1.32
a. ĐKXĐ: \(x>0;x\ne1\)
b.
\(P=\dfrac{3x+3\sqrt[]{x}-3}{\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+2\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}.\dfrac{\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(3x+3\sqrt{x}-3\right)-\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
c.
\(P=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
Do 1 nguyên nên P nguyên khi \(\dfrac{2}{\sqrt{x}-1}\) nguyên
\(\Rightarrow\sqrt{x}-1\inƯ\left(2\right)\)
Mà \(\sqrt{x}-1>-1;\forall x>0\)
\(\Rightarrow\sqrt{x}-1\in\left\{1;2\right\}\)
\(\Rightarrow x\in\left\{4;9\right\}\)
d.
\(P=\sqrt{x}\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\sqrt{x}\)
\(\Rightarrow\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(\Rightarrow\sqrt{x}+1=x-\sqrt{x}\)
\(\Rightarrow x-2\sqrt{x}-1=0\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1+\sqrt{2}\\\sqrt{x}=1-\sqrt{2}< 0\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}\)
1.33
\(A=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\left(x+2\right)+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
b.
- Với \(x=0\Rightarrow A=1\) (thỏa mãn)
- Với \(x\ne0\Rightarrow x\ge2\) (do x nguyên dương và \(x\ne1\))
\(\Rightarrow x+\sqrt{x}+1\ge2+\sqrt{2}+1=3+\sqrt{2}>2\)
\(\Rightarrow\dfrac{2}{x+\sqrt{x}+1}< 1\Rightarrow0< A< 1\Rightarrow A\) ko phải số nguyên do A nằm giữa 2 số nguyên liên tiếp
Vậy \(x=0\) là giá trị nguyên duy nhất thỏa mãn
1.34
a.
\(A=2\sqrt{2^2.2}-\sqrt{5^2.2}+\left|\sqrt{2}+1\right|=2.2\sqrt{2}-5\sqrt{2}+\sqrt{2}+1\)
\(=4\sqrt{2}-5\sqrt{2}+\sqrt{2}+1=1\)
\(B=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{1}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}\left(x-1\right)}=\dfrac{1}{\sqrt{x}}\)
b.
\(A=2B\Rightarrow1=\dfrac{2}{\sqrt{x}}\Rightarrow\sqrt{x}=2\)
\(\Rightarrow x=4\)
1.40
\(x=\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{2-1}}\)
\(=\dfrac{\left|\sqrt{2}-1\right|}{2}=\dfrac{\sqrt{2}-1}{2}\)
\(\Rightarrow2x=\sqrt{2}-1\)
\(\Rightarrow2x+1=\sqrt{2}\)
\(\Rightarrow\left(2x+1\right)^2=2\)
\(\Rightarrow4x^2+4x+1=2\)
\(\Rightarrow4x^2+4x-1=0\)
Do đó:
\(P=\left[\left(4x^5+4x^4-x^3\right)-\left(4x^3+4x^2-x\right)+\left(4x^2+4x-1\right)-1\right]^{2019}+2020\)
\(=\left[x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+0-1\right]^{2019}+2020\)
\(=\left(x^3.0-x.0+0-1\right)^{2019}+2020\)
\(=\left(-1\right)^{2019}+2020=2019\)
1.39
\(B=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{2\sqrt{x}+9}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-9+2\sqrt{x}+9}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
\(\dfrac{A}{B}>\dfrac{5}{3}\Rightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+3}>\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}>\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+3}{\sqrt{x}}>\dfrac{5}{3}\)
\(\Leftrightarrow1+\dfrac{3}{\sqrt{x}}>\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}}>\dfrac{2}{3}\)
\(\Rightarrow\sqrt{x}< \dfrac{9}{2}\) (do \(\sqrt{x}>0\) với mọi x thuộc TXĐ)
\(\Rightarrow x< \dfrac{81}{4}\)
Kết hợp ĐKXĐ \(\Rightarrow0< x< \dfrac{81}{4}\)
