a,   A\(=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x-1}{\sqrt{x}}\)  ĐK  x>0   ;\(x\ne1;x\ne-1\)

    \(A=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}}{x-1}\)

\(A=\frac{4x\sqrt{x}}{x-1}.\frac{\sqrt{x}}{x-1}\)=\(\frac{4x^2}{\left(x-1\right)^2}\)

b,  Để  A =2  \(\Rightarrow\frac{4x^2}{\left(x-1\right)^2}=2\Rightarrow4x^2=2\left(x-1\right)^2\)

                     <=>  \(4x^2=2x^2-4x+2\)

                      <=> \(2x^2+4x-2=0\)

                       <=> \(x^2+2x-1=0\)

                       \(\Delta=1^2-1.\left(-1\right)\) =  2

                => \(\orbr{\begin{cases}x_1=-1-\sqrt{2}\left(loại\right)\\x_2=-1+\sqrt{2}\left(nhận\right)\end{cases}}\)

Vậy x=\(-1+\sqrt{2}\)thì  A =2  

c, Thay   x =\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)=2

  =>A  =   \(\frac{4.2^2}{\left(2-1\right)^2}=16\)

Vậy  A=16  thì  x=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

mi tích tau tau tích mi xong tau trả lời nka

 việt nam nói là làm

a/ \(\sqrt{ab}+\sqrt{a}-\sqrt{b}\)

b/ \(\sqrt{ab}+\sqrt{a}-\sqrt{b}=2\)

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

Xong rồi nhá

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

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

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

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

\(A=\frac{4a\left(a+1\right)}{a-1}\)

ta có \(a=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(a=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(a=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)

\(a=2\left(16-15\right)\)

\(a=2\)

khi đó \(A=\frac{4.2.\left(2+1\right)}{2-1}=8.3=24\)

vậy.....

Câu C : Lần đầu làm dạng này :))

Xét hiệu A - 2 , ta có :

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

Ta thấy :

+) Do \(a\ge0\)\(\Rightarrow2\sqrt{a}\left(1-2\sqrt{a}\right)\le0\)

+) a khác 1 ; \(a\ge0\)=> 2a + 1 > 0

\(\Rightarrow\frac{2\sqrt{a}\left(1-2\sqrt{a}\right)}{2a+1}\le0\)

\(\Leftrightarrow A< 2\)

P/s : sai bỏ qua :))

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

ĐKXĐ : \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)

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

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

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

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

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

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

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

b) \(a=1-\frac{\sqrt{3}}{2}=\frac{2}{2}-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{2}\)( tmđk )

Rồi từ đây thế vô :)

c) Nhờ cao nhân làm tiếp chứ em mới lớp 8 thôi ạ :(

xin lỗi,giờ mình mới học lớp 6 thôi