BT1: Tính
A = (\(\frac{3}{2}\) .\(\sqrt{6}\)+ \(2\sqrt{\frac{2}{3}}\)- \(4\sqrt{\frac{3}{2}}\)) . (\(3\sqrt{\frac{2}{3}}\)- \(\sqrt{12}\)- \(\sqrt{6}\)
BT2: Rút gon
A = \(\frac{1}{\sqrt{1}+\sqrt{2}}\)+ \(\frac{1}{\sqrt{2}+\sqrt{3}}\)+ \(\frac{1}{\sqrt{3}+\sqrt{4}}\)+ ....... + \(\frac{1}{\sqrt{2024}+\sqrt{2025}}\)
CM: B = \(\frac{1}{\sqrt{1}}\)+ \(\frac{1}{\sqrt{2}}\)+ ...... + \(\frac{1}{\sqrt{2024}}\)> 88
BT3: Rút gọn
C = \(\sqrt{2a+\sqrt{4x-1}}\)+ \(\sqrt{2a-\sqrt{4a-1}}\)với \(\frac{1}{4}\)< a < \(\frac{1}{2}\)
BT4:
Hỏi M = \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}\)- \(\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)có phải là số tự nhiên không?
BT5:
Phân tích thành nhân tử: M = \(7\sqrt{x-1}\)- \(\sqrt{x^3-x^2}\)+ x - 1 (với x >=1)
GIÚP MÌNH VỚI Ạ MÌNH CẦN GẤP LẮM. CẨM ƠN NHIỀU ẠAAAAAAAA
Bài 2:
a) \(\frac{1}{\sqrt{1}+\sqrt{2}}=\frac{2-1}{\sqrt{1}+\sqrt{2}}=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{1}+\sqrt{2}}=\sqrt{2}-\sqrt{1}\)
Tương tự ta có: \(\frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2}\);
\(\frac{1}{\sqrt{3}+\sqrt{4}}=\sqrt{4}-\sqrt{3}\); ............. ; \(\frac{1}{\sqrt{2024}+\sqrt{2025}}=\sqrt{2025}-\sqrt{2024}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+......+\sqrt{2025}-\sqrt{2024}\)
\(=\sqrt{2025}-\sqrt{1}=45-1=44\)
Bài 4:
\(M=\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\frac{\sqrt{2-2\sqrt{2}+1}}{\sqrt{9-2.3.2\sqrt{2}+8}}-\frac{\sqrt{2+2\sqrt{2}+1}}{\sqrt{9+2.3.2\sqrt{2}+8}}\)
\(=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-\sqrt{8}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+\sqrt{8}\right)^2}}\)
\(=\frac{\left|\sqrt{2}-1\right|}{\left|3-\sqrt{8}\right|}-\frac{\left|\sqrt{2}+1\right|}{\left|3+\sqrt{8}\right|}=\frac{\sqrt{2}-1}{3-\sqrt{8}}-\frac{\sqrt{2}+1}{3+\sqrt{8}}\)
\(=\frac{\left(\sqrt{2}-1\right)\left(3+\sqrt{8}\right)}{\left(3-\sqrt{8}\right)\left(3+\sqrt{8}\right)}-\frac{\left(\sqrt{2}+1\right)\left(3-\sqrt{8}\right)}{\left(3+\sqrt{8}\right)\left(3-\sqrt{8}\right)}\)
\(=\left(3\sqrt{2}+\sqrt{16}-3-\sqrt{8}\right)-\left(3\sqrt{2}-\sqrt{16}+3-\sqrt{8}\right)\)
\(=3\sqrt{2}+4-3-\sqrt{8}-3\sqrt{2}+4-3+\sqrt{8}\)
\(=8-6=2\)là số tự nhiên
