Bài 1:Tính GT các biểu thức:
A= \(\left(\sqrt{5}+\sqrt{3}\right)^2\) - \(\left(\sqrt{5}-\sqrt{8}\right)^2\)
B= \(\sqrt{50}-3\sqrt{98}+2\sqrt{8}-3\sqrt{32}-5\sqrt{18}\)
C= \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
(mink đag cần gấp)
1
a. \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\) b.\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\) c. \(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)
d. \(\frac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\) e. \(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\) f. \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
Tính:
a) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
b) \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
A\(=\)\((3\sqrt{8}+2\sqrt{50}-4\sqrt{72})\)\(➗\)\(8\sqrt{2}\)
B\(=\)\((-4\sqrt{20}+5\sqrt{500}-3\sqrt{45})\div5 \)
C\(=(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\dfrac{\sqrt{3}-1}{\sqrt{3}+1})\div\sqrt{48}\)
Bài 1: Tìm các giá trị nguyên của x để các biểu thức sau có giá trị nguyên
a/C=\(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\) ; b/D=\(\dfrac{2\sqrt{x}-1}{\sqrt{x}+3}\)
Bài 2: Chứng minh
a/\(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}=\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=8\) b/\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
Bài 1: rút gọn rồi tính giá trị biểu thức:
A=\(\dfrac{2b\sqrt{x^2-1}-\sqrt{x+1}}{x-2\sqrt{x-1}}\) với x=3; y=\(\sqrt{2}\)
Bài 2: Trục căn thức ở mẫu
a/\(\dfrac{25}{5-2\sqrt{3}}\) b/\(\dfrac{8}{\sqrt{5}+2}\) c/\(\dfrac{6}{2\sqrt{3}-\sqrt{7}}\) d/\(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\) e/\(\dfrac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\)
tính
1\(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
2\(\left(2\sqrt{3}-3\right):5\sqrt{3}\)
3\(\left(2\sqrt{18}-3\sqrt{8}+6\right):\sqrt{2}\)
4\(\sqrt{27\left(1-\sqrt{3}\right)^2}:3\sqrt{15}\)
5\(\dfrac{a-\sqrt{b}}{\sqrt{b}}:\dfrac{\sqrt{b}}{a+\sqrt{b}}\)
1. Áp dụng quy tắc khai phương 1 thương, tính:
\(\frac{3\sqrt{128}}{\sqrt{2}}\)
2. Tính:
a. (\(\left(\sqrt{32}-\sqrt{50}+\sqrt{8}\right):\sqrt{2}\)
b. (\(5\sqrt{48}-3\sqrt{27}+2\sqrt{12}\)):\(\sqrt{3}\)
c. \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)
d. \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
e. \(\left(\sqrt{6}-\sqrt{2}\right)\sqrt{2+\sqrt{3}}\)
f. \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
Rút gọn:
a, \(\sqrt{\dfrac{4}{9-4\sqrt{5}}}\) -\(\sqrt{\dfrac{4}{9+4\sqrt{5}}}\)
b, \(\dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{2}}\)
c, \(\sqrt{14-8\sqrt{3}}\)-\(\sqrt{24-12\sqrt{3}}\)
d, \(\sqrt{2-\sqrt{3}}\)\(\times\)\(\left(\sqrt{5}+\sqrt{2}\right)\)