Sửa đề lại nhé
\(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=\left(\sqrt{19}\right)^2-3^2=19-9=10\)
Sửa đề :\(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)\)
⇔ \(\sqrt{19}^2-3^2\)
⇔ \(19-9=10.\)
Tính:
a,y=2\(+\sqrt{17-4\sqrt{9}+4\sqrt{5}}\)
b,t=\(\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right)\)
c,x=\(\sqrt{19+8\sqrt{3}}+\sqrt{19-8\sqrt{3}}\)
\(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\)
Tính:
\(\sqrt{19-6\sqrt{2}}\)
Phép 4:\(\sqrt{19-4\sqrt{ }15}\)( căn 19 tất cả -4 căn 15)
Phép 1: \(3\cdot\sqrt{7-4\sqrt{3}}\) ( 3 nhân căn 7 tất cả - 4 căn 3)
Phép 2:\(\sqrt{11+4\sqrt{7}}\)
Phép 3: \(2\cdot\sqrt{11-4\sqrt{ }7}\)( Căn 11 tất cả - 4 căn 7)
rút gọn biểu thức
\(\sqrt{19-8\sqrt{3}}+\left(2+\sqrt{3}\right)^2-\sqrt{27}\)
\(\frac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\frac{2\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
a.Rút gọn
b.GTNN
Chứng minh rằng \(19>1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}>18\)
1.Tính :
a.\(\sqrt{11+6\sqrt{2}}+\sqrt{19+6\sqrt{2}}\)
b.\(-\sqrt{2\left(2-\sqrt{3}\right)}+\sqrt{2\left(2+\sqrt{3}\right)}\)
c.\(\frac{\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}}{2\sqrt{3}}\)
1.Tính:
a.\(\sqrt{11+6\sqrt{2}-\sqrt{19+6\sqrt{2}}}\)
b.\(-\sqrt{2\left(2-\sqrt{3}\right)}+\sqrt{2\left(2+\sqrt{3}\right)}\)
c.\(\frac{\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}}{2\sqrt{3}}\)
