a) 16x-32=0

16x =0-32

16x=-32

x=-32:16

x=-2

Vậy x=-2 là nghiệm của đa thức

Cho mk xin yêu cầu của bài được ko vậy ???

a. 2x²-5x+1=0

△= b² - 4ac = (-5)² - 4*2*1 = 17 ⇒√△ = √17

\(\Rightarrow x_1=\frac{5+\sqrt{17}}{4};x_2=\frac{5-\sqrt{17}}{4}\)

Vậy .... S={\(\frac{5\pm\sqrt{17}}{4}\)}

b. 4x² +4x+1=0

⇔(2x+1)² = 0 ⇔ x=\(\frac{-1}{2}\)

c. -3x² +2x+8=0

△' = b'² - ac = 1² - (-3)*8 = 25 ⇒√△ = 5

\(\Rightarrow x_1=\frac{-1+5}{-3}=-\frac{4}{3};x_2=\frac{-1-5}{-3}=2\)

Vậy... S={-\(\frac{4}{3}\);2}

d. 5x2 6x1=0 (thiếu dấu nên mk chưa giải được)

e. -3x²+ 14x - 8=0

△' = b'² - ac = 7² - (-3)*(-8) = 25 ⇒ √△ = 5

⇒\(x_1=\frac{-7+5}{-3}=\frac{2}{3};x_2=\frac{-7-5}{-3}=4\)

Vậy .... S={\(\frac{2}{3};4\)}

g. -7x² +4x-3=0

△' = b'² - ac = 2² - (-7)*(-3) = -17<0

Vậy pt vô nghiệm , S=∅

\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)

trừ 55

\(3x^3-8x^2+4x\)

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

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

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

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

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

b: 

Sửa đề: 6x^4-4x^3+3x-2/3x-2

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

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

1.\(=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x+y\right)^2-\left(2z\right)^2\right]=5\left(x+y-2z\right)\left(x+y+2z\right)\)

2. \(=\left(-5x^2+15x\right)+\left(x-3\right)=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)

3. \(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)

4.\(=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)

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

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

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

\(1,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ 2,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ 3,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ 4,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=x^2+x+3x+3=\left(x+3\right)\left(x+1\right)\\ 6,=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ 7,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

\(a,=2x^3y+2x^2y^2-6xy^3\\ b,=3x^3+6x^2-4x-8\\ c,=\left(4x^2+16x-20x-80+76\right):\left(x+4\right)\\ =\left[\left(x+4\right)\left(4x-20\right)+76\right]:\left(x+4\right)\\ =4x-20\left(dư.76\right)\\ d,=\left(x^4-x^2-x^3+x-2x^2+2\right):\left(x^2-1\right)\\ =\left(x^2-1\right)\left(x^2-x-2\right):\left(x^2-1\right)\\ =x^2-x-2\)

em mới lp 7 =)))

a) (2x – 1)(4x2 + 2x + 1) – 4x(2x2 – 3) = 23

⇔ 8x3 – 1 – 8x3 + 12x = 23

⇔ 12x = 24 ⇔ x = 2.

Tập nghiệm của phương trình: S = {2}

b) ĐKXĐ : x + 1 ≠ 0 và x – 2 ≠ 0 (vì vậy x2 – x – 2 = (x + 1)(x – 2) ≠ 0)

⇔ x ≠ -1 và x ≠ 2

Quy đồng mẫu thức hai vế :

Khử mẫu, ta được : x2 – 4 – x – 1 = x2 – x – 2 – 3 ⇔ 0x = 0

Phương trình này luôn nghiệm đúng với mọi x ≠ -1 và x ≠ 2.

`A(x)+B(x)=(5x^4 -4x^2 +x-2)+(x^4 +3x^2 -4x)`

`=5x^4 -4x^2 +x-2+x^4 +3x^2 -4x`

`=5x^4 +x^4 -4x^2 +3x^2 +x-4x-2`

`=6x^4 -x^2 -3x-2`

a) A = x² - 4y² + 2x + 4y = (x-2y)(x+2y)+2(x+2y)=(x+2y)(x-2y+2)
b) A = 4x² - 9y² - 4x - 6y=(2x-3y)(2x+3y)-2(2x+3y)=(2x+3y)(2x-3y-2)
c) A = 3x² - 3xy - 5x + 5y=3x(x-y)-5(x-y)=(x-y)(3x-5)

a) \(A=x^2-4y^2+2x+4y=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)=\left(x+2y\right)\left(x-2y+2\right)\)

b) \(A=4x^2-9y^2-4x-6y=\left(2x-3y\right)\left(2x+3y\right)-2\left(2x+3y\right)=\left(2x+3y\right)\left(2x-3y-2\right)\)

c) \(A=3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)