Trục căn thức ở mẫu:
a) \(\dfrac{4+3\sqrt{5}}{\sqrt{5}}\); b) \(\dfrac{1}{\sqrt{5}-2}\);
c) \(\dfrac{3+\sqrt{3}}{1-\sqrt{3}}\); d) \(\dfrac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}\).
Viết số nghịch đảo của mỗi số sau dưới dạng không chứa dấu căn ở mẫu:
a) \(4\sqrt{3}\)
b) \(3\sqrt{2}+2\sqrt{3}\)
c) \(\dfrac{5+\sqrt{5}}{4\sqrt{2}}\)
a) \(\dfrac{1}{4\sqrt{3}}=\dfrac{\sqrt{3}}{12}\)
b) \(\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{3\sqrt{2}-2\sqrt{3}}{6}\)
c) \(\dfrac{4\sqrt{2}}{5+\sqrt{5}}=\dfrac{4\sqrt{2}\left(5-\sqrt{5}\right)}{20}=\dfrac{5\sqrt{2}-\sqrt{10}}{5}\)
\(a.\)
\(\dfrac{1}{4\sqrt{3}}=\dfrac{\sqrt{3}}{12}\)
\(b.\)
\(\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{3\sqrt{2}-2\sqrt{3}}{\left(3\sqrt{2}\right)^2-\left(2\sqrt{3}\right)^2}=\dfrac{3\sqrt{2}-2\sqrt{3}}{6}\)
\(c.\)
\(\dfrac{4\sqrt{2}}{5+\sqrt{5}}=\dfrac{4\sqrt{2}\cdot\left(5-\sqrt{5}\right)}{5^2-\left(\sqrt{5}\right)^2}=\dfrac{\sqrt{2}\cdot\left(5-\sqrt{5}\right)}{5}\)
có ai biết giải bài này k hộ mình vs ( chi tiết hộ mình nhé )
bài 1: trục căn thức ở mẫu và rút gọn
a, \(\dfrac{1}{2\sqrt{2}-3\sqrt{3}}\)
b, \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
bài 2: trục căn thức ở mẫu và rút gọn
a, \(\dfrac{\sqrt{8}}{\sqrt{5}-\sqrt{3}}\)
b, \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
bài 3: trục căn thức và thực hiện phép tính
a, M=\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
b, N= \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
Bài 1:
a.
\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)
b.
\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)
Bài 2.
a.
\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)
b.
\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)
Bài 3:
a.
\(M=\left[\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right](\sqrt{6}+11)\)
\(=\left[\frac{15(\sqrt{6}-1)}{6-1}+\frac{4(\sqrt{6}+2)}{6-2^2}-\frac{12(3+\sqrt{6})}{3^2-6}\right](\sqrt{6}+11)\)
\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)
b.
\(N=\left[1-\frac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{5}+1}\right].\left[\frac{\sqrt{5}(\sqrt{5}-1)}{1-\sqrt{5}}-1\right]\)
\(=(1-\sqrt{5})(-\sqrt{5}-1)=(\sqrt{5}-1)(\sqrt{5}+1)=5-1=4\)
Trục căn thức ở mẫu
a) \(\dfrac{7}{\sqrt{12}}\)
b)\(\dfrac{3}{2\sqrt{3}}\)
c)\(\dfrac{1}{5\sqrt{12}}\)
d)\(\dfrac{2\sqrt{3}+3}{4\sqrt{3}}\)
\(a,\dfrac{7}{\sqrt{12}}=\dfrac{7\sqrt{3}}{\sqrt{12}\cdot\sqrt{3}}\)
\(=\dfrac{7\sqrt{3}}{\sqrt{36}}=\dfrac{7\sqrt{3}}{6}\)
\(b,\dfrac{3}{2\sqrt{3}}=\dfrac{3\sqrt{3}}{2\sqrt{3}\cdot\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}}{2\cdot3}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)
\(c,\dfrac{1}{5\sqrt{12}}=\dfrac{\sqrt{3}}{5\cdot2\sqrt{3}\cdot\sqrt{3}}\)
\(=\dfrac{\sqrt{3}}{10\cdot3}=\dfrac{\sqrt{3}}{30}\)
\(d,\dfrac{2\sqrt{3}+3}{4\sqrt{3}}=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{4\sqrt{3}}\)
\(=\dfrac{2+\sqrt{3}}{4}\)
a) \(\dfrac{7}{\sqrt[]{12}}=\dfrac{7}{2\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{2\sqrt[]{3}.\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{6}\)
b) \(\dfrac{3}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}.\sqrt[]{3}}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{2}\)
c) \(\dfrac{1}{5\sqrt[]{12}}=\dfrac{1}{10\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{10\sqrt[]{3}.\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{30}\)
d) \(\dfrac{2\sqrt[]{3}+3}{4\sqrt[]{3}}=\dfrac{\sqrt[]{3}\left(2\sqrt[]{3}+3\right)}{4\sqrt[]{3}.\sqrt[]{3}}=\dfrac{3\left(2+\sqrt[]{3}\right)}{12}=\dfrac{2+\sqrt[]{3}}{4}\)
Trục căn thức ở mẫu và rút gọn
a,\(\dfrac{\sqrt{2}}{\sqrt{5}-\sqrt{3}}\) b,\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
c,\(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}\) d,\(\dfrac{2\sqrt{6}-\sqrt{10}}{4\sqrt{3}-2\sqrt{5}}\)
* Trục căn thức ở mẫu
a. \(\dfrac{7}{\sqrt{5}-\sqrt{3}-\sqrt{7}}\)
b. \(\dfrac{5}{2-\sqrt{3}-\sqrt{5}}\)
c. \(\dfrac{59}{\sqrt[3]{5}+\sqrt{3}-\sqrt{2}}\)
* Trục căn thức ở mẫu
a. \(\dfrac{7}{\sqrt{5}-\sqrt{3}-\sqrt{7}}\)
b. \(\dfrac{5}{2-\sqrt{3}-\sqrt{5}}\)
c. \(\dfrac{59}{\sqrt[3]{5}+\sqrt{3}-\sqrt{2}}\)
a) \(\dfrac{7}{\sqrt{5}-\sqrt{3}-\sqrt{7}}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}+\sqrt{7}\right)}{\left(\sqrt{5}-\sqrt{3}\right)^2-7}\)
\(=\dfrac{7\sqrt{5}-7\sqrt{3}+7\sqrt{7}}{8-2\sqrt{15}-7}\)
\(=\dfrac{7\sqrt{5}-7\sqrt{3}+7\sqrt{7}}{1-2\sqrt{15}}\)
\(=\dfrac{\left(7\sqrt{5}-7\sqrt{3}+7\sqrt{7}\right)\left(1+2\sqrt{15}\right)}{1-60}\)
\(=\dfrac{7\sqrt{5}+70\sqrt{3}-7\sqrt{3}-42\sqrt{5}+7\sqrt{7}+14\sqrt{105}}{-59}\)
\(=\dfrac{-35\sqrt{5}+63\sqrt{3}+7\sqrt{7}+14\sqrt{105}}{-59}\)
\(=\dfrac{35\sqrt{5}-63\sqrt{3}-7\sqrt{7}-14\sqrt{105}}{59}\)
* Trục căn thức ở mẫu
a. \(\dfrac{1}{\sqrt{5}-\sqrt{3}-\sqrt{2}}\)
b. \(\dfrac{2}{-1-\sqrt{2}+\sqrt{3}}\)
c. \(\dfrac{5}{\sqrt[3]{5}+\sqrt{3}}\)
* Trục căn thức ở mẫu
a. \(\dfrac{1}{\sqrt{5}-\sqrt{3}-\sqrt{2}}\)
b. \(\dfrac{2}{-1-\sqrt{2}+\sqrt{3}}\)
c. \(\dfrac{5}{\sqrt[3]{5}+\sqrt{3}}\)
a) Ta có: \(\dfrac{1}{\sqrt{5}-\sqrt{3}-\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{5-\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{5-5-2\sqrt{6}}\)
\(=\dfrac{-\sqrt{5}-\sqrt{3}-\sqrt{2}}{2\sqrt{6}}\)
\(=\dfrac{-\sqrt{6}\left(\sqrt{5}+\sqrt{3}+\sqrt{2}\right)}{12}\)
b) Ta có: \(\dfrac{2}{-1-\sqrt{2}+\sqrt{3}}\)
\(=\dfrac{2\left(-1-\sqrt{2}-\sqrt{3}\right)}{\left(-1-\sqrt{2}\right)^2-3}\)
\(=\dfrac{\left(-1-\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}}\)
\(=\dfrac{-\sqrt{2}-2-\sqrt{6}}{2}\)
* Trục căn thức ở mẫu
a. \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{7}}\)
b. \(\dfrac{5}{1-\sqrt{2}-\sqrt{3}}\)
c. \(\dfrac{59}{\sqrt[3]{5}+\sqrt{3}+\sqrt{2}}\)
* Trục căn thức ở mẫu
a. \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{7}}\)
b. \(\dfrac{5}{1-\sqrt{2}-\sqrt{3}}\)
c.\(\dfrac{59}{\sqrt[3]{5}+\sqrt{3}+\sqrt{2}}\)
a) Ta có: \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{7}}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)}{\left(\sqrt{5}-\sqrt{3}\right)^2-7}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)}{1-2\sqrt{15}}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)\left(1+2\sqrt{15}\right)}{1-60}\)
\(=\dfrac{-7\left(\sqrt{5}+10\sqrt{3}-\sqrt{3}-6\sqrt{5}-\sqrt{7}-2\sqrt{105}\right)}{59}\)
\(=\dfrac{-7\left(-5\sqrt{5}+9\sqrt{3}-\sqrt{7}-2\sqrt{105}\right)}{59}\)