\(=\sqrt{6-2\sqrt{6.5}+5}=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}=\left|\sqrt{6}-\sqrt{5}\right|=\sqrt{6}-\sqrt{5}\)
\(\sqrt{11-\sqrt{120}}=\sqrt{6}-\sqrt{5}\)
\(=\sqrt{6-2\sqrt{6.5}+5}=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}=\left|\sqrt{6}-\sqrt{5}\right|=\sqrt{6}-\sqrt{5}\)
\(\sqrt{11-\sqrt{120}}=\sqrt{6}-\sqrt{5}\)
\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)
Rút gọn giùm mình với ạ
Giải các phương trình vô tỉ (Phương trình có chứa căn thức)
1) \(\sqrt{x^2-20x+100}=10\)
2) \(\sqrt{x+2\sqrt{x}+1}=6\)
3) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)
4) \(\sqrt{3x+2\sqrt{3x}+1}=5\)
5) \(\sqrt{x^2+2x\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}\)
6) \(\sqrt{6x+4\sqrt{6x}+4}=7\)
7) \(\sqrt{2x^2-2x\sqrt{6}+3}-\sqrt{5-\sqrt{24}}=0\)
8) \(\sqrt{3-2\sqrt{2}}-\sqrt{x^2-2x\sqrt{2}+2}=0\)
9) \(\sqrt{11-\sqrt{120}}=\sqrt{5x^2+x\sqrt{120}+6}\)
Tính \(\left(\sqrt{7}+\sqrt{11}+\sqrt{13}\right)\left(\sqrt{11}+\sqrt{13}-\sqrt{7}\right)\left(\sqrt{7}+\sqrt{13}-\sqrt{11}\right)\left(\sqrt{7}+\sqrt{11}-\sqrt{13}\right)\)
\(a,\frac{2}{\sqrt{13}-\sqrt{11}}+\frac{5}{4+\sqrt{ }11}-\sqrt{52}
\)
b,\(\sqrt{6+2\sqrt{5}+\sqrt{9-4\sqrt{5}}-\sqrt{20}}\)
Rút gọn
\(C=\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
So sánh A=\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)
với B=\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)
\(11\sqrt{x}+11\sqrt{y}-x-\sqrt{xy}\)
Thu gọn biểu thức sau:
C=\(\sqrt{11-6\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(\left(\sqrt{6}-\sqrt{5}\right)^2-\sqrt{11+2\sqrt{30}}\)
\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\)
\(\sqrt{11+4\sqrt{7}}-\sqrt{14-6\sqrt{5}}\)
\(\sqrt{22-12\sqrt{2}}-\sqrt{19+6\sqrt{2}}\)
\(\sqrt{-6\sqrt{3}+12}+\sqrt{-12\sqrt{3}+21}\)