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

b: \(B=\dfrac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{2x+1}{x+2}\)

\(=\dfrac{8x-4}{2x-1}\cdot\dfrac{1}{x+2}=\dfrac{4}{x+2}\)

a, ĐK: \(\hept{\begin{cases}x+2\ne0\\x\ne0\end{cases}\Rightarrow}\hept{\begin{cases}x\ne-2\\x\ne0\end{cases}}\)

b, \(B=\left(1-\frac{x^2}{x+2}\right).\frac{x^2+4x+4}{x}-\frac{x^2+6x+4}{x}\)

\(=\frac{-x^2+x+2}{x+2}.\frac{\left(x+2\right)^2}{x}-\frac{x^2+6x+4}{x}\)

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

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

\(=\frac{-x^3-2x^2-2x}{x}=-x^2-2x-2\)

c, x = -3 thỏa mãn ĐKXĐ của B nên với x = -3 thì 

\(B=-\left(-3\right)^2-2.\left(-3\right)-2=-9+6-2=-5\)

d, \(B=-x^2-2x-2=-\left(x^2+2x+1\right)-1=-\left(x+1\right)^2-1\le-1\forall x\)

Dấu "=" xảy ra khi \(x+1=0\Rightarrow x=-1\)

Vậy GTLN của B là - 1 khi x = -1

Thanks bạn ;)

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

\(\Leftrightarrow x-2< 0\)  ( vì \(-1< 0\))

\(\Leftrightarrow x< 2\)

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

\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)

  \(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

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

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

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

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

theo câu a) \(A=\frac{-1}{x-2}\)  với ĐKXĐ: \(x\ne\pm2\)

b) \(\left|2x-1\right|=3\)

\(\Rightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)   \(\Rightarrow x=-1\)  ( vì \(x=2\)  ko TM ĐKXĐ )

+) khi \(x=-1\)thì \(A=\frac{-1}{-1-2}=\frac{-1}{-3}=\frac{1}{3}\)

vậy khi \(x=-1\)  thì \(A=\frac{1}{3}\)

a) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}=1+\dfrac{4}{\sqrt{x}-2}\)

Để A nguyên thì 4 ⋮ √x - 2

\(\Rightarrow\sqrt{x}-2\inƯ\left(4\right)\)

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

\(\Rightarrow\sqrt{x}\in\left\{3;1;4;0;6;-2\right\}\)

Mà x \(\sqrt{x}\ge0\)

=> x thuộc {9; 1; 16; 0; 36}

b) 

b, P=x+2x+3−5x2+3x−2x−6+12−xP=x+2x+3−5x2+3x−2x−6+12−x

=x+2x+3−5(x+3)(x−2)−1x−2=x+2x+3−5(x+3)(x−2)−1x−2

=(x+2)(x−2)(x+3)(x−2)−5(x+3)(x−2)−x+3(x+3)(x−2)=(x+2)(x−2)(x+3)(x−2)−5(x+3)(x−2)−x+3(x+3)(x−2)

=x2−4−5−x−3(x+3)(x−2)=x2−x−12(x+3)(x−2)=x2−4−5−x−3(x+3)(x−2)=x2−x−12(x+3)(x−2)

=x2−4x+3x−12(x+3)(x−2)=x2−4x+3x−12(x+3)(x−2)

=(x−4)(x+3)(x+3)(x−2)=x−4x−2=(x−4)(x+3)(x+3)(x−2)=x−4x−2

c, Để P=−34P=−34

⇔x−4x−2=−34⇔x−4x−2=−34

⇔4(x−4)=−3(x−2)⇔4(x−4)=−3(x−2)

⇔4x−16+3x−6=0⇔4x−16+3x−6=0

⇔7x−22=0⇔7x−22=0

⇔x=227⇔x=227

d, Để P có giá trị nguyên

⇔x−4⋮x−2⇔x−4⋮x−2

⇔(x−2)−2⋮x−2⇔(x−2)−2⋮x−2

⇔2⋮x−2⇔x−2∈Ư(2)={1;−1;2;−2}⇔2⋮x−2⇔x−2∈Ư(2)={1;−1;2;−2}

e,

x2−9=0x2−9=0

⇒x2=9⇒[x=3x=−3⇒x2=9⇒[x=3x=−3

Với x=3,có :

x−4x−2=3−43−2=−11=−1x−4x−2=3−43−2=−11=−1

Với x=-3,có :

x−4x−2=−3−4−3−2=75x−4x−2=−3−4−3−2=75