`M=\sqrt{a/b}+\sqrt{ab}-a\sqrt{1/(ab)}(a,b>0)`
`=\sqrt{a/b}+\sqrt{ab}-\sqrt{a^2*1/(ab)}`
`=\sqrt{a/b}+\sqrt{ab}-\sqrt{a*1/b}`
`=\sqrt{a/b}+\sqrt{ab}-\sqrt{a/b}`
`=(\sqrt{a/b}-\sqrt{a/b})+\sqrt{ab}`
`=0+\sqrt{ab}`
`=\sqrt{ab}`
Vậy `M=\sqrt{ab}` với `a>0,b>0`
`->` Chọn `bbB`
\(A=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{ab}-a}\right):\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)\)
Rut gon A
\(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}+a}-\frac{a+b}{\sqrt{ab}}\)
rut gon
Rut gon
\(Q=\frac{\left(\frac{a-b}{\sqrt{a}+\sqrt{b}}\right)^3+2a\sqrt{a}+b\sqrt{b}}{3a^2+3b\sqrt{ab}}+\frac{\sqrt{ab}-a}{a\sqrt{a}-b\sqrt{a}}\)
Cho biểu thức: \(B=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
a) Rút gon biểu thức
b) Tính giá trị của B nếu a=\(6+2\sqrt{5}\)
c) So sánh B với -1
Rut gon bieu thuc sau
\(\sqrt{\frac{a}{b}}+\sqrt{ab}+\frac{a}{b}\sqrt{\frac{b}{a}}\) voi a > 0 va b > 0
CÂU HỎI KHÓ NHẤT QUẢ ĐẤT
A)RÚT GON P
\(P=\left(\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right)\):\(\frac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\)
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left[\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1
+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}
+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a
+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a
+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
d) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
Câu 1 : Rút gọn biểu thức
a, \(\frac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\frac{2}{3}\sqrt{12}.\)b, \(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\frac{3}{3+3\sqrt{6}}.\)
c\(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}.\)Với a>0;b>0
\(P=\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right)\sqrt{\frac{1}{a}-\frac{1}{b}}\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{a-b}-\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{a-b}\right).\sqrt{\frac{b-a}{ab}}\)
\(=\frac{a-2\sqrt{ab}+b-a-2\sqrt{ab}-b}{a-b}.\sqrt{\frac{b-a}{ab}}\)
\(=\frac{-4\sqrt{ab}}{a-b}.\sqrt{\frac{b-a}{ab}}\)\(=\frac{-4\sqrt{ab}}{2017-2018}.\sqrt{\frac{2018-2017}{ab}}\)
\(=4\sqrt{ab}.\sqrt{\frac{1}{ab}}\)\(=\sqrt{\frac{16ab}{ab}}\)\(=4\)
Cho \(B=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
a. Rút gọn B
b. Tính giá trị của B khi \(a=6+2\sqrt{5}\)
c. So sanhs B với -1
