\(2\sqrt{a^2}=2\left|a\right|=2a\) (với \(a\ge0\) )

\(3\sqrt{\left(a-2\right)^2}=3\left|a-2\right|=3\left(2-a\right)=6-3a\) (\(a< 2\))

a: \(2\sqrt{a^2}=-2a\)

b: \(3\sqrt{\left(a-2\right)^2}=3\left|a-2\right|=3\left(2-a\right)\)

\(P=\dfrac{9\sqrt{a}-\sqrt{25a}+\sqrt{4a^3}}{a^2+2a}=\dfrac{9\sqrt{a}-5\sqrt{a}+2a\sqrt{a}}{a\left(a+2\right)}=\dfrac{4\sqrt{a}+2a\sqrt{a}}{a\left(a+2\right)}=\dfrac{2\sqrt{a}\left(2+a\right)}{a\left(2+a\right)}=\dfrac{2\sqrt{a}}{a}=\dfrac{2.\sqrt{a}}{\sqrt{a}.\sqrt{a}}=\dfrac{2}{\sqrt{a}}\)

3 nha !!!!!!!!!!!!!!!!!!

\(5\sqrt{a}-3\sqrt{25a}+2\sqrt{9a}\)\(=5\sqrt{a}-3.5\sqrt{a}+2.3\sqrt{a}\)\(=5\sqrt{a}-15\sqrt{a}+6\sqrt{a}\)\(=\left(5-15+6\right)\sqrt{a}=-4\sqrt{a}\)

Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.

\(=3\sqrt{a}+8\cdot\dfrac{1}{2}\sqrt{a}-\sqrt{\dfrac{9a^2}{a}}+\sqrt{3}\\ =3\sqrt{a}+4\sqrt{a}-3\sqrt{a}+\sqrt{3}\\ =4\sqrt{a}+\sqrt{3}\)

\(B=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{a-9}=\dfrac{11}{a-9}\)

a: Khi x=16 thì \(A=\dfrac{2\cdot\sqrt{16}}{\sqrt{16}+3}=\dfrac{2\cdot4}{4+3}=\dfrac{8}{7}\)

b: P=A+B

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{7\sqrt{x}+3}{9-x}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3x+5\sqrt{x}+6}{x-9}\)

a: Ta có: \(P=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right)\cdot\dfrac{\sqrt{a}-1}{a^2}\)

\(=\dfrac{4a-1}{\sqrt{a}-1}\cdot\dfrac{\sqrt{a}-1}{a^2}\)

\(=\dfrac{4a-1}{a^2}\)

b: Để P=3 thì \(4a-1=3a^2\)

\(\Leftrightarrow3a^2-4a+1=0\)

\(\Leftrightarrow\left(3a-1\right)\left(a-1\right)=0\)

hay \(a=\dfrac{1}{9}\)

a) ĐK: a>0; a≠1 

Ta có: \(P=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right).\dfrac{\sqrt{a}-1}{a^2}\)

                  \(=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}-1}\right).\dfrac{\sqrt{a}-1}{a^2}\)

                  \(=\dfrac{4a-1}{\sqrt{a}-1}.\dfrac{\sqrt{a}-1}{a^2}=\dfrac{4a-1}{a^2}\)

b) Ta có: \(P=3\Leftrightarrow\dfrac{4a-1}{a^2}=3\Leftrightarrow3a^2=4a-1\Leftrightarrow3a^2-4a+1=0\)

               \(\Leftrightarrow\left(a-1\right)\left(3a-1\right)=0\)

               \(\Leftrightarrow\left[{}\begin{matrix}a=1\left(loại\right)\\a=\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)