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a.
\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\left(x-\sqrt{x}-2\sqrt{x}+2\right)\\ =\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\left[\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\right]\\ =\frac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\left[\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\right]\\ =\frac{\sqrt{x}-1}{\sqrt{x}}\)
b.
\(A=\frac{\sqrt{x}-1}{\sqrt{x}}< \frac{1}{2}\\ \Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}-\frac{1}{2}< 0\\ \Leftrightarrow\frac{2\left(\sqrt{x}-1\right)-\sqrt{x}}{2\sqrt{x}}< 0\\ \Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}}< 0\\ \Leftrightarrow\sqrt{x}-2< 0\\ \Leftrightarrow x< 4\)
Vậy với 0<x<4 thì A < \(\frac{1}{2}\)
c. Ta có \(A=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\)
Để A đạt giá trị nguyên thì \(1⋮\sqrt{x}\Leftrightarrow\sqrt{x}\inƯ\left(1\right)\)
Mà \(\sqrt{x}>0\forall x>0\Rightarrow x=1\)
Vậy với x=1 thì A đạt giá trị nguyên
Sửa đề :
a) \(A=\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{4}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-5}{x-\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{x-\sqrt{x}+4\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-4-x+\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}\)
b) \(A=4\)
\(\Leftrightarrow\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}=4\)
\(\Leftrightarrow x+3\sqrt{x}+4=4\sqrt{x}+4\)
\(\Leftrightarrow x-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy \(A=4\Leftrightarrow x\in\left\{0;1\right\}\)
Rgọn : P= (\(\frac{x+2}{x\sqrt{x}+1}\) - \(\frac{1}{\sqrt{x}+1}\)) . \(\frac{4\sqrt{x}}{3}\)
= \(\frac{x+2-1\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\). \(\frac{4\sqrt{x}}{3}\)
= \(\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\).\(\frac{4\sqrt{x}}{3}\)
= \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
=>
\(Q=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
b.\(Q< 1\)
\(\Leftrightarrow x-\sqrt{x}-2< x-5\sqrt{x}+6\)
\(\Leftrightarrow4\sqrt{x}-8< 0\)
\(\Leftrightarrow0\le x< 4\)
Vay de Q<1 thi \(0\le0< 4\)
a)ĐKXĐ : x≠-3;2
b)A=x+1/x+3 - 10/(x^2+3x)-(2x+6) + 5/x-2
A=x+1/x+3 -10/x ×( x+3)-2 × (x+3) + 5/x-2
A= x+1/x+3 - 10/(x-2)(x+3). + .5/x-2
A= (x+1)(x-2) /(x-2)(x+3). - 10/(x-2)(x+3) + 5(x+3)/(x-2)(x+3)
A= x^2-2x+x-2-10+5x+15/(x-2)(x+3)
A= x^2+4x+3/(x-2)(x+3)
A= (x^2+x)+(3x+3)/ (x-2)(x+3)
A= x×(x+1) + 3×(x+1) / (x-2)(x+3)
A= (x+3)(x+1)/(x-2)(x+3)
A=x+1/x-2
c) để A>0 thì x+1/x-2>0
Để x+1/x-2>0 thì x+1 và x-2 phải cung dấu
Ta có hai trường hợp
TH1: x+1<0 suy ra x<-1
x-2<0. suy ra x<1
Đoi chiếu ĐKXĐ ta có x<1;x≠-3
TH2: x+1>0 suy ra x>-1
x-2>0 suy ra x>2
=) x>-1; x≠2
(Đây là toán lớp 8 chứ)
a)\(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}=\frac{2+5\sqrt{x}}{x-4}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-\left(2+5\sqrt{x}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b)\(A=2\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}=2\)
\(\Leftrightarrow2\left(\sqrt{x}+2\right)=3\sqrt{x}\)
\(\Leftrightarrow2\sqrt{x}+4=3\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
\(\Rightarrow A\le2\)