Đặt \(A=x^4-3x^3+6x^2-5x+3\)

Xét trường hợp \(A=\left(x^2+ax+1\right)\left(x^2+bx+3\right)\)

\(A=x^4+bx^3+3x^2+ax^3+abx^2+3ax+x^2+bx+3\)

\(A=x^4+x^3\left(b+a\right)+x^2\left(3+ab+1\right)+x\left(3a+b\right)+3\)

Đồng nhất hệ số ta có:

\(\Rightarrow\hept{\begin{cases}a+b=-3\\3+ab+1=6\\3a+b=-5\end{cases}\Rightarrow\hept{\begin{cases}a+3=-b\\ab=2\\3a+b=-5\end{cases}\Rightarrow}\hept{\begin{cases}a=-1\\b=-2\end{cases}}}\)

Vậy \(x^4-3x^3+6x^2-5x+3=\left(x^2-x+1\right)\left(x^2-2x+3\right)\)

Chúc bn hok tốt ##

\(x^4-3x^3+6x^2-5x+3\)

\(=x^4-2x^3+3x^2-x^3+2x^2-3x+x^2-2x+3\)

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

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

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

1) \(x^2-2x+1+x^2y-xy=\left(x-1\right)^2+xy\left(x-1\right)=\left(x-1\right)\left(x+xy-1\right)\)

2) \(x^2+6x+9+x^2y+3xy\)

\(=\left(x+3\right)^2+xy\left(x+3\right)\)

\(=\left(x+3\right)\left(x+xy+3\right)\)

a) x³ + 4x² - 29x + 24                                                           

= x³ - 3x² + 7x² - 21x - 8x + 24

= x²(x-3) + 7x(x-3) - 8(x-3)

= (x-3)(x²+7x-8)

=(x-3)(x²+8x-x-8)

= (x-3)[(x²+8x)-(x+8)]

= (x-3)[x(x+8)-(x+8)]

= (x-3)(x+8)(x-1)

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

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

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

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

b) \(3x^2-7x-10=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

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

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

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

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

b) \(3x^2-7x-10=\left(3x^2+3x\right)-\left(10x+10\right)\)

= \(3x\left(x+1\right)-10\left(x+1\right)\)

= \(\left(x+1\right)\left(3x-10\right)\)

1, \(x^3+8x^2+17x+10=\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)

\(=x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)\(=\left(x+1\right)\left(x^2+7x+10\right)=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)

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

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

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

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

4. \(81x^4+4=\left(9x^2\right)^2+2^2=\left(9x^2+2\right)^2-2.9x^2.2=\left(9x^2+2\right)^2-\left(6x\right)^2\)

\(=\left(9x^2+6x+2\right)\left(9x^2-6x+2\right)\)

* Trả lời:

\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)

\(\Leftrightarrow-3+6x-4-12x=-5x+5\)

\(\Leftrightarrow6x-12x+5x=3+4+5\)

\(\Leftrightarrow x=12\)

\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)

\(\Leftrightarrow6x-15-6+24x=-3x+7\)

\(\Leftrightarrow6x+24x+3x=15+6+7\)

\(\Leftrightarrow33x=28\)

\(\Leftrightarrow x=\dfrac{28}{33}\)

\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)

\(\Leftrightarrow1-3x-6x+12=-4x-5\)

\(\Leftrightarrow-3x-6x+4x=-1-12-5\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)

\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)

\(\Leftrightarrow-x-5x=-7\)

\(\Leftrightarrow-6x=-7\)

\(\Leftrightarrow x=\dfrac{7}{6}\)

\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)

\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)

\(\Leftrightarrow-15x+3x=4\)

\(\Leftrightarrow-12x=4\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

6:

a: ĐKXĐ: x<>0

\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)

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

b: ĐKXĐ: x<>1

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

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

c: ĐKXĐ: x<>-2

\(\dfrac{x^2+4x+4}{2x+4}\)

\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)

\(=\dfrac{x+2}{2}\)

d: ĐKXĐ: x<>-2

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

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

e: ĐKXĐ: x<>-y

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

g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)

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

7:

a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)

\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)

b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)

\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)

c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

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

d:

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

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

\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

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

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)