1: \(4x^2+4x+1=\left(2x+1\right)^2\)

2: \(x^2-20x+100=\left(x-10\right)^2\)

3: \(y^4-14y^2+49=\left(y^2-7\right)^2\)

4: \(125x^3-64y^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

dsssws

Bài 1: 

a: \(=\dfrac{2x^4-8x^3+2x^2+2x^3-8x^2+2x+18x^2-72x+18+56x-15}{x^2-4x+1}\)

\(=2x^2+2x+18+\dfrac{56x-15}{x^2-4x+1}\)

a. P(-1) = 5 . -1 - 1/2

             = -5 - 1/2

             = -11/2

Q(-3) = (-3)² - 9

         = 9 - 9

         = 0

R(-3/10) = 3 . (-3/10)² - 4 . -3/10

               = 3 . 9/100 - -12/10

               = 27/100 - -120/100

               = 147/100

b. P(x) = 5x - 1/2

Ta có: 5x - 1/2 = 0

           5x         = 1/2

             x         = 1/10

Vậy đa thức P(x) có nghiệm là {1/10}

Q(x) = x² - 9

Ta có: x² - 9 = 0

           x²      = 9

           x²      = (3)²

                        (-3)²

      => x = \(\pm\)3

Vậy nghiệm của đa thức Q(x) là {\(\pm\)3)

a,ta có P(-1)=5.(-1)-1/2

                    =(-5)-1/2=-11/2

             Q(-3)=(-3)^2-9

                      =9-9=0

             R(-3/10)=3.(-3/10)^2-4.(-3/10)

                           =147/100

ý R mik ko ghi cách làm sorry nhé 

cảm ơn vì bạn đã xem

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

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

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

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

b) \(P\left(x\right)+Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)+\left(x^4+x^3+2x^2+4x+2\right)\)

\(\Rightarrow P\left(x\right)+Q\left(x\right)=2x^4+4x^3+3x^2+6x+5\)

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

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

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

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

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

1. ( 3x + 2)² - 4

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

= 3x(3x+4)

2. 4x² - 25y²

= (2x-5y)(2x+5y)

3. 4x²- 49

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

4. 8z³ + 27

=(2z+3)(4x²-6z+9)

5. \(\dfrac{9}{25}x^4-\dfrac{1}{4}\)

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

6. x³²  - 1

=(x¹⁶-1)(x¹⁶+1)

7. 4x² + 4x + 1

=(2x+1)²

8. x² - 20x + 100

=(x-10)²

9. y⁴ -14y² + 49

=(y²-7)²

10.  125x³ - 64y³

= (5x-4y)(25x²+20xy+16y²)

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

2) \(4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

3) \(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)

4) \(8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)

5) \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)

6) \(x^{32}-1=\left(x^{16}-1\right)\left(x^{16}+1\right)\)

\(=\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

7) \(4x^2+4x+1=\left(2x+1\right)^2\)

8) \(x^2-20x+100=\left(x-10\right)^2\)

9) \(y^4-14y^2+49=\left(y^2-7\right)^2\)

\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Câu 1

a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)

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

Bài 2

a) \(7x^2+14xy=7x\left(x+2y\right)\)

b) \(3x+12-\left(x^2+4x\right)=-x^2-x+12=\left(-x+3\right)\left(x+4\right)\)

c) \(x^2-2xy+y^2=\left(x-y\right)^2\)

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

Câu b đề thiếu rồi bạn