Phép nhân và phép chia các đa thức

Bài 8:

a) \(x^2-25x=0\)

\(\Leftrightarrow x\left(x-25\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-25=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=25\end{matrix}\right.\)

b) \(7x\left(x-3\right)-5\left(3-x\right)=0\)

\(\Leftrightarrow7x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(7x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\7x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{7}\end{matrix}\right.\)

c) \(7x\left(x+4\right)-7x-28=0\)

\(\Leftrightarrow7x\left(x+4\right)-7\left(x+4\right)=0\)

\(\Leftrightarrow7\left(x+4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

d) \(\left(2x^2+5x\right)\left(x^2-x\right)+\left(2x^2+10\right)\left(x-x^2\right)=0\)

\(\Leftrightarrow\left(2x^2+5x\right)\left(x^2-x\right)-\left(2x^2+10\right)\left(x^2-x\right)=0\)

\(\Leftrightarrow\left(x^2-x\right)\left(2x^2+5x-2x^2-10\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(5x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\5x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

7:

a: \(4a^3b-12a^2b^2+8ab^3\)

\(=4ab\cdot a^2-4ab\cdot3ab+4ab\cdot2b^2\)

\(=4ab\left(a^2-3ab+2b^2\right)\)

\(=4ab\left(a^2-ab-2ab+2b^2\right)\)

\(=4ab\left[a\left(a-b\right)-2b\left(a-b\right)\right]\)

\(=4ab\left(a-b\right)\left(a-2b\right)\)

b: \(\dfrac{5}{2}x^4+\dfrac{3}{4}x^3-\dfrac{1}{5}x^2\)

\(=x^2\cdot\dfrac{5}{2}x^2+x^2\cdot\dfrac{3}{4}x-x^2\cdot\dfrac{1}{5}\)

\(=x^2\left(\dfrac{5}{2}x^2+\dfrac{3}{4}x-\dfrac{1}{5}\right)\)

c: \(3x^2\left(x+9\right)-2\left(x+9\right)\)

\(=\left(x+9\right)\cdot3x^2-\left(x+9\right)\cdot2\)

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

d: \(3x^2\left(7x+y\right)-5x\left(7x+y\right)+14x+2y\)

\(=\left(7x+y\right)\left(3x^2-5x\right)+2\left(7x+y\right)\)

\(=\left(7x+y\right)\left(3x^2-5x+2\right)\)

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

\(=\left(7x+y\right)\left[3x\left(x-1\right)-2\left(x-1\right)\right]\)

\(=\left(7x+y\right)\left(x-1\right)\left(3x-2\right)\)

Bài 12:

a) \(4x^2+4x+1=0\)

\(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot1+1^2=0\)

\(\Leftrightarrow\left(2x+1\right)^2=0\)

\(\Leftrightarrow2x+1=0\)

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

b) \(x^2-2x=-1\)

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

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

c) \(-x^2+10x=25\)

\(\Leftrightarrow-x^2+10x-25=0\)

\(\Leftrightarrow-\left(x^2-10x+25\right)=0\)

\(\Leftrightarrow-\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

d) \(9=12x-4x^2\)

\(\Leftrightarrow4x^2-12x+9=0\)

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

\(\Leftrightarrow2x-3=0\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Bài 7:

a) \(4a^3b-12a^2b^2+8ab^3\)

\(=4ab\left(a^2-3ab+2b^2\right)\)

\(=4ab\left(a^2-ab-2ab+2b^2\right)\)

\(=4ab\left(a-b\right)\left(a-2b\right)\)

b) \(\dfrac{5}{4}x^4+\dfrac{3}{4}x^3-\dfrac{1}{5}x^2\)

\(=x^2\cdot\left(\dfrac{5}{4}x^2+\dfrac{3}{4}x-\dfrac{1}{5}\right)\)

\(=x^2\cdot\dfrac{1}{4}\cdot\left(5x^2+3x-\dfrac{4}{5}\right)\)

\(=\dfrac{x^2}{4}\cdot\left(5x^2+3x-\dfrac{4}{5}\right)\)

c) \(3x^2\left(x+9\right)-2\left(x+9\right)\)

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

d) \(3x^2\left(7x+y\right)-5x\left(7x+y\right)+14x+2y\)

\(=3x^2\left(7x+y\right)-5x\left(7x+y\right)+2\left(7x+y\right)\)

\(=\left(7x+y\right)\left(3x^2-5x+2\right)\)

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

\(=\left(7x+y\right)\left(3x-2\right)\left(x-1\right)\)

Bài 8:

a: \(x^2-25x=0\)

=>\(x\left(x-25\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-25=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=25\end{matrix}\right.\)

b: \(7x\left(x-3\right)-5\left(3-x\right)=0\)

=>\(7x\left(x-3\right)+5\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(7x+5\right)=0\)

=>\(\left[{}\begin{matrix}x-3=0\\7x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{7}\end{matrix}\right.\)

c: \(7x\left(x+4\right)-7x-28=0\)

=>\(7x\left(x+4\right)-7\left(x+4\right)=0\)

=>\(x\left(x+4\right)-\left(x+4\right)=0\)

=>\(\left(x+4\right)\left(x-1\right)=0\)

=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

d: \(\left(2x^2+5x\right)\left(x^2-x\right)+\left(2x^2+10\right)\left(x-x^2\right)=0\)

=>\(\left(2x^2+5x\right)\left(x^2-x\right)-\left(2x^2+10\right)\left(x^2-x\right)=0\)

=>\(\left(x^2-x\right)\left(2x^2+5x-2x^2-10\right)=0\)

=>\(5\left(x-2\right)\cdot x\left(x-1\right)=0\)

=>\(x\left(x-1\right)\left(x-2\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

Bài 13:

a) \(x^3-49x=0\)

\(\Leftrightarrow x\left(x^2-49\right)=0\)

\(\Leftrightarrow x\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=7\end{matrix}\right.\)

b) \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

c) \(x^3+6x^2+9x=0\)

\(\Leftrightarrow x\left(x^2+6x+9\right)=0\)

\(\Leftrightarrow x\left(x^2+2\cdot3\cdot x+3^2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

d) \(\left(x+5\right)^2-16x^2=0\)

\(\Leftrightarrow\left(x+5\right)^2-\left(4x\right)^2=0\)

\(\Leftrightarrow\left(x+5-4x\right)\left(x+5+4x\right)=0\)

\(\Leftrightarrow\left(5-3x\right)\left(5x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5-3x=0\\5x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=5\\5x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

12:

a: \(4x^2+4x+1=0\)

=>\(\left(2x\right)^2+2\cdot2x\cdot1+1^2=0\)

=>\(\left(2x+1\right)^2=0\)

=>2x+1=0

=>2x=-1

=>\(x=-\dfrac{1}{2}\)

b: \(x^2-2x=-1\)

=>\(x^2-2x+1=0\)

=>\(x^2-2\cdot x\cdot1+1^2=0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1

c: \(-x^2+10x=25\)

=>\(-x^2+10x-25=0\)

=>\(x^2-10x+25=0\)

=>\(x^2-2\cdot x\cdot5+5^2=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5

d: \(12x-4x^2=9\)

=>\(-4x^2+12x-9=0\)

=>\(4x^2-12x+9=0\)

=>\(\left(2x\right)^2-2\cdot2x\cdot3+3^2=0\)

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

=>2x-3=0

=>2x=3

=>x=3/2

\(2^2\cdot2^2=2^{2+2}=2^4=16\)

\(\left(8\cdot3\right)^2=8^2\cdot3^2=64\cdot9=576\)

[9x³(x² - 1) - 6x²(x² - 1) + 12x(x² - 1)] : 3x(x² - 1)

= [9x³(x² - 1) : 3x(x² - 1)] - [6x²(x² - 1) : 3x(x² - 1) + [12x(x² - 1) : 3x(x² - 1)]

= 3x² - 2x + 4

2:

a: \(\left(3x-1\right)\left(3x-1\right)+9\left(1-x\right)\left(x+1\right)=12\)

=>\(\left(3x-1\right)^2-9\left(x-1\right)\left(x+1\right)=12\)

=>\(9x^2-6x+1-9\left(x^2-1\right)=12\)

=>\(9x^2-6x+1-9x^2+9=12\)

=>-6x+10=12

=>-6x=2

=>x=-1/3

b: \(4\left(x+1\right)^2-\left(2x-1\right)^2=0\)

=>\(4\left(x^2+2x+1\right)-4x^2+4x-1=0\)

=>\(4x^2+8x+4-4x^2+4x-1=0\)

=>12x+3=0

=>x=-1/4

3:

a: Xét ΔODC có AB//CD

nên \(\dfrac{OA}{AD}=\dfrac{OB}{BC}\)

mà AD=BC

nên OA=OB

=>ΔOAB cân tại O

b: Xét ΔABD và ΔBAC có

BA chung

BD=AC

AD=BC

Do đó: ΔABD=ΔBAC

c: ΔABD=ΔBAC

=>\(\widehat{EAB}=\widehat{EBA}\)

=>EA=EB

EA+EC=AC

EB+ED=BD

mà AC=BD và EA=EB

nên EC=ED
=>ΔECD cân tại E

d: OA+AD=OD

OB+BC=OC

mà OA=OB và AD=BC

nên OD=OC

OD=OC

ED=EC

Do đó: OE là đường trung trực của CD

OA=OB

EA=EB

Do đó: OE là đường trung trực của AB

a) P + Q = (x² + 2x³ - xy² + 5) + (x³ + xy² - 2x²y - 6)

= x² + 2x³ - xy² + 5 + x³ + xy² - 2x²y - 6

= (2x³ + x³) + x² + (-xy² + xy²) - 2x²y + (5 - 6)

= 3x³ + x² - 2x²y - 1

b) Q = P + N

N = Q - P

= (x³ + xy² - 2x²y - 6) - (x² + 2x³ - xy² + 5)

= x³ + xy² - 2x²y - 6 - x² - 2x³ + xy² - 5

= (x³ - 2x³) + (xy² + xy²) - 2x²y - x² + (-6 - 5)

= -x³ + 2xy² - 2x²y - x² - 11

Vậy N = -x³ + 2xy² - 2x²y - x² - 11

Tính tổng hai đa thức P và Q rồi tìm bậc của đa thức tổng 

Bài 5.5:

\(\left(2x-3\right)\left(x+1\right)+\left(4x^3-6x^2-6x\right):\left(-2x\right)=18\)

\(\Leftrightarrow\left(2x^2+2x-3x-3\right)+2x\cdot\left(2x^2-3x-3\right):\left(-2x\right)=18\)

\(\Leftrightarrow2x^2-x-3-2x^2+3x+3=18\)

\(\Leftrightarrow2x=18\)

\(\Leftrightarrow x=\dfrac{18}{2}\)

\(\Leftrightarrow x=9\) 

Bài 7:

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

= x² - 2x - 2x + 4 - x² - x + x + 1

= (x² - x²) + (-2x - 2x - x + x) + (4 + 1)

= -4x + 5

Tại x = 21 thì

A = -4.21 + 5

= -77

b) B = (x - 1)(x - 7) - (2x - 6)(x - 1)

= x² - 7x - x + 7 - 2x² + 2x + 6x - 6

= (x² - 2x²) + (-7x - x + 2x + 6x) + (7 - 6)

= -x² + 1

Tại x = 0 thì

B = -0² + 1

= 1

c) C = (2x + y)(2 + y) + (3x + y)(y - 2)

= 4x + 2xy + 2y + y² + 3xy - 6x + y² - 2y

= (4x - 6x) + (2xy + 3xy) + (2y - 2y) + (y² + y²)

= -2x + 5xy + 2y²

Tại x = 1; y = -1 thì

C = -2.1 + 5.1.(-1) + 2.(-1)²

= -5

d) D = (x - 1)(x + 2) - x(x - 2)

= x² + 2x - x - 2 - x² + 2x

= (x² - x²) + (2x - x + 2x) - 2

= 3x - 2

Tại x = 100 thì

D = 3.100 - 2

= 298

Bài 8

1) A = x³ - 30x² - 31x + 1

= x³ + x² - 31x² - 31x + 1

= (x³ + x²) - (31x² + 31x) + 1

= x²(x + 1) - 31x(x + 1) + 1

Tại x = 31 thì

A = 31².32 - 31².32 + 1

= 1

2) B = x³ - 17x² - 18x + 20

= x³ + x² - 18x² - 18x + 20

= (x³ + x²) - (18x² + 18x) + 20

= x²(x + 1) - 18x(x + 1) + 20

Tại x = 20 thì

 B = 18².19 - 18².19 + 20

= 20

3) C = x⁴ - 17x³ + 17x² - 17x + 20

= x⁴ - x³ - 16x³ + x² + 16x² - x + 16 - 16x

= (x⁴ - x³) - (16x³ - 16x²) + (x² - x) - (16x - 16)

= x³(x - 1) - 16x²(x - 1) + x(x - 1) - 16(x - 1)

Tại x = 16 thì

C = 16³.15 - 16³.15 + 16.15 - 16.15

= 0

4) D = x⁴ + 10x³ + 10x² + 10x + 10

= x⁴ + x³ + 9x³ + 9x² + x² + x + 9x + 9 + 1

= (x⁴ + x³) + (9x³ + 9x²) + (x² + x) + (9x + 9) + 1

= x³(x + 1) + 9x²(x + 1) + x(x + 1) + 9(x + 1) + 1

Tại x = -9 thì

D = (-9)³.(-8) + 9³.(-8) - 9.(-8) + 9.(-8) + 1

= 1