt3+3}+\sqrt{3-2\sqrt3+1}\)
\(=\sqrt{\left(4-\sqrt3\right)^2}+\sqrt{\left(\sqrt3-1\right)^2}=\left|4-\sqrt3\right|+\left|\sqrt3-1\right|\)
\(=4-\sqrt3+\sqrt3-1=3\)
b.
\(\sqrt{12+3\sqrt3+\sqrt{4+2\sqrt3}}-2\sqrt3=\sqrt{12+3\sqrt3+\sqrt{3+2\sqrt3+1}}-2\sqrt3\)
\(=\sqrt{12+3\sqrt3+\sqrt{\left(\sqrt3+1\right)^2}}-2\sqrt3=\sqrt{12+3\sqrt3+\sqrt3+1}-2\sqrt3\)
\(=\sqrt{13+4\sqrt3}-2\sqrt3=\sqrt{12+2.\sqrt{12}+1}-2\sqrt3=\sqrt{\left(\sqrt{12}+1\right)^2}-2\sqrt3\)
\(=\left|\sqrt{12}+1\right|-2\sqrt3=\sqrt{12}+1-2\sqrt3=2\sqrt3+1-2\sqrt3=1\)
