a) đk: x khác 1; \(\dfrac{3}{2}\)

 \(P=\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5}{2x-3}\right]:\left(\dfrac{3-3x+2}{1-x}\right)\)

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

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

b) Có \(\left|3x-2\right|+1=5\)

<=> \(\left|3x-2\right|=4\)

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

TH1: Thay x = 2 vào P, ta có:

P = \(\dfrac{-1}{2.2-3}=-1\)

TH2: Thay x = \(\dfrac{-2}{3}\)vào P, ta có:

P = \(\dfrac{-1}{2.\dfrac{-2}{3}-3}=\dfrac{3}{13}\)

c) Để P > 0

<=> \(\dfrac{-1}{2x-3}>0\)

<=> 2x - 3 <0

<=> x < \(\dfrac{3}{2}\) ( x khác 1)

d) P = \(\dfrac{1}{6-x^2}\)

<=> \(\dfrac{-1}{2x-3}=\dfrac{1}{6-x^2}\)

<=> \(\dfrac{-1}{2x-3}=\dfrac{-1}{x^2-6}\)

<=> 2x - 3 = x² - 6

<=> x² - 2x - 3 = 0

<=> (x-3)(x+1) = 0

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

\(\Leftrightarrow2\left(2x-3\right)-9=5-3x-2\)

=>4x-6-9=-3x+3

=>7x=18

hay x=18/7

\(\Leftrightarrow\dfrac{2x-3}{3}-\dfrac{3}{2}-\dfrac{5-3x}{6}+\dfrac{1}{3}=0\)

\(\Leftrightarrow\dfrac{2\left(2x-3\right)-9-5+3x+2}{6}=0\)

\(\Leftrightarrow4x-6-9-5+3x+2=0\)

\(\Leftrightarrow7x-18=0\)

\(\Leftrightarrow7x=18\)

\(\Leftrightarrow x=\dfrac{18}{7}\)

tham khaot

⇔2(2x−3)−9=5−3x−2

=>4x-6-9=-3x+3

=>7x=18

hay x=18/7

3x/6x+3=x/2x+2

⇒6x²+6x=6x²+3x

⇒ 6x²+6x-6x²-3x=0

⇒ 3x=0

⇒ x=0

`(3x)/(6x+3)=x(2x+2)`      ĐKXĐ : `x \ne -1/2 ; x \ne -1`

`->3x(2x+2)=x(6x+3)`

`->6x^2+6x=6x^2+3x`

`->3x=0`

`->x=0` ( tm )

Vậy `x=0`

Ta có: \(\dfrac{3x}{6x+3}=\dfrac{x}{2x+2}\)

\(\Leftrightarrow3x\left(2x+2\right)=x\left(6x+3\right)\)

\(\Leftrightarrow6x^2+6x-6x^2-3x=0\)

\(\Leftrightarrow3x=0\)

hay x=0

cứuuuuuuuuuuuu

ĐK: \(y\ne0\)

\(\dfrac{x+3}{6}=\dfrac{2x+1}{7}\Leftrightarrow7x+21=12x+6\\ \Leftrightarrow5x=15\Leftrightarrow x=3\\ \Leftrightarrow\dfrac{3+3}{6}=\dfrac{2+3\cdot3}{y}\Leftrightarrow\dfrac{11}{y}=1\Leftrightarrow y=11\)

Vậy \(x=3;y=11\)

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

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

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

\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)

\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)

a.

\(y'=\dfrac{3}{cos^2\left(3x-\dfrac{\pi}{4}\right)}-\dfrac{2}{sin^2\left(2x-\dfrac{\pi}{3}\right)}-sin\left(x+\dfrac{\pi}{6}\right)\)

b.

\(y'=\dfrac{\dfrac{\left(2x+1\right)cosx}{2\sqrt{sinx+2}}-2\sqrt{sinx+2}}{\left(2x+1\right)^2}=\dfrac{\left(2x+1\right)cosx-4\left(sinx+2\right)}{\left(2x+1\right)^2}\)

c.

\(y'=-3sin\left(3x+\dfrac{\pi}{3}\right)-2cos\left(2x+\dfrac{\pi}{6}\right)-\dfrac{1}{sin^2\left(x+\dfrac{\pi}{4}\right)}\)