+) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\) ( đk : a > 0 ; b >0 ; a khác b )

\(=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

\(=a-b\)

+) \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\) ( đk : a \(\ge\)0 ; a khác 1 )

\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right).\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)

\(=1-a\)

`\sqrt{9x^2+12x+4}=4`

`<=>\sqrt{(3x+2)^2}=4`

`<=>|3x+2|=4`

`<=>` $\left[\begin{matrix} 3x+2=4\\ 3x+2=-4\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{2}{3}\\ x=-2\end{matrix}\right.$

 anh có thể giái giúp em bài trong trang em được 0 ạ?nếu được em cảm ơn

1

\(T=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{a-4}.\left(\dfrac{a-4}{\sqrt{a}}\right)\)

\(=\dfrac{a-2\sqrt{a}+4-a-2\sqrt{a}-4}{a-4}.\dfrac{a-4}{\sqrt{a}}\)

\(=\dfrac{-4\sqrt{a}}{a-4}.\dfrac{a-4}{\sqrt{a}}=-4\)

`a)`\(P=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{2\sqrt{x}+1}{x+\sqrt{x}-2}\)

\(P=\left(\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(P=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)

\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

\(P=\dfrac{\sqrt{x}+1+1}{\sqrt{x}+1}=1+\dfrac{1}{\sqrt{x}+1}\)

`b)` Để `P` nguyên thì \(\dfrac{1}{\sqrt{x}+1}\in Z\) \(\Rightarrow\sqrt{x}+1\inƯ\left(1\right)=\left\{\pm1\right\}\)

`@`\(\sqrt{x}+1=1\rightarrow x=0\)

`@`\(\sqrt{x}+1=-1\) `->` vô lý

Vậy `x=0` thì `P` nguyên

`c)`\(P=1+\dfrac{1}{\sqrt{x}+1}\)

Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\)

\(\Rightarrow P\le1+\dfrac{1}{1}=2\)

Vậy \(Max_P=2\) khi `x=0`

\(N=\left[\left(a-3b\right)-\left(a+3b\right)\right]\left[\left(a-3b\right)+\left(a+3b\right)\right]-\left(a-1\right)\left(b-2\right)=\)

\(=\left(-6b\right).2a-\left(ab-2a-b+2\right)=\)

\(=2a+b-13ab-2=\) thay a;b vào để tính N