Trục căn thức ở mẫu:
a) \(\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{6}}\)
b)\(\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Trục căn thức ở mẫu
a) A= \(\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{6}}\)
b) B = \(\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
c) C = \(\frac{12}{\sqrt{7}+2\sqrt{6}}\)
trục căn thức ở mẫu :
a,\(\frac{3}{\sqrt{5}};\frac{2\sqrt{3}}{\sqrt{2}};\frac{a}{\sqrt{b}};\frac{x+1}{\sqrt{x^2-1}}\)
b,\(\frac{1}{\sqrt{3}+\sqrt{2}};\frac{2}{2-\sqrt{3}};\frac{\sqrt{2}+1}{\sqrt{2}-1};\frac{3\sqrt{2}}{\sqrt{3}+1}\)
c,\(\frac{1}{1+\sqrt{2}+\sqrt{3}}\)
d,\(\frac{1}{\sqrt{2\sqrt{3}-\sqrt{2}}.\sqrt{2}.\sqrt{\sqrt{2}+\sqrt{3}}}\)
a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)
\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)
\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)
\(\frac{x+1}{\sqrt{x^2-1}}=\frac{\left(x+1\right)\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2-1}\right)}\) = \(\frac{\left(\sqrt{x^2-1}\right)\left(x+1\right)}{x^2-1}\)
câu c chắc là như này
\(\frac{1}{1+\sqrt{2}+\sqrt{3}}=1+\frac{1}{\sqrt{2}+\sqrt{3}}\) = \(1+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}\)
= \(1+\frac{\sqrt{2}-\sqrt{3}}{2-3}=1+\frac{\sqrt{2}-\sqrt{3}}{-1}\) = \(1-\sqrt{2}+\sqrt{3}\)
Trục căn ở mẫu:
\(a)\frac{5}{\sqrt{10}}\\ b)\frac{-2}{1-\sqrt{5}}\\ c)\frac{4}{\sqrt{3}+\sqrt{2}}\\ d)\frac{1}{3-2\sqrt{2}}\\ e)\frac{6-\sqrt{6}}{1-\sqrt{6}}\\ g)\frac{3\sqrt{2}-2\sqrt{3}}{2\left(\sqrt{3}-\sqrt{2}\right)}\\ h)\frac{\sqrt{3}-3}{\sqrt{3}-1}\\ i)\frac{\sqrt{15}}{5\sqrt{3}+3\sqrt{5}}\)
Trục căn ở mẫu các biểu thức sau:
A=\(\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
2.trục căn thức ở mẫu sau:
a\(\frac{2xy}{2\sqrt{x}+3\sqrt{y}}\)
b\(\frac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}}\)
c\(\frac{2}{\sqrt{3}+1}\)
d\(\frac{6}{2\sqrt{3}+\sqrt{2}}\)
\(a,\frac{2xy}{2\sqrt{x}+3\sqrt{y}}=\frac{2xy.\left(2\sqrt{x}-3\sqrt{y}\right)}{\left(2\sqrt{x}+3\sqrt{y}\right)\left(2\sqrt{x}-3\sqrt{y}\right)}=\frac{4x\sqrt{x}y-6xy\sqrt{y}}{2x-3y}\)
\(b,\frac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\sqrt{x}}{2\sqrt{x}\sqrt{x}}=\frac{x+\sqrt{xy}}{2x}\)
\(c,\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{2.\left(\sqrt{3}-1\right)}{2}=\sqrt{3}-1\)
\(d,\frac{6}{2\sqrt{3}+\sqrt{2}}=\frac{6\left(2\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{3}+\sqrt{2}\right)\left(2\sqrt{3}-\sqrt{2}\right)}=\frac{6\left(2\sqrt{3}-\sqrt{2}\right)}{10}=\frac{6\sqrt{3}-3\sqrt{2}}{5}\)
Trục căn thức ở mẫu:
a,\(\frac{1}{\sqrt{2}-1}\)
b,\(\frac{2}{\sqrt{3}+1}\)
c,\(\frac{5}{\sqrt{7}-\sqrt{2}}\)
d,\(\frac{6}{2\sqrt{3}+\sqrt{2}}\)
e,\(\frac{1}{2\sqrt{a}+1}\)
g,\(\frac{2xy}{2\sqrt{x}+3\sqrt{y}}\)
h,\(\frac{x\sqrt{x}-1}{\sqrt{x}-1}\)
i,\(\frac{a-9b}{\sqrt{a}-3\sqrt{b}}\)
k,\(\frac{15-2\sqrt{5}}{3\sqrt{15}-2\sqrt{3}}\)
1,Trục căn thức ở mẫu, rút gọn: ( với \(x\ge0;x\ne1\))
a,\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
b,\(\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
2,Chứng minh các đẳng thức sau:
a,\(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}=1\)
b,\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)
c,\(\left(\frac{\sqrt{a}}{\sqrt{a}+2}+\frac{\sqrt{a}}{\sqrt{a}-2}+\frac{4\sqrt{a}-1}{a-4}\right):\frac{1}{a-4}=-1\)
d,\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
Bài 1:
a)
\(\frac{\sqrt{2.3}+\sqrt{2.7}}{2\sqrt{3}+2\sqrt{7}}=\frac{\sqrt{2}(\sqrt{3}+\sqrt{7})}{2(\sqrt{3}+\sqrt{7})}=\frac{\sqrt{2}}{2}\)
b)
\(\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)(\sqrt{2}+1)}=\frac{3+2\sqrt{2}}{2-1}=3+2\sqrt{2}\)
Bài 2:
a)
\(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}=\sqrt{4}-\sqrt{1}=1\) (đpcm)
b)
\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\)
\(=\sqrt{\frac{(\sqrt{3}+1)^2}{2}}+\sqrt{\frac{(\sqrt{3}-1)^2}{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\) (đpcm)
c) Sửa đề:
\(\left(\frac{\sqrt{a}}{\sqrt{a}+2}-\frac{\sqrt{a}}{\sqrt{a}-2}+\frac{4\sqrt{a}-1}{a-4}\right):\frac{1}{a-4}=\left[\frac{a-2\sqrt{a}-(a+2\sqrt{a})}{(\sqrt{a}+2)(\sqrt{a}-2)}+\frac{4\sqrt{a}-1}{a-4}\right].(a-4)\)
\(=\left(\frac{-4\sqrt{a}}{a-4}+\frac{4\sqrt{a}-1}{a-4}\right).(a-4)=-4\sqrt{a}+4\sqrt{a}-1=-1\)
d)
\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{(\sqrt{a}+\sqrt{b})^2-(\sqrt{a}-\sqrt{b})^2}{2(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}+\frac{2b}{a-b}=\frac{4\sqrt{ab}}{2(a-b)}+\frac{2b}{a-b}\)
\(=\frac{2\sqrt{ab}+2b}{a-b}=\frac{2\sqrt{b}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
Trục căn thức ở mẫu
a)\(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}}\) b)\(\frac{2\sqrt{5}}{\sqrt{3}}\) c)\(\frac{1}{2\sqrt{32}}\) d)\(\frac{x+\sqrt{y}}{x\sqrt{y}}\) e)\(\frac{3\sqrt{2}-1}{3-\sqrt{8}}\) f)\(\frac{3-3\sqrt{2}}{2\sqrt{2}-3\sqrt{6}}\) g \(\frac{xy}{\sqrt{x}-\sqrt{y}}\)
giúp mik vs
trục căn thức ở mẫu
\(\frac{1\sqrt{2}}{\sqrt{2}.\sqrt{2}}\) \(\frac{1\sqrt[]{3}}{\sqrt{3}.\sqrt{3}}\)\(\frac{2\sqrt{5}-5}{\sqrt{5}-2}\)\(\frac{6-2\sqrt{6}}{2-\sqrt{6}}\)\(\frac{5\sqrt{6}-6\sqrt{5}}{\sqrt{5}-\sqrt{6}}\)\(\frac{1}{\sqrt{5}-1}\)\(\frac{1}{4-2\sqrt{a}+a}\)\(\frac{1}{b-2\sqrt{b}+4}\)\(\frac{1}{b-3\sqrt{b}+9}\)