a: \(B=\dfrac{10x}{\left(x+4\right)\left(x-1\right)}-\dfrac{2x-3}{x+4}-\dfrac{x+1}{x-1}\)

\(=\dfrac{10x-\left(2x^2-2x-3x+3\right)-\left(x^2+5x+4\right)}{\left(x+4\right)\left(x-1\right)}\)

\(=\dfrac{10x-2x^2+5x-3-x^2-5x-4}{\left(x+4\right)\left(x-1\right)}\)

\(=\dfrac{-3x^2+10x-7}{\left(x+4\right)\left(x-1\right)}\)

\(=\dfrac{-\left(3x^2-10x+7\right)}{\left(x-1\right)\left(x+4\right)}=-\dfrac{\left(x-1\right)\left(3x-7\right)}{\left(x-1\right)\left(x+4\right)}\)

\(=\dfrac{-3x+7}{x+4}\)

b: \(B+3=\dfrac{-3x+7+3x+12}{x+4}=\dfrac{19}{x+4}>0\)

=>B>-3

\(A=x\left(x+4\right)-6\left(x-1\right)\left(x+1\right)+\left(2x-1\right)^2\)

\(A=x^2+4x-6\left(x^2-1\right)+\left(4x^2-4x+1\right)\)

\(A=x^2+4x-6x^2+6+4x^2-4x+1\)

\(A=-x^2+7\)

Để A có giá trị bằng 3 thì :

\(-x^2+7=3\)

\(-x^2=-4\)

\(x^2=4\)

\(x\in\left\{\pm2\right\}\)

Vậy..........

dkxd  \(\hept{\begin{cases}\\\end{cases}}x-2=0;x+2=0\Leftrightarrow\hept{\begin{cases}\\\end{cases}x=+2;x=-2}\)

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

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

tới khúc này bí rồi ^^

a,ĐKXĐ của A là:\(x\ne+2;-2\)

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

c,Để A\(\in\)Z=> (x+2)(x-2)\(\inƯ\)(4) hay \(x^2-4\inƯ\)(4)=\(\left(4;-4;2;-2;1;-1\right)\)

Ta có bảng

Vậy A\(Z=>x\in\)( 0;\(\sqrt{8};\sqrt{6};\sqrt{2};\sqrt{5}\))

\(\frac{-4}{\left(x+2\right)\left(x-2\right)}\) chơ bn

a:

ĐKXĐ: x<>2

|2x-3|=1

=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Thay x=1 vào A, ta được:

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

b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)

\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)

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

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

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

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

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

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

Để P lớn nhất thì \(\dfrac{2}{x-2}\) max

=>x-2=1

=>x=3(nhận)

a) \(B=\left[\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{x-4}{x-3}-\frac{\left(x-1\right)}{x+3}\right]:\left(\frac{x+3-1}{x+3}\right)\)

ĐK: \(\hept{\begin{cases}x\ne3\\x\ne-3\end{cases}}\)

\(=\left[\frac{21+x-4-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\left(\frac{x+2}{x+3}\right)\)

\(=\left[\frac{21+x-4-x^2+3x+x-3}{\left(x+3\right)\left(x-3\right)}\right]\times\left(\frac{x+3}{x+2}\right)\)

\(=\left(\frac{-x^2+5x+14}{x-3}\right)\left(\frac{1}{x+2}\right)\)

\(=\frac{-\left(x^2+2x-7x-14\right)}{\left(x-3\right)\left(x+2\right)}\)

\(=\frac{-\left(x+2\right)\left(x-7\right)}{\left(x-3\right)\left(x+2\right)}\)

\(=\frac{7-x}{x-3}\)

b) \(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Mà \(x\ne-3\)

\(\Rightarrow x=2\)

Thế \(x=2\)vào B ta được:

\(B=\frac{7-2}{2-3}=-5\)

c) \(B=\frac{7-x}{x-3}=\frac{-3}{5}\)

\(\Leftrightarrow5\left(7-x\right)=-3\left(x-3\right)\)

\(\Leftrightarrow35-5x+3x-9=0\)

\(\Leftrightarrow-2x=-26\)

\(\Leftrightarrow x=13\)

Vậy để \(B=\frac{-3}{5}\)thì \(x=13\)

d) B<0\(\Rightarrow\frac{7-x}{x-3}< 0\)

TH1: \(\hept{\begin{cases}7-x< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x>7\\x>3\end{cases}\Rightarrow}x>7}\)

TH2: \(\hept{\begin{cases}7-x>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 7\\x< 3\end{cases}\Rightarrow}x< 3}\)

Để B<0 thì x>7 hoặc x<3

a) \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)         ĐKXĐ: x khác =-3; x khác -2

\(B=\frac{21+x^2-x-12-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}:\frac{x+2}{x+3}\)

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

\(B=\frac{3\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x+2}\)

\(B=\frac{3}{x-3}\)

b) bước đầu tiên ta phải tìm x:

 \(\left|2x+1\right|=5\)

TH1: 2x+1=5                      TH2: 2x+1=-5

            2x=4                                 2x=-6

          x=2 (nhận)                             x=-3 (loại)

thay x=2 vào biểu thức B, ta được:

\(B=\frac{3}{2-3}=\frac{3}{-1}=-3\)

vậy B=-3 tại x=2

c) Để \(B=-\frac{3}{5}\)thì \(\frac{3}{x-3}=-\frac{3}{5}\)

\(\Leftrightarrow-3\left(x-3\right)=15\)

\(\Leftrightarrow x-3=-5\)

\(\Leftrightarrow x=-2\)

vậy \(x=-2\)thì \(B=-\frac{3}{5}\)

d) để B<0 thì \(\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)

vậy để B<0 thì x phải < 3 và x khác -3

khong co may tinh sao lam

a: ĐKXĐ: x<>2; x<>-2

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

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

c: Khi x=-3 thì \(A=\dfrac{3\cdot\left(-3\right)^2-4\cdot3+6}{2\left(-3-2\right)\left(-3+2\right)}=\dfrac{21}{10}\)