a) Đk \(x\ne\pm1\), sau khi rút gọn ta được: (bạn tư làm)

   \(P=\frac{x}{x+1}\)

b) Khi \(\left|x-\frac{2}{3}\right|=\frac{1}{3}\) thì hoặc \(x-\frac{2}{3}=\frac{1}{3}\) hoặc \(x-\frac{2}{3}=-\frac{1}{3}\)

Hay là \(x=1\) hoặc \(x=\frac{1}{3}\)

Do để P có nghĩa thì \(x\ne\pm1\) nên \(x=\frac{1}{3}\), khi đó: 

 \(P=\frac{\frac{1}{3}}{\frac{1}{3}+1}=\frac{1}{4}\)

c) P > 1 khi \(\frac{x}{x+1}>1\)

   \(\Leftrightarrow1-\frac{1}{x+1}>1\)

   \(\Leftrightarrow\frac{1}{x+1}< 0\)

   \(\Leftrightarrow x< -1\)

e) Đề không rõ ràng

dễ mà ko bt lm à

1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)

2:

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

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

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

b: \(2P=2\sqrt{x}+5\)

=>\(P=\sqrt{x}+\dfrac{5}{2}\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}=\dfrac{2\sqrt{x}+5}{2}\)

=>\(\sqrt{x}\left(2\sqrt{x}+5\right)=2\sqrt{x}+2\)

=>\(2x+3\sqrt{x}-2=0\)

=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)

=>\(2\sqrt{x}-1=0\)

=>x=1/4

a

\(ĐKXĐ:x\in R\)

\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\left(x^4+\frac{1-x^4}{1+x^2}\right)\)

\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\left(x^4-x^2+1\right)\)

\(=\frac{\left(x^2-1\right)\left(x^4-x^2+1\right)}{x^4-x^2+1}-\frac{x^4-x^2+1}{x^2+1}\)

\(=x^2-1-\frac{x^4-x^2+1}{x^2+1}\)

\(=-1+\frac{x^4+x^2-x^4+x^2+1}{x^2+1}\)

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

b

Xét \(x>0\Rightarrow M>0\)

Xét \(x=0\Rightarrow M=0\)

Xét \(x< 0\Rightarrow M>0\)

Vậy \(M_{min}=0\) tại \(x=0\)

a) \(p=\left(\frac{x^2-x}{x+1}\right)\left(\frac{4x-2x+2}{x\left(x-1\right)}\right)\)

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

b)\(m=\frac{x+2-\left(x-2\right)+x^2+4x}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x+2}{x-2}=1+\frac{4}{x-2}\)

Để m nguyên thì \(4⋮x-2\)

\(\Rightarrow x-2\in\left\{1,2,4,-1,-2,-4\right\}\)

\(\Leftrightarrow x\in\left\{3,4,6,1,0,-2\right\}\)

\(M=\frac{1}{x-2}-\frac{1}{x+2}+\frac{x^2+4x}{x^2-4}\left(x\ne\pm2\right)\)

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

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

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

Để M có giá trị nguyên thì x+2 chia hết cho x-2

Ta có x+2=x-2+4

=> x-2+4 chia hết cho x-2

=>4 chia hết cho x-2

Vì x nguyên => x-2 nguyên

=> x-2 thuộc Ư (4)={-4;-2;-1;1;2;4}

Ta có bảng

Bạn giải hộ tui thêm 1 í câu a với...tìm các giá trị của x để P xác đihj

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

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

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

b, x=1 

M = \(\frac{3}{2}\)

c, M= 0 

=> x +\(\sqrt{x}\)+1= 0

mặt khác x+\(\sqrt{x}\)+1 = (\(\sqrt{x}\)+0,5)²+0,75 >0

=> x vô nghiệm........