ĐKXĐ : \(x\ne0;x\ne\pm1\)

a) Bạn ghi lại rõ đề.

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

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

c) \(P=A.B=\dfrac{x^2+x-2}{x.\left(x-1\right)}=\dfrac{\left(x-1\right).\left(x+2\right)}{x\left(x-1\right)}=\dfrac{x+2}{x}=1+\dfrac{2}{x}\)

Không tồn tại Min P \(\forall x\inℝ\)

a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)

a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)

= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)

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

Có vài bước mình làm tắc á nha :>

a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x+2}{6}\)

\(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-6}{6}\cdot\dfrac{1}{x-2}=\dfrac{-1}{x-2}\)

b: x=2 ko thỏa mãn ĐKXĐ

=>Loại

Khi x=3 thì A=-1/(3-2)=-1

c: A=2

=>x-2=-1/2

=>x=3/2

1,

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

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

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

2.

\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)

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

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

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

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

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

\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)

1) Sửa đề: x=0,09

Thay x=0,09 vào A, ta được:

\(A=\dfrac{\sqrt{0.09}}{\sqrt{0.09}-1}=\dfrac{0.3}{0.3-1}=\dfrac{0.3}{-0.7}=\dfrac{-3}{7}\)

\(a,A=\dfrac{1}{2}x^3-2x^2-4x-\dfrac{1}{2}x^3-x+1\\ =-2x^2-5x+1\)

b, Thay x=2 vào A ta có:
\(A=-2.2^2-5.2+1=-8-10+1=-17\)

a)  Ta có:   \(A=\dfrac{x}{x+2}-\dfrac{2x}{x-2}+\dfrac{x^2+12}{x^2-4}\left(x\ne\pm2\right)\)

\(A=\dfrac{x\left(x-2\right)-2x\left(x+2\right)+x^2+12}{\left(x-2\right)\left(x+2\right)}\)

\(A=\dfrac{x^2-2x-2x^2-4x+x^2+12}{\left(x-2\right)\left(x+2\right)}\)

\(A=\dfrac{-6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(A=\dfrac{-6}{x+2}\)

b) Để A có giá trị nguyên thì \(x+2\inƯ\left(6\right)\)

Mà \(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Từ đó, ta có:

\(x+1=1\Leftrightarrow x=0\) ( nhận )

\(x+1=-1\Leftrightarrow x=-2\)  ( loại )

\(x+1=2\Rightarrow x=1\) ( nhận )

\(x+1=-2\Rightarrow x=-3\) ( nhận )

\(x+1=3\Rightarrow x=2\) ( loại )

\(x+1=-3\Rightarrow x=-4\) ( nhận )

\(x+1=6\Rightarrow x=5\) ( nhận )

\(x+1=-6\Rightarrow x=-7\) ( nhận )

Vậy để A nhận giá trị nguyên thì \(x\in\left\{-7;-4;-3;0;1;5\right\}\)

\(a,\dfrac{x}{x+2}-\dfrac{2x}{x-2}+\dfrac{x^2+12}{x^2-4}\)

\(=\dfrac{x}{x+2}-\dfrac{2x}{x-2}+\dfrac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2-2x-2x^2-4x+x^2+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-6x+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-6}{x-2}\)

\(b,\) Để \(A\in Z\) thì \(\dfrac{-6}{x-2}\in Z\)

\(\Rightarrow x-2\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Vậy \(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)