Khử mẫu của biểu thức lấy căn
1 600 ; 11 540 ; 3 50 ; 5 98 ; 1 - 3 2 27
Khử mẫu của biểu thức lấy căn:
\(\sqrt{\dfrac{1}{600}};\sqrt{\dfrac{11}{540}};\sqrt{\dfrac{3}{50}};\sqrt{\dfrac{5}{98}};\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}.\)
Bài 48 (trang 29 SGK Toán 9 Tập 1)
Khử mẫu của biểu thức lấy căn
$\sqrt{\dfrac{1}{600}}; \sqrt{\dfrac{11}{540}}$ ; $\sqrt{\dfrac{3}{50}} ; \sqrt{\dfrac{5}{98}}$ ; $\sqrt{\dfrac{(1-\sqrt{3})^{2}}{27}}$
\(\sqrt{\dfrac{1}{600}}\)=\(\sqrt{\dfrac{1}{10^2\cdot6}}\)=\(\sqrt{\dfrac{1\cdot6}{10^2\cdot6\cdot6}}\)=\(\dfrac{\sqrt{6}}{60}\)
\(\sqrt{\dfrac{11}{540}}\)=\(\sqrt{\dfrac{11\cdot540}{540\cdot540}}\)=\(\dfrac{\sqrt{5940}}{540}\)=\(\dfrac{\sqrt{165}}{90}\)
\(\sqrt{\dfrac{3}{50}}\)=\(\sqrt{\dfrac{3\cdot50}{50\cdot50}}\)=\(\dfrac{\sqrt{150}}{50}\)=\(\dfrac{\sqrt{6}}{10}\)
\(\sqrt{\dfrac{5}{98}}\)=\(\sqrt{\dfrac{5\cdot98}{98\cdot98}}=\dfrac{\sqrt{490}}{98}=\dfrac{\sqrt{10}}{14}\)
\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)
\(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{6}}{60}\)
\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{165}}{90}\)
\(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{6}}{10}\)
\(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{10}}{14}\)
\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)
Khử mẫu của biểu thức lấy căn
a) \(\sqrt{\dfrac{1}{600}}\) b) \(\sqrt{\dfrac{11}{540}}\) c) \(\sqrt{\dfrac{3}{50}}\) d) \(\sqrt{\dfrac{5}{98}}\) e)\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}\)
a) \(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{1}}{10\sqrt{6}}=\dfrac{\sqrt{1}.\sqrt{6}}{10\sqrt{6}.\sqrt{6}}=\dfrac{\sqrt{6}}{60}\)
b) \(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{11}}{6\sqrt{15}}=\dfrac{\sqrt{11}.\sqrt{15}}{6\sqrt{15}.\sqrt{15}}=\dfrac{\sqrt{165}}{90}\)
c) \(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{3}}{5\sqrt{2}}=\dfrac{\sqrt{3}.\sqrt{2}}{5\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6}}{10}\)
d) \(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{5}}{7\sqrt{2}}=\dfrac{\sqrt{5}.\sqrt{2}}{7\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{10}}{14}\)
e) \(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{\sqrt{\left(1-\sqrt{3}\right)^2}}{3\sqrt{3}}=\dfrac{\sqrt{3}-1}{3\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{3\sqrt{3}.\sqrt{3}}=\dfrac{3-\sqrt{3}}{9}\)
\(\sqrt{\dfrac{1}{600}}=\sqrt{\dfrac{1\cdot6}{600\cdot6}}=\sqrt{\dfrac{6}{60^2}}=\dfrac{\sqrt{6}}{60}\)
\(\sqrt{\dfrac{11}{540}}=\sqrt{\dfrac{11\cdot15}{540\cdot15}}=\sqrt{\dfrac{165}{90^2}}=\dfrac{\sqrt{165}}{90}\)
\(\sqrt{\dfrac{3}{50}}=\sqrt{\dfrac{3\cdot2}{50\cdot2}}=\sqrt{\dfrac{6}{10^2}}=\dfrac{\sqrt{6}}{10}\)
\(\sqrt{\dfrac{5}{98}}=\sqrt{\dfrac{5\cdot2}{98\cdot2}}=\sqrt{\dfrac{10}{12^2}}=\dfrac{\sqrt{10}}{12}\)
\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\sqrt{\dfrac{3\left(1-\sqrt{3}\right)^2}{27\cdot3}}\)
\(=\dfrac{\sqrt{3\left(1-\sqrt{3}\right)^2}}{\sqrt{9^2}}=\dfrac{\left|1-\sqrt{3}\right|\cdot\sqrt{3}}{9}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\sqrt{3}}{9}\)
Khử mẫu của bthuc lấy căn
a)√3/2a^2
b)√1/600
√11/540
√3/50
√5/98
c)√(1-√3)^2/27
d)√2/3
e)√x^2/5
f) √3/x
g)√x^2- x^2/7
h)ab√a/b
i)a/b√a/b
√1/b +1/b^2
√9a^3/36b
3ab√2/ab
a: \(\sqrt{\dfrac{3}{2}a^2}=\left|a\right|\cdot\dfrac{\sqrt{6}}{2}\)
b: \(\sqrt{\dfrac{1}{600}}=\dfrac{1}{10\sqrt{6}}=\dfrac{\sqrt{6}}{60}\)
\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{165}}{90}\)
\(\sqrt{\dfrac{3}{50}}=\sqrt{\dfrac{6}{100}}=\dfrac{\sqrt{6}}{10}\)
\(\sqrt{\dfrac{5}{98}}=\sqrt{\dfrac{10}{196}}=\dfrac{1}{14}\cdot\sqrt{10}\)
c: \(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{\sqrt{3}-1}{3\sqrt{3}}=\dfrac{3-\sqrt{3}}{9}\)
d: căn 2/3=căn 6/9=1/3*căn 6
e: \(\sqrt{\dfrac{x^2}{5}}=\sqrt{\dfrac{5x^2}{25}}=\pm\dfrac{x\sqrt{5}}{5}\)
f: \(\sqrt{\dfrac{3}{x}}=\sqrt{\dfrac{3x}{x^2}}=\dfrac{\sqrt{3x}}{\left|x\right|}\)
Khử mẫu của biểu thức lấy căn
\(\sqrt{\frac{1}{600}}\) ; \(\sqrt{\frac{11}{540}}\) ; \(\sqrt{\frac{3}{50}}\) ; \(\sqrt{\frac{5}{98}}\) ; \(\sqrt{\frac{\left(1-\sqrt{3}\right)^2}{27}}\)
ab\(\sqrt{\frac{a}{b}}\) ; \(\frac{a}{b}\)\(\sqrt{\frac{b}{a}}\) ; \(\sqrt{\frac{1}{b}+\frac{1}{b^2}}\) ; \(\sqrt{\frac{9a^3}{36b}}\) ; 3xy\(\sqrt{\frac{2}{xy}}\)
(Gỉa thiế các biểu thức có nghĩa
\(\sqrt{\frac{1}{600}}=\sqrt{\frac{6}{3600}}=\frac{\sqrt{6}}{\sqrt{3600}}=\frac{\sqrt{6}}{60}\)
\(\sqrt{\frac{11}{540}}=\sqrt{\frac{11}{36.15}}=\frac{1}{6}\sqrt{\frac{165}{15^2}}=\frac{1}{6}.\frac{\sqrt{165}}{15}=\frac{\sqrt{165}}{90}\)
\(\sqrt{\frac{3}{50}}=\sqrt{\frac{3}{25.2}}=\frac{1}{5}\sqrt{\frac{3}{2}}=\frac{1}{5}\sqrt{\frac{6}{4}}=\frac{1}{5}.\frac{\sqrt{6}}{2}=\frac{\sqrt{6}}{10}\)
\(\sqrt{\frac{5}{98}}=\sqrt{\frac{5}{49.2}}=\frac{1}{7}\sqrt{\frac{5}{2}}=\frac{1}{7}.\sqrt{\frac{10}{4}}=\frac{\sqrt{10}}{14}\)
\(\sqrt{\frac{\left(1-\sqrt{3}\right)^2}{27}}=\frac{\left|1-\sqrt{3}\right|}{\sqrt{9.3}}=\frac{\sqrt{3}-1}{3\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{9}\)
1) thực hiện phép tính
\(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)
2) trục căn thức ở mẫu : \(\dfrac{2}{\sqrt{3}-5}\)
3) khử mẫu của biểu thức lấy căn: \(\sqrt{\dfrac{2}{5}}\)
1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)
\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)
\(=6\sqrt{3}+2\sqrt{3}-3\sqrt{3}\)
\(=5\sqrt{3}\)
2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)
\(=\dfrac{2\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}\)
\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)
\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)
\(=\dfrac{-\sqrt{3}-5}{11}\)
3) Ta có: \(\sqrt{\dfrac{2}{5}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{5}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)
\(=\dfrac{\sqrt{10}}{5}\)
Khử mẫu của biểu thức lấy căn \(\sqrt{\dfrac{3}{\left(-4\right)^2}}\)
\(\sqrt{\dfrac{3}{\left(-4\right)^2}}=\dfrac{\sqrt{3}}{\sqrt{\left(-4\right)^2}}=\dfrac{\sqrt{3}}{4}\)
\(\sqrt{\dfrac{3}{\left(-4\right)^2}}=\dfrac{\sqrt{3}}{4}\)
Khử mẫu của biểu thức lấy căn
4 5
Khử mẫu của biểu thức lấy căn
\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}\)
(giả thiết các biểu thức có nghĩa)
\(=\sqrt{\dfrac{b+1}{b^2}}=\left[{}\begin{matrix}\dfrac{\sqrt{b+1}}{b}\left(b>0\right)\\-\dfrac{\sqrt{b+1}}{b}\left(-1\le b< 0\right)\end{matrix}\right.\)