Sửa đề: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)
a) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(x=4-2\sqrt{3}\)
\(=3-2\cdot\sqrt{3}\cdot1+1\)
\(=\left(\sqrt{3}-1\right)^2\)(nhận)
Thay \(x=\left(\sqrt{3}-1\right)^2\) vào biểu thức \(A=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\), ta được:
\(A=\frac{2\cdot\sqrt{\left(\sqrt{3}-1\right)^2}-1}{\sqrt{\left(\sqrt{3}-1\right)^2}+1}\)
\(=\frac{2\cdot\left|\sqrt{3}-1\right|-1}{\left|\sqrt{3}-1\right|+1}\)
\(=\frac{2\left(\sqrt{3}-1\right)-1}{\sqrt{3}-1+1}\)(Vì \(\sqrt{3}>1\))
\(=\frac{2\sqrt{3}-2-1}{\sqrt{3}}\)
\(=\frac{2\sqrt{3}-3}{\sqrt{3}}\)
\(=\frac{\sqrt{3}\left(2-\sqrt{3}\right)}{\sqrt{3}}\)
\(=2-\sqrt{3}\)
Vậy: Khi \(x=4-2\sqrt{3}\) thì \(A=2-\sqrt{3}\)
c) Để \(A=\frac{1}{2}\) thì \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{2}\)
\(\Leftrightarrow2\cdot\left(2\sqrt{x}-1\right)=\sqrt{x}+1\)
\(\Leftrightarrow4\sqrt{x}-2-\sqrt{x}-1=0\)
\(\Leftrightarrow3\sqrt{x}-3=0\)
\(\Leftrightarrow3\sqrt{x}=3\)
\(\Leftrightarrow\sqrt{x}=1\)
hay x=1(loại)
Vậy: Không có giá trị nào của x để \(A=\frac{1}{2}\)
d) Để A<1 thì A-1<0
hay \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}-1< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}-1-\left(\sqrt{x}+1\right)}{\sqrt{x}+1}< 0\)
mà \(\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ
nên \(2\sqrt{x}-1-\left(\sqrt{x}+1\right)< 0\)
\(\Leftrightarrow2\sqrt{x}-1-\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Leftrightarrow x< 4\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 4\\x\ne1\end{matrix}\right.\)
Vậy: Khi \(\left\{{}\begin{matrix}0\le x< 4\\x\ne1\end{matrix}\right.\)thì A<1
e) Để A là số nguyên thì \(2\sqrt{x}-1⋮\sqrt{x}+1\)
\(\Leftrightarrow2\sqrt{x}+2-3⋮\sqrt{x}+1\)
mà \(2\sqrt{x}+2⋮\sqrt{x}+1\forall x\)
nên \(-3⋮\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(-3\right)\)
\(\Leftrightarrow\sqrt{x}+1\in\left\{1;-1;3;-3\right\}\)
mà \(\sqrt{x}+1\ge1\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}+1\in\left\{1;3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\)
hay \(x\in\left\{0;4\right\}\)(nhận)
Vậy: Khi \(x\in\left\{0;4\right\}\) thì A là số nguyên
có sai đề k bạn nếu sai mk xin sửa
đkxđ \(x\ne1;x\ge0\)
a,\(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{x-1}\)
\(=\frac{2x+3-3\sqrt{x}-1}{x-1}=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b, khi x = \(4+2\sqrt{3}\)
\(A=\frac{2\left(3+2\sqrt{3}\right)+1}{4+2\sqrt{3}-2}=\frac{9+4\sqrt{5}}{3+2\sqrt{3}}\)
c, để A = 1/2
\(\frac{2\sqrt{x}+1}{\sqrt{x}-1}=\frac{1}{2}\Leftrightarrow4\sqrt{x}+2=\sqrt{x}-1\Leftrightarrow3\sqrt{x}=3\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(ktmđk\right)\)
d, để A<1\(\Leftrightarrow\frac{2\sqrt{x}+1}{\sqrt{x}-1}< 1\left(1\right)\)
th1 với \(\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
Ta có (1)\(\Leftrightarrow2\sqrt{x}+1< \sqrt{x}-1\Leftrightarrow\sqrt{x}< -2\left(loại\right)\)
th2 \(\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow0\le x< 1\)
ta có (1) \(\Leftrightarrow2\sqrt{x}+1>\sqrt{x}-1\Leftrightarrow\sqrt{x}>-2\left(luônđúng\right)\)
\(\Rightarrow0\le x< 1\)
e, \(A=\frac{2\sqrt{x}+1}{\sqrt{x}-1}=\frac{2\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=2+\frac{3}{\sqrt{x}-1}\)
để A là số nguyen thì \(\frac{3}{\sqrt{x}-1}\)cũng là số nguyên
\(\left[{}\begin{matrix}\sqrt{x}-1=1\\\sqrt{x}-1=-1\\\sqrt{x}-1=3\\\sqrt{x}-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=0\\\sqrt{x}=4\\\sqrt{x}=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tmđk\right)\\x=0\left(tmđk\right)\\x=16\left(tmđk\right)\end{matrix}\right.\)
