x>9

a: Ta có: \(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\cdot\sqrt{x}}\)

\(=\left(2x+\sqrt{x}-1\right)\left(\frac{1}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}\right)\)

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

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

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

Ta có: \(M=1-\left(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\cdot\sqrt{x}}\right)\cdot\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\)

\(=1-\frac{\left(2\sqrt{x}-1\right)}{\left.\left(1-\sqrt{x}\right)\right.\left(1-\sqrt{x}+x\right)}\cdot\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}=1-\frac{x-\sqrt{x}}{1-\sqrt{x}+x}\)

\(=\frac{x-\sqrt{x}+1-x+\sqrt{x}}{x-\sqrt{x}+1}=\frac{1}{x-\sqrt{x}+1}\)

b: Để M là số nguyên thì \(x-\sqrt{x}+1\inƯ\left(1\right)\)

=>\(x-\sqrt{x}+1\in\left\lbrace1;-1\right\rbrace\)

mà \(x-\sqrt{x}+1=\left(\sqrt{x}-\frac12\right)^2+\frac34>0\forall x\) thỏa mãn ĐKXĐ
nên \(x-\sqrt{x}+1=1\)

=>\(x-\sqrt{x}=0\)

=>\(\sqrt{x}\left(\sqrt{x}-1\right)=0\)

=>x=0(nhận) hoặc x=1(loại)

Câu a:

​\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)

\(=\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{2x+2-3\sqrt{x}-1}{x-1}=\frac{2x-3\sqrt{x}+1}{x-1}\)

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

A nguyên khi và chỉ khi \(3⋮\left(\sqrt{x}+1\right)\)

• TH1 : \(\left(\sqrt{x}+1\right)=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
• TH2 : \(\left(\sqrt{x}-1\right)=3\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

Câu b : \(\frac{m\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)}=\sqrt{x}-2\Leftrightarrow2m\sqrt{x}-m-x+\sqrt{x}+2=0\)

\(\Leftrightarrow x-\left(2m+1\right)\sqrt{x}+m-2=0\)phương trình có hai nghiệm phân biệt khi 

\(\Delta>0\)hay \(\Delta=\left(2m+1\right)^2-\left(m-2\right)4=m^2+9>0\forall m\)

Câu C: để \(A=2-\frac{3}{\sqrt{x}+1}\ge2-\frac{3}{0+1}=-1\)\(\Rightarrow A_{Min}=-1\)khi \(x=0\)

mấy câu trả lời trước đâu rồi