\(\sqrt{\dfrac{1}{600}};\,\,\sqrt{\dfrac{11}{540}};\,\,\sqrt{\dfrac{3}{50}};\,\,\sqrt{\dfrac{5}{98}}; \,\,\sqrt{\dfrac{(1-\sqrt{3})^{2}}{27}}.\) Lời giải: +\(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt 1}{\sqrt{600}}\)\(=\dfrac{ 1}{\sqrt{6.100}}\)\(=\dfrac{1}{\sqrt{6.10^2}}\) \(=\dfrac{ 1}{\sqrt{6}.\sqrt{10^2}}\)\(=\dfrac{ 1}{10\sqrt{6}}\)\(=\dfrac{ 1.\sqrt 6}{10.6}\)\(=\dfrac{ \sqrt 6}{60}\) +\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{11}}{\sqrt{540}}=\dfrac{\sqrt{11}}{\sqrt{36.15}}\) \(=\dfrac{\sqrt{11}}{\sqrt{36}.\sqrt{15}}=\dfrac{\sqrt{11}}{\sqrt{6^2}.\sqrt{15}}\) \(=\dfrac{\sqrt{11}}{6\sqrt{15}}=\dfrac{\sqrt{11}.\sqrt{15}}{6.15}\) \(=\dfrac{\sqrt{11.15}}{90}=\dfrac{\sqrt{165}}{90}\). + \(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt 3}{\sqrt{50}}=\dfrac{\sqrt 3}{\sqrt{25.2}}=\dfrac{\sqrt{3}}{\sqrt{25}.\sqrt{2}}\) \(=\dfrac{\sqrt{3}}{\sqrt{5^2}.\sqrt{2}}=\dfrac{\sqrt{3}}{5\sqrt{2}}=\dfrac{\sqrt{3}.\sqrt 2}{5.2}\) \(=\dfrac{\sqrt{3.2}}{10}=\dfrac{\sqrt{6}}{10}\) + \(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt 5}{\sqrt {98}}=\dfrac{\sqrt 5}{\sqrt{49.2}}=\dfrac{\sqrt 5}{\sqrt{49}\sqrt{2}}\) \(=\dfrac{\sqrt 5}{\sqrt{7^2}.\sqrt 2}=\dfrac{\sqrt 5}{7\sqrt 2}=\dfrac{\sqrt 5 . \sqrt 2}{7. 2}\) \(=\dfrac{\sqrt {5. 2}}{14}=\dfrac{\sqrt{10}}{14}\). +\(\sqrt{\dfrac{(1-\sqrt{3})^{2}}{27}}=\dfrac{\sqrt{(1-\sqrt 3)^2}}{\sqrt {27}}=\dfrac{\sqrt{(1-\sqrt 3)^2}}{\sqrt {9.3}}\) \(=\dfrac{\sqrt{(1-\sqrt 3)^2}}{\sqrt {3^2.3}}\)\(=\dfrac{|1-\sqrt{3}|}{3\sqrt {3}}\) Vì \(1< 3 \Leftrightarrow \sqrt 1 < \sqrt 3 \Leftrightarrow 1< \sqrt 3\) \( \Leftrightarrow 1- \sqrt 3 < 0\) \(\Leftrightarrow |1- \sqrt 3|=-(1-\sqrt 3)=-1 + \sqrt 3 = \sqrt 3 -1.\) Do đó: \(\dfrac{|1-\sqrt{3}|}{3\sqrt {3}}=\dfrac{\sqrt{3}-1}{3\sqrt {3}}=\dfrac{\sqrt 3(\sqrt{3}-1)}{9}=\dfrac{3-\sqrt 3}{9}.\) Bài 49 trang 29 SGK Toán lớp 9 tập 1 Câu hỏi: Khử mẫu của biểu thức lấy căn \(ab\sqrt{\dfrac{a}{b}};\,\,\, \dfrac{a}{b}\sqrt{\dfrac{b}{a}};\,\,\, \sqrt{\dfrac{1}{b}+\dfrac{1}{b^{2}}};\,\,\,\ \sqrt{\dfrac{9a^{3}}{36b}};\,\,\, 3xy\sqrt{\dfrac{2}{xy}}.\) (Giả thiết các biểu thức có nghĩa).
Lời giải: Theo đề bài các biểu thức đều có nghĩa. + Ta có \(ab\sqrt{\dfrac{a}{b}}=ab\sqrt{\dfrac{a.b}{b.b}}=ab\sqrt{\dfrac{ab}{b^2}}=ab\dfrac{\sqrt{ab}}{\sqrt{b^2}}=ab\dfrac{\sqrt{ab}}{\left | b \right |}.\) *) Nếu \( b > 0\) thì \(|b|=b \Rightarrow ab\dfrac{\sqrt{ab}}{\left | b \right |}=ab\dfrac{\sqrt{ab}}{b}=a\sqrt{ab}\). *) Nếu \( b < 0\) thì \(|b|=-b \Rightarrow ab\dfrac{\sqrt{ab}}{\left | b \right |}=-ab\dfrac{\sqrt{ab}}{b}=-a\sqrt{ab}\). + Ta có: \( \dfrac{a}{b}\sqrt{\dfrac{b}{a}}=\dfrac{a}{b}\sqrt{\dfrac{b.a}{a.a}}=\dfrac{a}{b}\sqrt{\dfrac{ab}{a^2}}\) \(=\dfrac{a}{b}.\dfrac{\sqrt{ab}}{\sqrt{a^2}}\)\(=\dfrac{a}{b}.\dfrac{\sqrt{ab}}{|a|}\)\(=\dfrac{a\sqrt{ab}}{b|a|}\) *) Nếu \(a> 0\) thì \( |a|=a \Rightarrow \dfrac{a\sqrt{ab}}{b|a|}=\dfrac{a\sqrt{ab}}{ab}=\dfrac{\sqrt{ab}}{b} .\) *) Nếu \(a<0\) thì \(|a|=-a \Rightarrow \dfrac{a\sqrt{ab}}{b|a|}=-\dfrac{a\sqrt{ab}}{ab}=-\dfrac{\sqrt{ab}}{b} .\) + Ta có: \(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\sqrt{\dfrac{b}{b^2}+\dfrac{1}{b^2}}=\sqrt{\dfrac{b+1}{b^2}}\) \(=\dfrac{\sqrt{b+1}}{\sqrt{b^2}}=\dfrac{\sqrt{b+1}}{|b|}\). *) Nếu \(b> 0\) thì \(|b|=b \Rightarrow \dfrac{\sqrt{b+1}}{|b|}=\dfrac{\sqrt{b+1}}{b}\). *) Nếu \(-1 \le b < 0\) thì \(|b|=-b \Rightarrow \dfrac{\sqrt{b+1}}{|b|}=-\dfrac{\sqrt{b+1}}{b}\). + Ta có: \(\sqrt{\dfrac{9a^3}{36b}}=\sqrt{\dfrac{9}{36}}.\sqrt{\dfrac{a^3}{b}}=\sqrt{\dfrac{1}{4}}.\sqrt{\dfrac{a^3.b}{b.b}}\) \(=\dfrac{1}{2}.\sqrt{\dfrac{a^2.ab}{b^2}}\)\(=\dfrac{1}{2}.\dfrac{\sqrt{a^2}.\sqrt{ab}}{\sqrt{b^2}}\) \(=\dfrac{1}{2}.\dfrac{|a|\sqrt{ab}}{|b|}=\dfrac{|a|\sqrt{ab}}{2|b|}\). *) Nếu \(a \ge 0,\ b > 0\) thì \(|a|=a,\ |b| =b \Rightarrow \dfrac{|a|\sqrt{ab}}{2|b|}=\dfrac{a\sqrt{ab}}{2b}\). *) Nếu \(a < 0,\ b < 0\) thì \(|a|=-a,\ |b| =-b \Rightarrow \dfrac{|a|\sqrt{ab}}{2|b|}=\dfrac{a\sqrt{ab}}{2b}\). (Chú ý: Theo đề bài \(\sqrt{\dfrac{9a^3}{36b}}\) có nghĩa nên \(a,\ b\) cùng dấu, do đó chỉ cần xét 2 trường hợp \(a,\ b\) cùng âm hoặc cùng dương). + Ta có: \(3xy\sqrt{\dfrac{2}{xy}}=3xy.\sqrt{\dfrac{2.xy}{xy.xy}}=3xy.\dfrac{\sqrt{2xy}}{\sqrt{(xy)^2}}\) \(=3xy.\dfrac{\sqrt{2xy}}{|xy|}\) \(=\dfrac{3xy.\sqrt{2xy}}{xy}=3\sqrt{2xy}\). (Vì theo đề bài \(\sqrt{\dfrac{2}{xy}}\) có nghĩa nên \(\dfrac{2}{xy} > 0 \Leftrightarrow xy > 0 \Rightarrow |xy|=xy\).) Bài 50 trang 30 SGK Toán lớp 9 tập 1 Câu hỏi: Trục căn thức ở mẫu với giả thiết các biểu thức chữ đều có nghĩa: \(\dfrac{5}{\sqrt{10}};\,\,\, \dfrac{5}{2\sqrt{5}};\,\,\, \dfrac{1}{3\sqrt{20}};\,\,\, \dfrac{2\sqrt{2}+2}{5\sqrt{2}};\,\,\, \dfrac{y+b.\sqrt{y}}{b. \sqrt{y}}.\) Phương pháp: + \( (\sqrt{a})^2=a\), với \(a \ge 0\). + \(\dfrac{a}{\sqrt b}=\dfrac{a\sqrt b}{b}\), \((b > 0)\). + \( \sqrt{A^2 B}=A\sqrt{B}\), nếu \(A,\ B \ge 0\). + \( \sqrt{A^2 B}=-A\sqrt B\), nếu \(A < 0,\ B \ge 0\). Lời giải: + Ta có: \(\dfrac{5}{\sqrt{10}}=\dfrac{5.\sqrt{10}}{\sqrt{10}.\sqrt{10}}=\dfrac{5\sqrt{10}}{(\sqrt{10})^2}=\dfrac{5\sqrt{10}}{10}\) \(=\dfrac{5.\sqrt{10}}{5.2}\)\(=\dfrac{\sqrt{10}}{2}\). + Ta có: \(\dfrac{5}{2\sqrt{5}}=\dfrac{5.\sqrt 5}{2\sqrt 5.\sqrt 5}=\dfrac{5\sqrt{5}}{2.(\sqrt 5.\sqrt 5)}=\dfrac{5\sqrt{5}}{2(\sqrt 5)^2}\) \(=\dfrac{5\sqrt 5}{2.5}=\dfrac{\sqrt 5}{2}\). + Ta có: \(\dfrac{1}{3\sqrt{20}}=\dfrac{1.\sqrt{20}}{3\sqrt{20}.\sqrt{20}}=\dfrac{\sqrt{20}}{3.(\sqrt{20}.\sqrt{20})}=\dfrac{\sqrt{20}}{3.(\sqrt{20})^2}\) \(=\dfrac{\sqrt{20}}{3.20}=\dfrac{\sqrt{2^2.5}}{60}=\dfrac{2\sqrt 5}{60}=\dfrac{2\sqrt 5}{2.30}=\dfrac{\sqrt 5}{30}\). + Ta có: \(\dfrac{(2\sqrt{2}+2)}{5.\sqrt 2}=\dfrac{(2\sqrt 2+2).\sqrt 2}{5\sqrt 2. \sqrt 2}=\dfrac{2\sqrt 2.\sqrt 2+2.\sqrt 2}{5.(\sqrt 2)^2}\) \(=\dfrac{2.2+2\sqrt 2}{5.2}=\dfrac{2(2+\sqrt 2)}{5.2}=\dfrac{2+\sqrt 2}{5}\). + Ta có: \(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}=\dfrac{(y+b\sqrt y).\sqrt y}{b\sqrt y .\sqrt y}=\dfrac{y\sqrt y+b\sqrt y.\sqrt y}{b.(\sqrt y)^2}\) \(= \dfrac{y\sqrt y+b(\sqrt y)^2}{by}=\dfrac{y\sqrt y+by}{by}\) \(=\dfrac{y(\sqrt y+b)}{b.y}=\dfrac{\sqrt y+b}{b}\). Cách khác: \(\dfrac{{y + b\sqrt y }}{{b\sqrt y }} = \dfrac{{{{\left( {\sqrt y } \right)}^2} + b\sqrt y }}{{b\sqrt y }} \)\(= \dfrac{{\sqrt y \left( {\sqrt y + b} \right)}}{{b\sqrt y }} = \dfrac{{\sqrt y + b}}{b}\) Bài 51 trang 30 SGK Toán lớp 9 tập 1 Câu hỏi: Trục căn thức ở mẫu với giả thiết các biểu thức chữ đều có nghĩa: \(\dfrac{3}{\sqrt{3}+1};\,\,\,\dfrac{2}{\sqrt{3}-1};\,\,\,\dfrac{2+\sqrt{3}}{2-\sqrt{3}};\,\,\,\dfrac{b}{3+\sqrt{b}};\,\,\,\dfrac{p}{2\sqrt{p}-1}.\) Lời giải: + Ta có: \(\dfrac{3}{\sqrt{3}+1}=\dfrac{3(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\dfrac{3\sqrt 3 - 3.1}{(\sqrt 3)^2-1^2}\) \(=\dfrac{3\sqrt 3 -3}{3-1}=\dfrac{3\sqrt{3}-3}{2}\). + Ta có: \(\dfrac{2}{\sqrt{3}-1}=\dfrac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\dfrac{2(\sqrt 3 + 1)}{(\sqrt 3)^2-1^2}\) \(=\dfrac{2(\sqrt 3 + 1)}{3-1}=\dfrac{2(\sqrt{3}+1)}{2}=\sqrt{3}+1\). + Ta có: \(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=\dfrac{(2+\sqrt{3}).(2+\sqrt 3)}{(2-\sqrt{3})(2+\sqrt{3})}=\dfrac{(2+\sqrt{3})^2}{2^2-(\sqrt{3})^2}\) \(=\dfrac{2^2+2.2.\sqrt 3+(\sqrt{3})^2}{4-3}\) \(=\dfrac{7+4\sqrt 3}{1}=7+4\sqrt{3}\). + Ta có: \(\dfrac{b}{3+\sqrt{b}}=\dfrac{b(3-\sqrt{b})}{(3+\sqrt{b})(3-\sqrt{b})}\) \(=\dfrac{b(3-\sqrt{b})}{3^2-(\sqrt b)^2}=\dfrac{b(3-\sqrt{b})}{9-b};(b\neq 9)\). + Ta có: \(\dfrac{p}{2\sqrt{p}-1}=\dfrac{p(2\sqrt{p}+1)}{(2\sqrt{p}-1)(2\sqrt{p}+1)}\) \(=\dfrac{2p\sqrt{p}+p}{(2\sqrt{p})^2-1^2}\) \(=\dfrac{2p\sqrt{p}+p}{4p-1}\) Bài 52 trang 30 SGK Toán lớp 9 tập 1 Câu hỏi: Trục căn thức ở mẫu với giả thiết các biểu thức chữ đều có nghĩa: \(\dfrac{2}{\sqrt{6}-\sqrt{5}};\,\,\ \dfrac{3}{\sqrt{10}+\sqrt{7}};\,\,\, \dfrac{1}{\sqrt{x}-\sqrt{y}};\,\,\, \dfrac{2ab}{\sqrt{a}-\sqrt{b}}\). Lời giải: + Ta có: \(\dfrac{2}{\sqrt{6}-\sqrt{5}}=\dfrac{2(\sqrt{6}+\sqrt{5})}{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})}\) \(=\dfrac{2(\sqrt{6}+\sqrt{5})}{(\sqrt{6})^2-(\sqrt{5})^2}=\dfrac{2(\sqrt{6}+\sqrt{5})}{6-5}\) \(=\dfrac{2(\sqrt{6}+\sqrt{5})}{1}=2(\sqrt{6}+\sqrt{5})\). + Ta có: \(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\dfrac{3(\sqrt{10}-\sqrt{7})}{(\sqrt{10}+\sqrt{7})(\sqrt{10}-\sqrt{7})}\) \(=\dfrac{3(\sqrt{10}-\sqrt{7})}{(\sqrt{10})^2-(\sqrt{7})^2}\)\(=\dfrac{3(\sqrt{10}-\sqrt{7})}{10-7}\) \(=\dfrac{3(\sqrt{10}-\sqrt{7})}{3}=\sqrt{10}-\sqrt{7}\). + Ta có: \(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{1.(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}\) \(=\dfrac{\sqrt x + \sqrt y}{(\sqrt x)^2-(\sqrt y)^2}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\) + Ta có: \(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}=\dfrac{2ab(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}\) |
