\(\text{a) 2(x+3)-3(x-1)=2}\)

\(2x+6-3x+3=2\)

\(2x-3x=2-3-6\)

\(-x=-7\)

\(x=7\)

\(\text{b) 7-(x-2)=5(2x-3)}\)

\(7-x+2=10x-15\)

\(-x-10x=-15-2-7\)

\(-11x=-24\)

\(x=-24:\left(-11\right)\)

\(x=\frac{24}{11}\)

\(\text{c) 32-4(0,5y-5)=3y+2}\)

\(32-2y+20=3y+2\)

\(-2y-3y=2-20-32\)

\(-y=-50\)

\(y=50\)

\(\text{d) 3(x-1)-x=2x-3}\)

\(3x-3-x=2x-3\)

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

\(0=0\)( vô nghiệm )

a) 2(x + 3) - 3(x - 1) = 2

<=> 2x + 6 - 3x + 3 = 2

<=> -x + 9 = 2

<=> -x = -2 - 9

<=> -x = -7

<=> x = 7

b) 7 - (x - 2) = 5(2x - 3)

<=> 7 - x + 2 = 10x - 15

<=> 9 - x = 10x - 15

<=> 9 - x - 10 = -15

<=> 9 - 11x = -15

<=> -11x = -15 - 9

<=> -11x = -24

<=> x = 24/11

c) 32 - 4(0,5y - 5) = 3y + 2

<=> 32 - 2y + 20 = 3y + 2

<=> 52 - 2y = 3y + 2

<=> 52 - 2y - 3y = 2

<=> 52 - 5y = 2

<=> -5y = 2 - 52

<=> -5y = -50

<=> y = 10

x(x + 1)(x - 1)(x + 2) = 24

<=> x^4 + 2x^3 - x^2 - 2x = 24

<=> x^4 + 2x^3 - x^2 - 2x - 24 = 0

<=> (x - 2)(x + 3)(x^2 + x + 4) = 0

<=> x - 2 = 0 hoặc x + 3 = 0 hoặc x^2 + x + 4 khác 0

<=> x = 2 hoặc x = -3

\(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\)\(\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\)
\(\Leftrightarrow\) \(\text{ }\left(x^2+x\right)\left(x^2+x-2\right)=24\)

Đặt \(x^2+x=a\), ta có:  \(a\left(a-2\right)=24\)
\(\Leftrightarrow\) \(a^2-2a-24=0\)
\(\Leftrightarrow\) \(\left(a-6\right)\left(a+4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}a-6=0\\a+4=0\end{cases}}\) \(\Leftrightarrow\) \(\orbr{\begin{cases}a=6\\a=-4\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x^2+x=6\\x^2+x=-4\end{cases}}\)\(\Leftrightarrow\) \(\orbr{\begin{cases}x^2+x-6=0\\x^2+x+4=0\end{cases}}\)\(\Leftrightarrow\) \(\orbr{\begin{cases}\left(x+3\right)\left(x-2\right)=0\\x^2+x+\frac{1}{4}+\frac{11}{4}=0\end{cases}}\) (1)

Có : \(x^2+x+\frac{1}{4}+\frac{11}{4}=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\ge0+\frac{11}{4}>0\forall x\) (2)

(1); (2)\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}}\)

Vậy PT có tập nghiệm: S = {-3; 2}

D max thì \(\frac{6}{X^2+2}\)max

mà \(\frac{6}{X^2+2}\) thì X²+2 min   (1)

Ta có X² \(\ge0\)\(\forall X\)

=>X²+2\(\ge2\forall X\)(2)

Từ (1),(2)=> X²+2=2 <=> X =0

Thay X=0 ta có D = 3

Vậy D max =3 <=> X=0

Ta có: x² + 2 \(\ge\)2 \(\forall\) x=> \(\frac{6}{x^2+2}\le\frac{6}{2}=3\forall x\)

Dấu "=" xảy ra <=> x = 0

Vậy MaxD = 3 khi x = 0

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

\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x};x\ne2;x\ne0\)

\(\Leftrightarrow\frac{x+2}{x-2}-\frac{1}{x}-\frac{2}{x^2-2x}=0\)

\(\Leftrightarrow\frac{x+2}{x-2}-\frac{1}{x}-\frac{2}{x\times\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{x\times\left(x+2\right)-\left(x-2\right)-2}{x\times\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{x\times\left(x+2\right)-x+2-2}{x\times\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{x^2+2x-x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x^2+x}{x\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{x\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{x+1}{x-2}=0\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

1) y/(y + 2) - 3/(y - 2) = (y^2 + 8)/(y^2 - 4)

<=> y/(y + 2) - 3/(y - 2) = (y^2 + 8)/((y - 2)(y + 2))

<=> y(y - 2) - 3(y + 2) = y^2 + 8

<=> y^2 - 2y - 3y - 6 = y^2 + 8

<=> y^2 - 5y - 6 = y^2 + 8

<=> -5y - 6 = 8

<=> -5y = 8 + 6

<=> -5y = 14

<=> y = -14/5

2) 7/(2x - 3) + 1/(2x - 2) = 3/(x - 1)

<=> 14(x - 1) + 2x - 3 = 6(2x - 3)

<=> 14x - 14 + 2x - 3 = 12x - 18

<=> 16x - 17 = 12x - 18

<=> 16x - 17 - 12x = -18

<=> 4x - 17 = -18

<=> 4x = -18 + 17

<=> 4x = -1

<=> x = -1/4