đk: \(x\ge0\)và      \(x\ne1\)

\(\Leftrightarrow P=\frac{x-1+\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-1}\right)}-\frac{2x-10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow P=\frac{x-1+x+\sqrt{x}-6-2x+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow P=\frac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)

để P > 0

\(\Leftrightarrow1>\sqrt{x}-1\)

\(\Leftrightarrow-\sqrt{x}>-2\)

\(\Leftrightarrow\sqrt{x}< 2\)

\(\Leftrightarrow x< 4\)

có sai xót mong m.n bỏ qa cho ♥

1/ \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\)

\(x=\left(1+\frac{\sqrt{10}\left(\sqrt{10}+1\right)}{1+\sqrt{10}}\right)\left(\frac{\sqrt{10}\left(\sqrt{10}-1\right)}{\sqrt{10}-1}-1\right)\)

\(x=\left(1+\sqrt{10}\right)\left(\sqrt{10}-1\right)\)

\(x=10-1=9\)

Thay \(x=9\) vào A:

\(A=\frac{2\sqrt{9}+1}{9+\sqrt{9}}=\frac{7}{12}\)

Vậy với \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\Leftrightarrow A=\frac{7}{12}\)

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

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

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

\(\Leftrightarrow B=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)

3/ \(P=A.B=\frac{2\sqrt{x}+1}{x+\sqrt{x}}\cdot\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{2\sqrt{x}+1}{3\sqrt{x}-1}\)

Để \(P\in Z\Leftrightarrow2\sqrt{x}+1⋮3\sqrt{x}-1\)

\(\Leftrightarrow6\sqrt{x}+2⋮3\sqrt{x}-1\)

\(\Leftrightarrow2\left(3\sqrt{x}-1\right)+4⋮3\sqrt{x}-1\)

\(\Leftrightarrow4⋮3\sqrt{x}-1\)

\(\Leftrightarrow3\sqrt{x}-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Leftrightarrow3\sqrt{x}\in\left\{0;2;-1;3;-3;5\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;\frac{2}{3};-\frac{1}{3};1;-1;\frac{5}{3}\right\}\)

\(\Leftrightarrow x\in\left\{0;\frac{4}{9};\frac{1}{9};1;\frac{25}{9}\right\}\)

Loại bỏ những giá trị x < 0 , x \(x\notin Z\)và x không thỏa mãn ĐKXĐ

Vậy để \(P\in Z\Leftrightarrow x\in\left\{1\right\}\)

1: Ta có: \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\cdot\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\)

\(=\left(\frac{1+\sqrt{10}+10+\sqrt{10}}{1+\sqrt{10}}\right)\cdot\left(\frac{10-\sqrt{10}-\left(\sqrt{10}-1\right)}{\sqrt{10}-1}\right)\)

\(=\frac{1+2\sqrt{10}\cdot1+\left(\sqrt{10}\right)^2}{1+\sqrt{10}}\cdot\frac{\left(\sqrt{10}\right)^2-2\cdot\sqrt{10}\cdot1+1}{\sqrt{10}-1}\)

\(=\left(1+\sqrt{10}\right)\cdot\left(\sqrt{10}-1\right)\)

\(=10-1=9\)

Thay x=9 vào biểu thức \(A=\frac{2\sqrt{x}+1}{x+\sqrt{x}}\), ta được:

\(A=\frac{2\cdot\sqrt{9}+1}{9+\sqrt{9}}=\frac{2\cdot3+1}{9+3}=\frac{7}{12}\)

Vậy: \(\frac{7}{12}\) là giá trị của biểu thức \(A=\frac{2\sqrt{x}+1}{x+\sqrt{x}}\) tại \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\cdot\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\)

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

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

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

\(=\frac{3x+3\sqrt{x}+2}{9\sqrt{x}-3}\)

mk làm luôn.

a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

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

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

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

mk làm phần rút gọn xong mk bận nên bn tự làm câu b nha ^^

\(A=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{6\sqrt{x}}{3\sqrt{x}+1}\)

\(A=\left[\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right].\frac{3\sqrt{x}+1}{6\sqrt{x}}\)

\(A=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{6\sqrt{x}}\)

\(A=\frac{3\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}.\frac{1}{6\sqrt{x}}\)

\(A=\frac{\sqrt{x}+1}{6\sqrt{x}-2}\)

\(A=\frac{5}{6}\Leftrightarrow\frac{\sqrt{x}+1}{6\sqrt{x}-2}=\frac{5}{6}\)

\(\Leftrightarrow6\sqrt{x}+6=30\sqrt{x}-10\)

\(\Leftrightarrow24\sqrt{x}=16\)

\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\Leftrightarrow x=\frac{4}{9}\)

\(A=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]\div\frac{6\sqrt{x}}{3\sqrt{x}+1}\)

\(A=\left[\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]\times\frac{3\sqrt{x}+1}{6\sqrt{x}}\)

\(A=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}\times\frac{1}{6\sqrt{x}}\)

\(A=\frac{3\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}\times\frac{1}{6\sqrt{x}}\)

\(A=\frac{\sqrt{x}+1}{6\sqrt{x}-2}\)

\(A=\frac{5}{6}\)

\(\Leftrightarrow\frac{\sqrt{x}+1}{6\sqrt{x}-2}=\frac{5}{6}\)

\(\Leftrightarrow6\sqrt{x}+6=30\sqrt{x}-10\)

\(\Leftrightarrow24\sqrt{x}=16\)

\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{4}{9}\)

a) ĐK: x > 1

 \(P=\left(\frac{\sqrt{x-1}}{3+\sqrt{x-1}}+\frac{x+8}{9-\left(x-1\right)}\right):\left(\frac{3\sqrt{x-1}+1}{\left(x-1\right)-3\sqrt{x-1}}-\frac{1}{\sqrt{x-1}}\right)\)

\(P=\frac{\sqrt{x-1}\left(3-\sqrt{x-1}\right)+x+8}{9-\left(x-1\right)}:\frac{3\sqrt{x-1}+1-\left(\sqrt{x-1}-3\right)}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}\)

\(P=\frac{3\sqrt{x-1}-x+1+x+8}{10-x}:\frac{2\sqrt{x-1}+4}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}\)

\(P=\frac{3\left(\sqrt{x-1}+3\right)}{10-x}.\frac{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}{2\sqrt{x-1}+4}\)

\(P=\frac{-3\sqrt{x-1}}{2\sqrt{x-1}+4}\)

b) \(x=\sqrt[4]{\frac{17+12\sqrt{2}}{1}}-\sqrt[4]{\frac{17-12\sqrt{2}}{1}}=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)

Vậy \(P=\frac{-3\sqrt{2-1}}{2\sqrt{2-1}+4}=-\frac{1}{2}\)

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