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

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

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

\(=\frac{2a}{\sqrt{a-4}}\) 

bài này không phải của lớp 7

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)

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

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

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

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

     \(=\frac{2a}{\sqrt{a-4}}\)

mk ko bt 123

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 \(G = 4 \)

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

\(P=\frac{\sqrt{\left(\sqrt{a-4}\right)^2+2.2.\sqrt{a-4}+4}+\sqrt{\left(\sqrt{a-4}\right)^2-2.2.\sqrt{a-4}+4}}{\sqrt{1^2-2.\frac{4}{a}}+\frac{4^2}{a^2}}\)

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

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

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

=\(\frac{2a-8}{\frac{a-4}{a}}\)

=\(\frac{2.\left(a-4\right)}{\frac{a-4}{a}}\)

=\(2.\left(a-4\right).\frac{a}{a-4}\)

=2a

(ĐKXĐ: a khác 4)

Bài 1:

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

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

b: A<0

=>\(\sqrt{a}-1<0\)

=>\(\sqrt{a}<1\)

=>0<a<1

Bài 2:

\(A=\left(\frac{3\sqrt{x}+6}{x-4}+\frac{\sqrt{x}}{\sqrt{x}-2}\right):\frac{x-9}{\sqrt{x}-3}\)

\(=\left(\frac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}}{\sqrt{x}-2}\right):\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}-3}\)

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

\(B=3\sqrt8-\sqrt{50}-\sqrt{\left(\sqrt2-1\right)^2}\)

\(=3\cdot2\sqrt2-5\sqrt2-\left|\sqrt2-1\right|\)

\(=\sqrt2-\left(\sqrt2-1\right)=1\)

\(C=\frac{2}{x-1}\cdot\sqrt{\frac{x^2-2x+1}{4x^2}}\)

\(=\frac{2}{x-1}\cdot\sqrt{\frac{\left(x-1\right)^2}{\left(2x\right)^2}}=\frac{2}{x-1}\cdot\frac{\left(1-x\right)}{2x}=\frac{-1}{x}\)

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

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

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

=1