cho biểu thức A=\(\frac{2\sqrt{x}+1}{x+\sqrt{x}}\) và B=\(\left(1-\frac{2\sqrt{x}}{3\sqrt{x}+1}+\frac{\sqrt{x}+1}{9x-1}\right):\frac{3}{3\sqrt{x}+1}\) với x>0, x≠\(\frac{1}{9}\)
1, tính giá trị của A khi x=\(\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\)
2, rút gọn biểu thức B
3, đặt P=A.B. tìm các giá trị nguyên của x để P có giá trị nguyên
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: \(B=\left(1-\frac{2\sqrt{x}}{3\sqrt{x}+1}+\frac{\sqrt{x}+1}{9x-1}\right):\frac{3}{3\sqrt{x}+1}\)
\(=\left(\frac{9x-1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}-\frac{2\sqrt{x}\left(3\sqrt{x}-1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right)\cdot\frac{3\sqrt{x}+1}{3}\)
\(=\frac{9x-1-6x+2\sqrt{x}+\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}+2}{9\sqrt{x}-3}\)
