Câu 2:

a: \(=\sqrt{\left(37-35\right)\left(37+35\right)}=\sqrt{72\cdot2}=12\)

b: \(=\sqrt{\left(65-63\right)\left(65+63\right)}=\sqrt{128\cdot2}=16\)

c: \(=\sqrt{\left(221-220\right)\left(221+220\right)}=\sqrt{441}=21\)

d: \(=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{225\cdot9}=3\cdot15=45\)

6: \(=3\cdot2\sqrt{3}-4\cdot3\sqrt{3}+5\cdot4\sqrt{3}=14\sqrt{3}\)

7: \(=2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=3\sqrt{3}\)

8: \(=2\cdot4\sqrt{2}+4\cdot2\sqrt{2}-5\cdot3\sqrt{2}=\sqrt{2}\)

9: \(=3\cdot2\sqrt{5}-2\cdot3\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)

10: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)

1) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}\)

\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)

\(=-6\sqrt{2}\)

2) \(\sqrt{50}-\sqrt{18}+\sqrt{200}-\sqrt{162}\)

\(=5\sqrt{2}-3\sqrt{2}+10\sqrt{2}-9\sqrt{2}\)

\(=3\sqrt{2}\)

3) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)

\(=5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)

\(=-2\sqrt{5}\)

4) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)

\(=20\sqrt{3}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)

\(=4\sqrt{3}\)

5) \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)

\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{3}\sqrt{3}\)

\(=-\dfrac{17}{3}\sqrt{3}\)

bn ghi đề rõ hơn ik bn

a) Ta có: \(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{12}{3-\sqrt{3}}\)

\(=\dfrac{2\left(\sqrt{3}+1\right)}{2}-\dfrac{3\left(2+\sqrt{3}\right)}{1}+\dfrac{12\left(3+\sqrt{3}\right)}{6}\)

\(=\sqrt{3}+1-6-3\sqrt{3}+6+2\sqrt{3}\)

\(=1\)

b) Ta có: \(\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{2}{\sqrt{7}+\sqrt{5}}-\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{7}+\sqrt{3}}\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{7}+\sqrt{5}-\sqrt{5}-\sqrt{2}+\sqrt{7}-\sqrt{3}\)

=0

a) Ta có: \(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{12}{3-\sqrt{3}}\)

\(=\dfrac{2\left(\sqrt{3}+1\right)}{2}-\dfrac{3\left(2+\sqrt{3}\right)}{1}+\dfrac{12\left(3+\sqrt{3}\right)}{6}\)

\(=\sqrt{3}+1-6-3\sqrt{3}+2\left(3+\sqrt{3}\right)\)

\(=-2\sqrt{3}-5+6+2\sqrt{3}\)

=1

b) Ta có: \(\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{2}{\sqrt{7}+\sqrt{5}}-\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{7}+\sqrt{3}}\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{7}+\sqrt{5}-\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{3}\)

\(=\sqrt{2}-\sqrt{3}\)

Ứng dụng giải toán đã được review rất hay bởi trang báo uy tín https://www.facebook.com/docbaoonlinethayban/videos/467035000526358/?v=467035000526358 Cả nhà tải ngay bằng link dưới đây nhé. https://giaingay.com.vn/downapp.html

Đề thi thử + tính điểm với những đề mới nhất cả nhà tải app dùng thử nhé https://giaingay.com.vn/downapp.html

Ứng dụng giải toán đã được review rất hay bởi trang báo uy tín https://www.facebook.com/docbaoonlinethayban/videos/467035000526358/?v=467035000526358 Cả nhà tải ngay bằng link dưới đây nhé. https://giaingay.com.vn/downapp.html

\(\sqrt{\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}+\sqrt{\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}\)

\(=\sqrt{\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2}{3-2}}+\sqrt{\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2}{3-2}}\)

\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}=2\sqrt{3}\)

no biết