Chụp rõ đề hơn đi bạn.

a) \(\dfrac{A}{x-2}=\dfrac{x^2+3x+2}{x^2-4}\)

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

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

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

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

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

a) x(x + 4) - 3x - 12 = 0

x(x + 4) - 3(x + 4) = 0

(x + 4)(x - 3) = 0

x = -4 hoặc x = 3

b) 2x(x - 4) - x + 4 = 0

2x(x - 4) - (x - 4) = 0

(x - 4)(2x - 1) = 0

x = 4 hoặc x = 1/2

c) 5x(x - 3) - x + 3 = 0

5x(x - 3) - (x - 3) = 0

(x - 3)(5x - 1) = 0

x = 3 hoặc x = 1/5.

a) x(x + 4) - 3x - 12 = 0

x(x + 4) - 3(x + 4) = 0

(x + 4)(x - 3) = 0

x = -4 hoặc x = 3

b) 2x(x - 4) - x + 4 = 0

2x(x - 4) - (x - 4) = 0

(x - 4)(2x - 1) = 0

x = 4 hoặc x = 1/2

c) 5x(x - 3) - x + 3 = 0

5x(x - 3) - (x - 3) = 0

(x - 3)(5x - 1) = 0

x = 3 hoặc x = 1/5.

a: \(\left(2x-1\right)^3-4x^2\left(2x-3\right)=5\)

=>\(8x^3-12x^2+6x-1-8x^3+12x^2=5\)

=>6x-1=5

=>6x=6

=>x=1

b: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

=>\(x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\)

=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)

=>12x-4=-10

=>12x=-6

=>\(x=-\frac{6}{12}=-\frac12\)

c: \(\left(x+2\right)^3-x^2\left(x+6\right)=4\)

=>\(x^3+6x^2+12x+8-x^3-6x^2=4\)

=>12x+8=4

=>12x=-4

=>\(x=-\frac{4}{12}=-\frac13\)

d: \(\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x^2=5\)

=>\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x^2=5\)

=>\(x^3+12x-8-x^3+x=5\)

=>13x-8=5

=>13x=13

=>x=1

e: \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x+2\right)\left(x^2-2x+4\right)\)

=>\(x^3-3x^2+3x-1+3\left(x^2+2x+1\right)=x^3+8\)

=>\(-3x^2+3x-1+3x^2+6x+3=8\)

=>9x+2=8

=>9x=6

=>\(x=\frac69=\frac23\)

f: \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)+6\left(x-2\right)\left(x+2\right)=60\)

=>\(x^3-6x^2+12x-8-\left(x^3-8\right)+6\left(x^2-4\right)=60\)

=>\(-6x^2+12x+6x^2-24=60\)

=>12x-24=60

=>12x=84

=>x=7

a: \(\left(x-3\right)^2-\left(x-3\right)\left(3-x^2\right)=0\)

=>\(\left(x-3\right)\left(x-3-3+x^2\right)=0\)

=>\(\left(x-3\right)\left(x^2+x-6\right)=0\)

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

=>\(\left[\begin{array}{l}x-3=0\\ x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-3\\ x=2\end{array}\right.\)

b: \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)

=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)

=>4x+13=11

=>4x=-2

=>\(x=-\frac24=-\frac12\)

c: \(\left(x-3\right)^2-\left(x+3\right)\left(x-5\right)=3x+4\)

=>\(x^2-6x+9-\left(x^2-5x+3x-15\right)=3x+4\)

=>\(x^2-6x+9-\left(x^2-2x-15\right)=3x+4\)

=>\(x^2-6x+9-x^2+2x+15=3x+4\)

=>-4x+24=3x+4

=>-7x=-20

=>\(x=\frac{20}{7}\)

d: \(3\left(x-5\right)^2+2x\left(x-5\right)=0\)

=>\(\left(x-5\right)\left\lbrack3\left(x-5\right)+2x\right\rbrack=0\)

=>(x-5)(3x-15+2x)=0

=>(x-5)(5x-15)=0

=>5(x-5)(x-3)=0

=>(x-5)(x-3)=0

=>\(\left[\begin{array}{l}x-5=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=3\end{array}\right.\)

e: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

=>(2x-1-x-3)(2x-1+x+3)=0

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

=>\(\left[\begin{array}{l}x-4=0\\ 3x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-\frac23\end{array}\right.\)

f: \(4\left(3x+2\right)\left(3x-2\right)-\left(6x+1\right)^2=7\)

=>\(4\left(9x^2-4\right)-36x^2-12x-1=7\)

=>\(36x^2-16-36x^2-12x-1=7\)

=>-12x-17=7

=>-12x=24

=>x=-2

g: \(3\left(2x-1\right)^2-6x\left(2x-3\right)=6\)

=>\(3\left(4x^2-4x+1\right)-6x\left(2x-3\right)=6\)

=>\(12x^2-12x+3-12x^2+18x=6\)

=>6x+3=6

=>6x=3

=>\(x=\frac36=\frac12\)

h: \(\left(2x+3\right)^2-4\left(x-1\right)^2=16\)

=>\(4x^2+12x+9-4\left(x^2-2x+1\right)=16\)

=>\(4x^2+12x+9-4x^2+8x-4=16\)

=>20x+5=16

=>20x=11

=>\(x=\frac{11}{20}\)

a: \(\frac{x}{x-y}+\frac{y}{x+y}+\frac{2y^2}{x^2-y^2}\)

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

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

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

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

b: \(\frac{x^3+2x}{x^3+1}+\frac{2x}{x^2-x+1}+\frac{1}{x+1}\)

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

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

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

c: \(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}+\frac{1}{1-x}\)

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

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

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

d: \(\frac{x}{x+3}+\frac{2x}{x-3}+\frac{9-3x^2}{x^2-9}\)

\(=\frac{x}{x+3}+\frac{2x}{x-3}+\frac{9-3x^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x\left(x-3\right)+2x\left(x+3\right)+9-3x^2}{\left(x-3\right)\left(x+3\right)}=\frac{x^2-3x+2x^2+6x+9-3x^2}{\left(x-3\right)\left(x+3\right)}=\frac{3x+9}{\left(x-3\right)\left(x+3\right)}=\frac{3}{x-3}\)

e: \(\frac{2}{x-3}+\frac{2x}{x^2-4x+3}+\frac{x}{x-1}\)

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

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

f: \(\frac{2}{x-1}+\frac{2x-1}{x^2+x+1}+\frac{x^2+6x+2}{x^3-1}\)

\(=\frac{2}{x-1}+\frac{2x-1}{x^2+x+1}+\frac{x^2+6x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{2\left(x^2+x+1\right)+\left(2x-1\right)\left(x-1\right)+x^2+6x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{2x^2+2x+2+2x^2-3x+1+x^2+6x+2}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{5x^2+5x+5}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{5}{x-1}\)

g: \(\frac{x}{x^2-4}+\frac{2\left(x^2+2x+4\right)}{8-x^3}+\frac{1}{x+2}\)

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

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

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

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

h: \(\frac{x+2}{x-2}+\frac{1}{x}+\frac{-8}{x^2-2x}\)

\(=\frac{x\left(x+2\right)+x-2-8}{x\left(x-2\right)}=\frac{x^2+2x+x-10}{x\left(x-2\right)}\)

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

a: \(\frac{3x}{x+1}-\frac{1}{1+x}=\frac{3x}{x+1}-\frac{1}{x+1}=\frac{3x-1}{x+1}\)

b: \(\frac{3y-2x}{x-2y}-\frac{x-y}{2y-x}\)

\(=\frac{-2x+3y}{x-2y}+\frac{x-y}{x-2y}=\frac{-2x+3y+x-y}{x-2y}=\frac{-x+2y}{x-2y}\)

=-1

c: \(\frac{1}{x-2}-\frac{1}{x+1}\)

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

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

d: \(\frac{12}{x^2-9}-\frac{2}{x-3}\)

\(=\frac{12}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x-3}\)

\(=\frac{12-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{12-2x-6}{\left(x-3\right)\left(x+3\right)}\)

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

e: \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}\)

\(=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)

\(=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)

f: \(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\)

\(=\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\)

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

g: \(\frac{2x^3+x^2-x}{x^3-1}-\frac{2x-1}{x-1}\)

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

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

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

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

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

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

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

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

i: \(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\)

\(=\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\)

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

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

j: \(\frac{x^3+1}{x^2-9}-\frac{3}{x+3}-x\)

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

\(=\frac{x^3+1-3\left(x-3\right)-x\left(x^2-9\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x^3+1-3x+9-x^3+9x}{\left(x-3\right)\left(x+3\right)}=\frac{6x+10}{\left(x-3\right)\cdot\left(x+3\right)}\)

k: \(\frac{2x}{x^2-1}-\frac{3}{2x+2}+\frac{1}{2-2x}\)

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

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

l: \(x+\frac{1}{x+1}-1=x-1+\frac{1}{x+1}=\frac{\left(x-1\right)\left(x+1\right)+1}{x+1}\)

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

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

\(=\frac{x^2}{2x}-5x^2-\frac{2x}{2x}+10x+\frac{3}{2x}-15\)

\(=\frac{x^2}{2x}-5x^2-16+10x+\frac{3}{2x}\)

\(=-5x^2+\frac{x^2}{2x}+\frac{20x^2}{2x}+\frac{3}{2x}-16\)

\(=-5x^2+\frac{x^2+20x+3}{2x}-16\)

học tốt

(x^2-2x+3)(1/2x-5)=1/2x^3-5x^2-x^2+10x+3/2x-15=1/2x^3-6x^2+11,5x-15

Lời giải:
Để $x^4+2x^3-ax^2+5x+b$ chia $x^2+x-2$ dư $3x+4$ thì:

$x^4+2x^3-ax^2+5x+b=(x^2+x-2)Q(x)+3x+4$ với $Q(x)$ là đa thức thương.

$\Leftrightarrow x^4+2x^3-ax^2+5x+b=(x-1)(x+2)Q(x)+3x+4$

Cho $x=1$ thì:

$8-a+b=7\Leftrightarrow a-b=1(1)$
Cho $x=-2$ thì:
$-10-4a+b=-2\Leftrightarrow -4a+b=8(2)$

Từ $(1); (2)\Rightarrow a=-3; b=-4$