a: ĐKXĐ: x<>1; x<>-1

b: \(A=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)

c: Để A nguyên thì x+1-2 chia hết cho x+1

=>\(x+1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{0;-2;-3\right\}\)

\(a,ĐK:x\ne3;x\ge1\\ b,A=\dfrac{\left(\sqrt{x-1}+\sqrt{2}\right)\left(\sqrt{x-1}-\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x-1}+\sqrt{2}\\ b,A=4\left(2-\sqrt{3}\right)\\ \Leftrightarrow\sqrt{x-1}+\sqrt{2}=8-4\sqrt{3}\\ \Leftrightarrow\sqrt{x-1}=8-4\sqrt{3}-\sqrt{2}\\ \Leftrightarrow x-1=\left(8-4\sqrt{3}-\sqrt{2}\right)^2\\ \Leftrightarrow x=\left(8-4\sqrt{3}-\sqrt{2}\right)^2+1=...\\ d,A=\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\\ A_{min}=\sqrt{2}\Leftrightarrow x-1=0\Leftrightarrow x=1\)

a: ĐKXĐ: x<>1; x<>-1

b: \(P=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2\left(x+1\right)}\)

c: Để P=1/2 thì 1/2(x+1)=1/2

=>x+1=1

=>x=0

a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(A=\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\)

\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-4x^2-8x}{\left(x-2\right)\left(x+2\right)}=\dfrac{-4x}{x-2}\)

Bài 1: 

a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)

\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)

\(=\dfrac{2x}{x-1}\)

Mn ơi giúp mình với. Please!!!!!!!!!!!!!!!!!!!!!!!!!!

\(a,ĐK:x\ge1;x\ne3\\ b,A=\dfrac{\left(\sqrt{x-1}+\sqrt{2}\right)\left(\sqrt{x-1}-\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x-1}+\sqrt{2}\)

\(P=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}+\frac{1}{2-x}\)

a) P xác định \(\Leftrightarrow\hept{\begin{cases}x+3\ne0\\x-2\ne0\\2-x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}}\)

Vậy P xác định khi \(x\ne-3;x\ne2\)

b) \(P=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{1}{2-x}\left(x\ne-3;x\ne2\right)\)

\(\Leftrightarrow P=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)

\(\Leftrightarrow P=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)

\(\Leftrightarrow P=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)

\(\Leftrightarrow P=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)

Vậy \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2\right)\)

c) \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2\right)\)

Để P\(=-\frac{3}{4}\Rightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)

\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)

\(\Leftrightarrow4x-16=-3x+6\)

\(\Leftrightarrow4x+3x=6+16\)

\(\Leftrightarrow7x=22\)

\(\Leftrightarrow x=\frac{22}{7}\left(tm\right)\)

Vậy \(=\frac{22}{7}\)thì \(P=\frac{-3}{4}\)

d) \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2\right)\)

\(\Leftrightarrow P=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)

Để P nhận giá trị nguyên thì \(1-\frac{2}{x-2}\)nhận giá trị nguyên 

\(\Leftrightarrow\frac{2}{x-2}\)nhận giá trị nguyên (1)

\(x\inℤ\Rightarrow x-2\inℤ\)(2)

\(\left(1\right)\left(2\right)\Rightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Ta có bảng giá trị

Vậy \(x\in\left\{0;-1;3;4\right\}\)