Ta có: \(\sqrt{9-4\sqrt{2}}\)
\(=\sqrt{8-2\cdot2\sqrt{2}\cdot1+1}\)
\(=\sqrt{\left(2\sqrt{2}-1\right)^2}\)
\(=\left|2\sqrt{2}-1\right|\)
\(=2\sqrt{2}-1\)(Vì \(2\sqrt{2}>1\))
Ta có: \(\sqrt{9-4\sqrt{2}}\)
\(=\sqrt{8-2\cdot2\sqrt{2}\cdot1+1}\)
\(=\sqrt{\left(2\sqrt{2}-1\right)^2}\)
\(=\left|2\sqrt{2}-1\right|\)
\(=2\sqrt{2}-1\)(Vì \(2\sqrt{2}>1\))
rút gọn
a, \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
b\(\sqrt{3+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
thankyou các bạn trước
Rút gọn biểu thức sau;\(\sqrt{9-4}\sqrt{5}-\sqrt{5}\)
rút gọn K
K = \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}-\sqrt{\left(-8\right)^2}\)
\(\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{3-2\sqrt{2}}}}\)
Rút gọn hộ mik vs <:
Mong mng giúp ạ
câu1 rút gọn
a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
b)\(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\left(x\ne\sqrt{2},x\ne-\sqrt{2}\right)\)
c)\(\sqrt{9\text{x}^2}-2\text{x}\left(x< 0\right)\)
d)\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)
e)\(\dfrac{x^2-5}{x+\sqrt{5}}\left(x\ne-\sqrt{5}\right)\)
M = \(\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)
a) Rút gọn M
b) Tìm x để M = \(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
Rút gọn rồi tính:
a) \(5\sqrt{\left(-2\right)^4}\) c)\(\sqrt{\sqrt{\left(-5\right)^8}}\)
b)\(-4\sqrt{\left(-3\right)^6}\)
2.4 Rút gọn biểu thức
\(a,\dfrac{3-\sqrt{x}}{x-9}\) ( vs x ≥ 0, x≠ 9)
b, \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}\)( vs x ≥ 0 ; x ≠ 9)
c, \(6-2x-\sqrt{9-6x+x^2}\left(x< 3\right)\)
Rút gọn biểu thức sau:
\(\dfrac{\left(4+\sqrt{7}\right).\sqrt{4-\sqrt{7}}}{\sqrt{4+\sqrt{7}}}\)
Rút gọn biểu thức:
\(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)
\(B=3\sqrt{\left(1,5\right)^2}-4\sqrt{\left(3-\sqrt{2}\right)^2}\)