a: \(B=\dfrac{3x\left(2x-3\right)-4\left(2x+3\right)-4x^2+23x+12}{\left(2x-3\right)\left(2x+3\right)}\cdot\dfrac{2x+3}{x+3}\)

\(=\dfrac{6x^2-9x-8x-12-4x^2+23x+12}{2x-3}\cdot\dfrac{1}{x+3}\)

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

b: 2x^2+7x+3=0

=>(2x+3)(x+2)=0

=>x=-3/2(loại) hoặc x=-2(nhận)

Khi x=-2 thì \(A=\dfrac{2\cdot\left(-2\right)}{-2-3}=\dfrac{-4}{-7}=\dfrac{4}{7}\)

d: |B|<1

=>B>-1 và B<1

=>B+1>0 và B-1<0

=>\(\left\{{}\begin{matrix}\dfrac{2x+2x-3}{2x-3}>0\\\dfrac{2x-2x+3}{2x-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\\dfrac{4x-3}{2x-3}>0\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{4}\)

Đề không hiển thị biểu thức B. Bạn xem lại.

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

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

b: |x|=3

=>x=3 hoặc x=-3

Khi x=3 thì \(B=\dfrac{-4\cdot3}{2\cdot3-1}=\dfrac{-12}{5}\)

Khi x=-3 thì \(B=\dfrac{-4\cdot\left(-3\right)}{2\cdot\left(-3\right)-1}=\dfrac{12}{-7}=\dfrac{-12}{7}\)

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

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

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

\(\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}\)

Để biểu thức trên nhận giá trị âm khi \(\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}< 0\)

\(\Rightarrow x^3-2x^2-4x+8< 0\)do \(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-2x\left(x+2\right)< 0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2< 0\Leftrightarrow x< -2\)

a: \(P=\dfrac{8+5x-2x-8}{x\left(x+4\right)}=\dfrac{3x}{x\left(x+4\right)}=\dfrac{3}{x+4}\)

b: Khi x=1/2 thì P=3/(1/2+4)=3:9/2=3*2/9=6/9=2/3

a) ĐKXĐ: \(\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\\2-x\ne0\end{cases}}\) => \(\hept{\begin{cases}x\ne-2\\x\ne\pm2\\x\ne2\end{cases}}\) => \(x\ne\pm2\)

Ta có:Q = \(\frac{x-1}{x+2}+\frac{4x+4}{x^2-4}+\frac{3}{2-x}\)

Q = \(\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4x+4}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

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

b) ĐKXĐ P: x - 3 \(\ne\)0 => x \(\ne\)3

Ta có: P = 3 => \(\frac{x+2}{x-3}=3\)

=> x + 2 = 3(x - 3)

=> x + 2 = 3x - 9

=> x - 3x = -9 - 2

=> -2x = -11

=> x = 11/2 (tm)

Với x = 11/2 thay vào Q => Q = \(\frac{\frac{11}{2}}{\frac{11}{2}+2}=\frac{11}{15}\)

c) Với x \(\ne\)\(\pm\)2; x \(\ne\)3

Ta có: M = PQ = \(\frac{x+2}{x-3}\cdot\frac{x}{x+2}=\frac{x}{x-3}=\frac{x-3+3}{x-3}=1+\frac{3}{x-3}\)

Để M \(\in\)Z <=> 3 \(⋮\)x - 3

=> x - 3 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng:

Vậy ...