Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

a) \(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

b) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)

\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)

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

Kết hợp đk:

\(\Rightarrow x\in\left\{0\right\}\)

d) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)

\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}\in Z\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow2⋮\sqrt{x}+1\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}+1\in\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;1\right\}\\ \Leftrightarrow x\in\left\{0;1\right\}\)

\(d,P=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

Có \(\dfrac{2}{\sqrt{x}+1}>0\left(2>0;\sqrt{x}+1>0\right)\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}< 1\Leftrightarrow P< 1\)

\(e,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

Có \(\sqrt{x}+1\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\le2\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)

\(P_{min}=-1\Leftrightarrow x=0\)

1: \(B=\dfrac{6x+x^2-3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x^2+3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x}{x-3}\)

Bài 1 :

a) \(ĐKXĐ:x\ne1\)

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

\(\Leftrightarrow A=\frac{3+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\left(x+1\right)\)

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

b) Thay x = \(\frac{2}{5}\)vào A ta được :

\(A=\frac{\frac{2}{5}+2}{\frac{2}{5}-1}=\frac{\frac{12}{5}}{-\frac{3}{5}}=-4\)

c) Để \(A=\frac{5}{4}\)

\(\Leftrightarrow\frac{x+2}{x-1}=\frac{5}{4}\)

\(\Leftrightarrow4x+8=5x-5\)

\(\Leftrightarrow x=13\)

d) Để \(A>\frac{1}{2}\)

\(\Leftrightarrow\frac{x+2}{x-1}>\frac{1}{2}\)

\(\Leftrightarrow\frac{x+2}{x-1}-\frac{1}{2}>0\)

\(\Leftrightarrow2x+4-x+1>0\)

\(\Leftrightarrow x+5>0\)

\(\Leftrightarrow x>-5\)

Bài 2 :

a) \(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne0\end{cases}}\)

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

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

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

b) Để \(A=1\)

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

\(\Leftrightarrow2x-1=x+1\)

\(\Leftrightarrow x=2\)

b) Để \(A< 2\)

\(\Leftrightarrow\frac{2x-1}{x+1}< 2\)

\(\Leftrightarrow\frac{2x-1}{x+1}-2< 0\)

\(\Leftrightarrow2x-1-2x-1< 0\)

\(\Leftrightarrow-2< 0\)(luôn đúng)

Vậy A < 2 <=> mọi x

câu c la tim gì vậy bn

bạn ghi lại biểu thức cho rõ đi...

Q=(\(\frac{1}{x+1}\)-\(\frac{6x+3}{x^3+1}\)+\(\frac{2}{x^2-x+1}\)) : (x+2)

`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`

`b)` Với `x ne -1;x ne -5` có:

`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`

`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`

`A=[x^2-3x-4]/[(x+1)(x+5)]`

`A=[(x+1)(x-4)]/[(x+1)(x+5)]`

`A=[x-4]/[x+5]`

`c)` Với `x ne -5; x ne -1; x ne 4` có:

`P=A.B=[x-4]/[x+5].[-10]/[x-4]`

           `=[-10]/[x+5]`

Để `P` nguyên `<=>[-10]/[x+5] in ZZ`

    `=>x+5 in Ư_{-10}`

Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`

`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)