a: 

Sửa đề: a+2căn a+8

\(=\dfrac{5a+10\sqrt{a}-3\sqrt{a}-6+3a-6\sqrt{a}-a-2\sqrt{a}-8}{\left(a-4\right)}\)

\(=\dfrac{7a-\sqrt{a}-14}{\left(a-4\right)}\)

b: A>0

=>(7a-căn a-14)/(a-4)>0

=>a>4 hoặc 0<a<(1+căn 393)/14

\(G=\sqrt{a-4+4\sqrt{a-4}+4}+\sqrt{a-4-4\sqrt{a-4}+4}\)

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

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

\(G = \sqrt{a + 4 \sqrt{a – 4}} + \sqrt{a – 4\sqrt{a – 4}} \) \(= \sqrt{a – 4 + 4 + 4\sqrt{a – 4}} + \sqrt{a – 4 + 4 – 4\sqrt{a – 4}}\)

                                                 \(= \sqrt{\sqrt{a - 4}^2 + 2^2 + 4\sqrt{a – 4}} + \sqrt{\sqrt{a - 4}^2 + 2^2 - 4\sqrt{a – 4}}\)

                                                 \(= \sqrt{(\sqrt{(a – 4)} + 2)^2} + \sqrt{(\sqrt{(a – 4)} - 2)^2}\)

                                                  \(= \sqrt{a – 4} + 2 +|\sqrt{a – 4} – 2|\)

+) Với \(4 < a < 8 ⇔ 0 < a – 4 < 4 ⇔ \sqrt{0} < \sqrt{a – 4} < \sqrt{4} ⇔ 0 <\sqrt{a – 4} < 2 \)

Do đó, ta có: \(G = \sqrt{a – 4} + 2 + 2 - \sqrt{a – 4} \) (vì \(2 > \sqrt{a – 4}\))

                          \(=4\)

➤Với \(4 < a < 8 \) thì \(G = 4 \)

1 like cho Ng.Hữu Minh và nhiều like cho H.Việt Tân.

a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\)

\(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}-2}\right)\cdot\dfrac{a-4}{\sqrt{4a}}\)

\(=\dfrac{2\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{2a}\)

\(=\sqrt{a}+2\)

b: A-2<0

=>\(\sqrt{a}+2-2< 0\)

=>\(\sqrt{a}< 0\)

=>\(a\in\varnothing\)

c: Bạn ghi đầy đủ đề đi bạn

Bài 1

a) \(P=\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a}-2}{1-\sqrt{a}}\)    (ĐK : x\(\ge0\) ; x\(\ne\) 1)

        \(=\frac{3a+\sqrt{9a}-3}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\)

         \(=\frac{3a+\sqrt{9a}-3-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)

         \(=\frac{3a+\sqrt{9a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)

         \(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)

         \(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)

         \(=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)

b) \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=\frac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)

Vậy để P là số nguyên thì: \(\sqrt{a}-1\inƯ\left(2\right)\)

Mà Ư(2)={-1;1;2;-1}

=> \(\sqrt{a}-1\in\left\{1;-1;2;-2\right\}\)

Ta có bảng sau:

vậy a={0;4;9} thì P nguyên

Bài 2

  \(P=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)(ĐK:a\(\ge\)8)

      \(=\frac{\sqrt{\left(a-4\right)+4\sqrt{a-4}+4}+\sqrt{\left(a-4\right)-4\sqrt{a-4}+4}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)

     \(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{1-\frac{4}{a}}\)

      \(=\sqrt{a-4}+2+\sqrt{a-4}-2:\frac{a-4}{a}\)

     \(=2\sqrt{a-4}\cdot\frac{a}{a-4}\)

     \(=\frac{2a}{\sqrt{a-4}}\)

\(=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)

\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}\right)-2}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)

\(=\frac{\sqrt{a-4}+2+\sqrt{a-4}-2}{1-\frac{4}{a}}\)

\(=\frac{2a}{\sqrt{a-4}}\) 

bài này không phải của lớp 7

\(a,P=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}\\ P=\sqrt{a}+2+2+\sqrt{a}=2\sqrt{a}+4\\ b,P=a+1\Leftrightarrow a+1=2\sqrt{a}+4\\ \Leftrightarrow a-2\sqrt{a}-3=0\\ \Leftrightarrow\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)=0\\ \Leftrightarrow\sqrt{a}=3\left(\sqrt{a}\ge0\right)\\ \Leftrightarrow a=9\left(tm\right)\)