\(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)
Rút gọn
rút gọn pt
\(\left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)\sqrt{\dfrac{1}{b}-\dfrac{1}{a}}\)
\(=\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b}{a-b}\cdot\dfrac{\sqrt{ab}\cdot\sqrt{a-b}}{ab}\)
\(=\dfrac{4ab}{ab}\cdot\dfrac{1}{\sqrt{a-b}}=\dfrac{4}{\sqrt{a-b}}\)
Rút gọn: \(\dfrac{\sqrt{a}-\sqrt{b}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)
Biểu thức này có thể coi là ko rút gọn được, thứ duy nhất rút gọn được là ở phân thức đằng sau, \(\dfrac{a\sqrt[]{b}-b\sqrt[]{a}}{\sqrt[]{ab}}=\sqrt[]{a}-\sqrt[]{b}\)
Ngoài ra thì hết rồi, vẫn rất cồng kềnh
Rút gọn \(A=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{1}{a-b}\left(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\)
\(A=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{1}{a-b}\left(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\left(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{2\sqrt{b}-\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\dfrac{-\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
=-1
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) \(\dfrac{a-b}{\sqrt{a}+\sqrt{b}}-\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b\)
b: \(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
=0
\(\dfrac{a+b-\sqrt{2ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
Rút gọn biểu thức
\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}=0\)
Đk: \(a,b\ge0\) và \(a\ne b\)
\(\dfrac{a+b-\sqrt{2ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=\sqrt{a}-\sqrt{b}-\left(\sqrt{a}-\sqrt{b}\right)=0\)
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}`
\(\text{}\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}.\sqrt{b}}=\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}.\sqrt{ab}=a\sqrt{b}-b\sqrt{a}\)
Bài : Rút gọn
\(\dfrac{\sqrt{a}+\sqrt{ab}}{a-b}\) - \(\dfrac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\) với a,b ≥ 0 , a≠b
\(\dfrac{\sqrt{a}+\sqrt{ab}}{a-b}-\dfrac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\left(a,b\ge0;a\ne b\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{ab}-\sqrt{ab}-b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}-b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}-b}{a-b}\)
rút gọn P=\(\left(\dfrac{\sqrt{a}-b}{\sqrt{a}+b}-\dfrac{\sqrt{a}+b}{\sqrt{a}-b}\right).\left(\sqrt{a^3}-\dfrac{ab^2}{\sqrt{a}}\right)\)
\(P=\left(\dfrac{\sqrt{a}-b}{\sqrt{a}+b}-\dfrac{\sqrt{a}+b}{\sqrt{a}-b}\right)\cdot\left(\sqrt{a^3}-\dfrac{ab^2}{\sqrt{a}}\right)\)
\(=\dfrac{\left(\sqrt{a}-b\right)^2-\left(\sqrt{a}+b\right)^2}{\left(\sqrt{a}+b\right)\left(\sqrt{a}-b\right)}\cdot\dfrac{\sqrt{a^4}-ab^2}{\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}-b-b-\sqrt{a}\right)\left(\sqrt{a}-b+b+\sqrt{a}\right)}{\left(\sqrt{a}+b\right)\left(\sqrt{a}-b\right)}\cdot\dfrac{a^2-ab^2}{\sqrt{a}}\)
\(=\dfrac{\left(-2b\right)\cdot\left(2\sqrt{a}\right)}{a-b^2}\cdot\dfrac{a\left(a-b^2\right)}{\sqrt{a}}\)
\(=\dfrac{-4b\sqrt{a}}{\sqrt{a}}\cdot a=-4ba\)
Rút gọn biểu thức
a) \(\dfrac{9-a}{\sqrt{a}+3}-\dfrac{9-6\sqrt{a}+a}{\sqrt{a}-3}\)
b) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
HELP ME
\(a,\dfrac{9-a}{\sqrt{a}+3}-\dfrac{9-6\sqrt{a}+a}{\sqrt{a}-3}\left(dkxd:a\ne9,a\ge0\right)\)
\(=\dfrac{-\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}{\sqrt{a}+3}-\dfrac{\left(3-\sqrt{a}\right)^2}{3-\sqrt{a}}\)
\(=-\left(\sqrt{a}-3\right)+\left(3-\sqrt{a}\right)\)
\(=-\sqrt{a}+3+3-\sqrt{a}\)
\(=6-2\sqrt{a}\)
\(b,\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\left(dkxd:a\ne b,a\ge0,b\ge0\right)\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=\sqrt{a}-\sqrt{b}-\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
a) \(\dfrac{9-a}{\sqrt{a}+3}-\dfrac{9-6\sqrt{a}+a}{\sqrt{a}-3}\)
\(=\dfrac{\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)}{\sqrt{a}+3}-\dfrac{\left(\sqrt{a}-3\right)^2}{\sqrt{a}-3}\)
\(=\dfrac{3-\sqrt{a}}{1}-\dfrac{\sqrt{a}-3}{1}\)
\(=3-\sqrt{a}-\sqrt{a}+3\)
\(=-2\sqrt{a}+6\)
b) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{a}-\sqrt{b}}{1}-\dfrac{\sqrt{a}-\sqrt{b}}{1}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
rút gọn : với a,b dương, ab ≠ 0
\(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}.\dfrac{1}{\sqrt{a}+\sqrt{b}}\)
\(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}.\dfrac{1}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{ab}}.\dfrac{1}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{a-\sqrt{ab}+b}{\sqrt{ab}}\)