f) Tìm x để F>0

k mình mình giải chi tiết cho

có ai giải dc ko

Giải được nhưng mà lâu quá rồi giải làm gì nữa

\(A=\left(\frac{2x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{5-x^2}{x+2}\right)\) ĐKXĐ : \(x\ne\pm2\)

\(A=\left(\frac{2x}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4}{x+2}+\frac{5-x^2}{x+2}\right)\)

\(A=\left(\frac{2x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+5-x^2}{x+2}\right)\)

\(A=\frac{x-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{1}\)

\(A=\frac{x-6}{x-2}\)

b, ta có \(/\frac{1}{2}/=\frac{1}{2}=\frac{-1}{2}\)

TH1 : Thay x = 1/2 vào A 

.....

Th2 : Thay x = -1/2 vào A :

... 

Bn tự tính vào kết luận 

c, Để \(A< 0\) \(\Rightarrow\frac{x-6}{x-2}\)\(< 0\)

Trường hợp 1 : \(\hept{\begin{cases}x-6>0\\x-2< 0\end{cases}\Rightarrow\hept{\begin{cases}x>6\\x< 2\end{cases}\Rightarrow x\in}\varnothing}\)

Trường hợp 2 \(\hept{\begin{cases}x-6< 0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x< 6\\x>2\end{cases}\Rightarrow}2< x< 6}\)

Vậy để \(A< 0\)thì \(2< x< 6\)

\(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\)   ĐK đề bài

\(=\frac{x-5+2\left(x+5\right)-2x-10}{\left(x+5\right)\left(x-5\right)}=\frac{-\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=-\frac{1}{x-5}\)

b/ có A=-3 => \(-\frac{1}{x-5}=-3 \Rightarrow x-5=\frac{1}{3}\Rightarrow x=\frac{16}{3}\)

có \(9x^2-42x+49=\left(3x-7\right)^2=\left(\frac{3.16}{3}-7\right)^2=81\)

Bài làm

Như đã nhắn là mình sẽ làm theo quan điểm của mình là 5/(x^2 - 1) nha

\(A=\left[\frac{3\left(x+2\right)}{2x^3+2x+2x^2+2}+\frac{2x^2-x-10}{2x^3-2-2x^2+2x}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2x+2}-\frac{3}{2x-2}\right]\)

\(A=\left[\frac{3\left(x+2\right)}{2x^2\left(x+1\right)+2\left(x+1\right)}+\frac{2x^2+4x-5x-10}{\left(2x^3-2x^2\right)+\left(2x-2\right)}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)

\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{2x\left(x+2\right)-5\left(x+2\right)}{2x^2\left(x-1\right)+2\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)

\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{\left(2x-5\right)\left(x+2\right)}{\left(2x^2+2\right)\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)

\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}+\frac{\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3\left(x-1\right)}{2\left(x^2-1\right)}-\frac{3\left(x+1\right)}{2\left(x^2-1\right)}\right]\)

\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)+\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10}{2\left(x^2-1\right)}+\frac{3x-3}{2\left(x^2-1\right)}-\frac{3x+3}{2\left(x^2-1\right)}\right]\)

\(A=\left[\frac{\left(x+2\right)\left[3x-3+\left(2x-5\right)\left(x+1\right)\right]}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10+3x-3-3x-3}{2\left(x^2-1\right)}\right]\)

\(A=\left[\frac{\left(x+2\right)\left(3x-3+2x^2+2x-5x-5\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\frac{4}{2\left(x^2-1\right)}\)

\(A=\frac{\left(x+2\right)\left(2x^2-8\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\cdot\frac{\left(x^2-1\right)}{2}\)

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

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

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

:>>> Chả biết đúng không nữa nhưng số to quá :>>