thực hiện phép tính ( rút gọn biểu thức )
a) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\)
b) \(\dfrac{a-b}{\sqrt{a}+\sqrt{b}}-\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\)
Cho biểu thức I = \(\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right)\).\(\left[\left(\dfrac{1}{\sqrt{a}-\sqrt{b}}+\dfrac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)
Rút gọn I
a) Tính giá trị của I với a = 16, b = 4
\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)
\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)
Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)
\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)
c. rút gọn biểu thức
\(C=\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\dfrac{a}{\sqrt{ab}+b}-\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\right)\) vs \(a\ge0,a\ne4,a\ne9\)
Cho biểu thức M= \(\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\)nhau.với hai số a, b dương khác
a/ Rút gọn M
b/Tính giá trị của M khi a=\(\sqrt{6+2\sqrt{5}}\),b=\(\sqrt{6-2\sqrt{5}}\)
a: ta có: \(M=\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\)
\(=\dfrac{a\left(\sqrt{ab}-a\right)+b\left(\sqrt{ab}+b\right)}{\left(\sqrt{ab}+b\right)\left(\sqrt{ab}-a\right)}-\dfrac{a+b}{\sqrt{ab}}\)
\(=\dfrac{-\sqrt{ab}\left(a+b\right)+\left(a-b\right)\left(a+b\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)\cdot\sqrt{a}\cdot\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{a+b}{\sqrt{ab}}\)
\(=\dfrac{-\sqrt{ab}\left(a+b\right)+\left(a-b\right)\left(a+b\right)}{\sqrt{ab}\left(a-b\right)}-\dfrac{a^2-b^2}{\sqrt{ab}\left(a-b\right)}\)
\(=\dfrac{-\sqrt{ab}}{\sqrt{ab}\left(a-b\right)}\)
\(=-\dfrac{1}{a-b}\)
b: Thay \(a=\sqrt{5}+1\) và \(b=\sqrt{5}-1\) vào M, ta được:
\(M=\dfrac{-1}{\sqrt{5}+1-\sqrt{5}+1}=\dfrac{-1}{2}\)
rút gọn biểu thức B= \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}.\sqrt{b}}\)
\(B=\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}.\sqrt{b}}\)
\(B=\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}.\sqrt{ab}\)
\(B=a\sqrt{b}-b\sqrt{a}\)
Với `a,b > 0` có:
`B=[a\sqrt{b}-b\sqrt{a}]/\sqrt{ab} :1/[\sqrt{a}.\sqrt{b}]`
`B=[a\sqrt{b}-b\sqrt{a}]/[\sqrt{ab}] .\sqrt{ab}`
`B=a\sqrt{b}-b\sqrt{a}`
\(\text{}\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}.\sqrt{b}}=\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}.\sqrt{ab}=a\sqrt{b}-b\sqrt{a}\)
Cho biểu thức \(M=\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}-\dfrac{a}{\sqrt{a}+\sqrt{b}}-\dfrac{b}{\sqrt{b}-\sqrt{a}}\) với a,b>0 và \(a\ne b\) . Rút gọn M và tính giá trị biểu thức M biết \(\left(1-a\right).\left(1-b\right)+2\sqrt{ab}=1\)
\(M=\dfrac{a\sqrt{a}-b\sqrt{b}-a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\\ M=\dfrac{a\sqrt{b}+b\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ M=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
\(\left(1-a\right)\left(1-b\right)+2\sqrt{ab}=1\\ \Leftrightarrow1-a-b+ab+2\sqrt{ab}=1\\ \Leftrightarrow a+b-ab-2\sqrt{ab}=0\\ \Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2=ab\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{a}-\sqrt{b}=\sqrt{ab}\\\sqrt{a}-\sqrt{b}=-\sqrt{ab}\end{matrix}\right.\)
Với \(\sqrt{a}-\sqrt{b}=\sqrt{ab}\Leftrightarrow M=\dfrac{\sqrt{ab}}{\sqrt{ab}}=1\)
Với \(\sqrt{a}-\sqrt{b}=-\sqrt{ab}\Leftrightarrow M=\dfrac{\sqrt{ab}}{-\sqrt{ab}}=-1\)
\(M=\dfrac{a\sqrt{a}-b\sqrt{b}-a\left(\sqrt{a}-\sqrt{b}\right)+b\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a\sqrt{b}+b\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
\(\left(1-a\right)\left(1-b\right)+2\sqrt{ab}=1\)
\(\Leftrightarrow a+b-ab-2\sqrt{ab}=0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2=ab\Leftrightarrow\sqrt{a}-\sqrt{b}=\sqrt{ab}\)
\(M=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}}{\sqrt{ab}}=1\)
Cho biểu thức:
\(D=\left(\dfrac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\dfrac{a+b+2ab}{1-ab}\right)\)
a) Tìm đkxđ và rút gọn \(D\)
b) Tính \(D\) với \(a=\dfrac{2}{2+\sqrt{3}}\)
c) Tìm giá trị lớn nhất của \(D\)
Bài 1 : (2 điểm) : Thực hiện phép tính và rút gọn các biểu thức sau :
a)A\(=-\left(\dfrac{1}{3-\sqrt{5}}+\dfrac{1}{3+\sqrt{5}}\right):\sqrt{5}\)
b)\(B=\sqrt{48+\sqrt{5\dfrac{1}{3}+2\sqrt{75}-\sqrt[5]{1\dfrac{1}{3}}}}\)
\(A=-\dfrac{3+\sqrt{5}+3-\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\dfrac{\sqrt{5}}{5}\\ A=\dfrac{-6}{4}\cdot\dfrac{\sqrt{5}}{5}=\dfrac{-3\sqrt{5}}{10}\)
Rút gọn biểu thức:
A = (\(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}\) - \(\sqrt{xy}\)) + (\(\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\))
B = (\(\sqrt{a}\) + \(\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)) : (\(\dfrac{a}{\sqrt{ab}}\) + \(\dfrac{b}{\sqrt{ab-a}}\) - \(\dfrac{a+b}{\sqrt{ab}}\))
C = \(\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}\) + \(\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\)(\(\dfrac{\sqrt{b}}{a-\sqrt{ab}}\) + \(\dfrac{\sqrt{b}}{a+\sqrt{ab}}\))
D = (\(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\) - \(\dfrac{x\sqrt{x}-y\sqrt{y}}{x-y}\)) . \(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{x\sqrt{x}+y\sqrt{y}}\)
Thực hiện phép tính.
a) \(\left(\sqrt{ab}+2\sqrt{\dfrac{b}{a}}-\sqrt{\dfrac{a}{b}+\sqrt{\dfrac{1}{ab}}}\right)\sqrt{ab}\)
b) \(\left(\dfrac{am}{b}\sqrt{\dfrac{n}{m}}-\dfrac{ab}{n}\sqrt{mn}+\dfrac{a^2}{b^2}\sqrt{\dfrac{m}{n}}\right).a^2b^2.\sqrt{\dfrac{n}{m}}\)
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