ĐKXĐ: ...

\(\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{15\sqrt{x}-11-x-4\sqrt{x}-3+2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x+12\sqrt{x}-17}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

cái này là do đề bài sai hay sao ấy

C=\(\frac{7\sqrt[]{x}-2}{\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+3\right)}\)

bạn có thể làm từng bước được k ạ

pt tương đương 273√81x−8=(3x−2)3−462781x−83=(3x−2)3−46
Đặt 3√81x−8=a81x−83=a; 3x−2=b3x−2=b suy ra a3−27b=46a3−27b=46 pt trở thành 
27a=b3−a3+27b27a=b3−a3+27b (dễ thấy pt này (a;b)(a;b) có vai trò như nhau nên sẽ có nghiệm a=ba=b)
suy ra (b−a)(a2+b2+ab+27)=0(b−a)(a2+b2+ab+27)=0 suy ra a=ba=b mà a3−27b=46a3−27b=46
suy ra b3−27b=46b3−27b=46 pt này có 1 nghiệm bằng −2−2 pt còn lại là pt bậc 2 thì dễ rồi nhé
từ đó suy ra xx 

\(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(1-\sqrt{x}\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{3+\sqrt{x}}\)

\(N=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

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

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

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

Cái đó chỗ số 3 phải có mẫu nữa mà mình quên mất mẫu

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

ĐKXĐ: \(x\ge0;x\ne1\)

\(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-5x+8\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{-\left(\sqrt{x}-1\right)\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{3-5\sqrt{x}}{\sqrt{x}+3}\)