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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 \(P\in Z\Leftrightarrow x\in\left\{1\right\}\)
\(P=\sqrt{\frac{15}{2}}.\sqrt{\frac{10.\left(a-1\right)^2}{3}}\) ( ĐK a<1 )
\(\Leftrightarrow P=\frac{\sqrt{15}}{\sqrt{2}}.\frac{\sqrt{10}.\sqrt{\left(a-1\right)^2}}{\sqrt{3}}\)
\(\Leftrightarrow P=\frac{\sqrt{15}.\sqrt{2}}{2}.\frac{\sqrt{10}.\sqrt{3}.\left|a-1\right|}{3}\)
\(\Leftrightarrow P=\frac{\sqrt{30}}{2}.\frac{\sqrt{30}\left(1-a\right)}{3}\)( vì a-1<0)
\(\Leftrightarrow P=\frac{\sqrt{30}.\sqrt{30}\left(1-a\right)}{2.3}\)
\(\Leftrightarrow\frac{30\left(1-a\right)}{6}\)
\(\Leftrightarrow5\left(1-a\right)\)
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 ♥
\(P=\sqrt{\left(10^n+1\right)^2-2.10^n+\left(\frac{10^n}{10^n+1}\right)^2}+\frac{10^n}{10^n+1}\)
\(=\sqrt{\left(10^n+1-\frac{10^n}{10^n+1}\right)^2}+\frac{10^n}{10^n+1}\)
\(=10^n+1-\frac{10^n}{10^n+1}+\frac{10^n}{10^n+1}\left(\text{vì }10^n+1-\frac{10^n}{10^n+1}>0\text{ }\right)\)
\(=10^n+1\)
cut cho