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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 \(P\in Z\Leftrightarrow x\in\left\{1\right\}\)
Bài 1 :
a) \(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{x}\)
b) \(P>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-2x}{x}>0\)
\(\Leftrightarrow\sqrt{x}-2x+1>0\left(x>0\right)\)
\(\Leftrightarrow\sqrt{x}+x^2-2x+1-x^2>0\)
\(\Leftrightarrow\sqrt{x}+x^2+\left(x-1\right)^2>0\left(\forall x>0\right)\)
Vậy P > 1/2 với mọi x> 0 ; x khác 1
Bài 2 :
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+a}+\frac{2}{a-1}\right)\)
\(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{2}{a-1}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1+2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)\left(\sqrt{a}+1\right)}\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1+2a+2\sqrt{a}}\)
\(K=\frac{\left(a-1\right)^2}{3a+2\sqrt{a}-1}\)
b) \(a=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)( thỏa mãn ĐKXĐ )
Thay a vào biểu thức K , ta có :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)
a) Đkxđ : \(\left\{{}\begin{matrix}a\ge0\\a\ne9\end{matrix}\right.\)
A = \(\left(\frac{\sqrt{a}+3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\frac{\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right)\left(1-\frac{3}{\sqrt{a}}\right)\)
= \(\frac{2\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\frac{\sqrt{a}-3}{\sqrt{a}}\)
= \(\frac{2}{\sqrt{a}+3}\)
b) Để A > \(\frac{1}{2}\)
<=> \(\frac{2}{\sqrt{a}+3}>\frac{1}{2}\Leftrightarrow\frac{2}{\sqrt{a}+3}-\frac{1}{2}>0\)
<=> \(4-\sqrt{a}-3>0\Leftrightarrow1-\sqrt{a}>0\Leftrightarrow a< 1\)
Vậy để A >1/2 thì a <1
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 ^^