Trục căn thức ở mẫu
\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}\)
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NT
0
LV
0
6 tháng 7 2019
a) \(\frac{a-\sqrt{a}}{\sqrt{a}-1}=\frac{\left(a-\sqrt{a}\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{a\sqrt{a}-\sqrt{a}}{a-1}=\frac{\sqrt{a}\left(a-1\right)}{a-1}=\sqrt{a}\)
b) \(\frac{p-2\sqrt{p}}{p-\sqrt{2}}=\frac{\left(p-2\sqrt{p}\right)\left(p+\sqrt{2}\right)}{\left(p-\sqrt{2}\right)\left(p+\sqrt{2}\right)}=\frac{\left(p-2\sqrt{p}\right)\left(p+\sqrt{2}\right)}{p^2-2}\)
D
1
D
1
VD
23 tháng 10 2021
(\(\sqrt{a}\)+\(\sqrt{b}\)+1) /\(\sqrt{a}+\sqrt{B}-1\).\(\sqrt{a}+\sqrt{b}+1\)=
B
23 tháng 7 2019
\(a,\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{6}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{\left(\sqrt{2}+\sqrt{3}\right)^2-\sqrt{6}^2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{2\sqrt{6}-1}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{6}+1\right)}{2\sqrt{6}^2-1^2}=\frac{4\sqrt{3}+6\sqrt{2}+12+\sqrt{2}+\sqrt{3}+\sqrt{6}}{11}\)\(=\frac{\sqrt{6}+5\sqrt{3}+7\sqrt{2}+12}{11}\)
\(b,\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\left(\sqrt{z}+\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{z}\right)}=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{z}^2}\)
\(=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{x+2\sqrt{xy}+y-z}\)
LT
2
VD
1 tháng 10 2017
\(\frac{\left(\sqrt{a}\right)^3-\left(\sqrt{b}\right)^3}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}\\ \)
\(a+\sqrt{ab}+b\)
2 tháng 10 2017
Ta có:
\(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
\(\Leftrightarrow\frac{\sqrt{a}^3-\sqrt{b}^3}{\sqrt{a}-\sqrt{b}}\)
\(\Leftrightarrow\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}\)
\(\Rightarrow a+\sqrt{ab}+b\)
8 tháng 10 2018
\(A=\frac{\sqrt{a+b}-\sqrt{a}+\sqrt{b}}{\sqrt{a+b}-\sqrt{a}-\sqrt{b}}=1+\frac{2\sqrt{b}}{\sqrt{a+b}-\sqrt{a}-\sqrt{b}}=1+B\)
\(B=\frac{2\sqrt{b}\left(\sqrt{a+b}+\sqrt{a}+\sqrt{b}\right)}{-2\sqrt{ab}}=-\frac{\left(\sqrt{a+b}+\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}}=-\frac{\sqrt{a}\left(\sqrt{a+b}+\sqrt{a}+\sqrt{b}\right)}{a}\)
9 tháng 10 2018
\(=\frac{\left(\sqrt{a+b}-\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a+b}-\sqrt{a}\right)^2-b}\)
\(=\frac{2a+2b-2\sqrt{a\left(a+b\right)}+2\sqrt{b\left(a+b\right)}-2\sqrt{ab}}{2a-2\sqrt{a\left(a+b\right)}}\)
\(=\frac{a+b-\sqrt{a\left(a+b\right)}+\sqrt{b\left(a+b\right)}-\sqrt{ab}}{a-\sqrt{a\left(a+b\right)}}\)
\(=\frac{\left(a+b-\sqrt{a\left(a+b\right)}+\sqrt{b\left(a+b\right)}-\sqrt{ab}\right)\left(a+\sqrt{a\left(a+b\right)}\right)}{a^2-a\left(a+b\right)}\)
\(=\frac{b\sqrt{a\left(a+b\right)}+\sqrt{ab}\left(a+b\right)-a\sqrt{ab}}{-ab}\)
\(=\frac{-b\sqrt{a\left(a+b\right)}+b\sqrt{ab}}{ab}\)
\(=\frac{\sqrt{ab}-\sqrt{a\left(a+b\right)}}{a}\)
\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}=\frac{\left(1-a\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(=\frac{1-a^2+\sqrt{a}-a\sqrt{a}}{1-a}\)
\(=\frac{\left(1-a\right)\left(1+a\right)+\sqrt{a}\left(1-a\right)}{1-a}\)
\(=\frac{\left(1-a\right)\left(1+a+\sqrt{a}\right)}{1-a}\)
\(=1+a+\sqrt{a}\)
1+a=\(\sqrt{a}\)