Bài 7: Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức

a) 2x²-3x=0

\(\Rightarrow\)x(2x-3)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

c) (x+1)(x-2)-(x-3)²=x+1

⇒(x+1)(x-2)-(x-3)²-(x+1)=0

\(\Rightarrow\)x²-2x+x-2-(x²-6x+9)-x-1=0

⇒x²-2x+x-2-x²+6x-9-x-1=0

\(\Rightarrow\)4x-12=0

\(\Rightarrow\)4x=12

\(\Rightarrow\)x=3

d)(2x+3-x-1)(2x+3+x+1)=(x+2)(3x+4)=3x²+4x+6x+8=3x²+10x+8

c)4x²+12x+9=(2x+3)²

d)(x-y)²+3(x+y)²=x²-2xy+y²+3(x²+2xy+y²)=x²-2xy+y²+3x²+6xy+3y²=4x²+4xy+4y²

a) -x²+6x-9=-(x²-6x+9)=-(x-3)²

b)(x+1)²-(2x+3)²=(x+1-2x-3)(x+1+2x+3)=(-x-2)(3x+3)

h)x²-4x+4=(x-2)²

g)x³-125=(x-5)(x²+5x+25)

GIÚP E BÀI 1

TRÌNH BÀY ĐẦY ĐỦ CÁC BƯỚC. KO BỎ BƯỚC

Bài 1:

a) \(x^2+4x+4=\left(x+2\right)^2\)

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

c) \(x^4-16y^2=\left(x^2-4y\right)\left(x^2+4y\right)\)

d) \(81-y^4=\left(9-y^2\right)\left(9+y^2\right)=\left(y^2+9\right)\left(3-y\right)\left(3+y\right)\)

e) \(125x^3-8=\left(5x-2\right)\left(25x^2+10x+4\right)\)

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

g) \(1+9x+27x^2+27x^3=\left(1+3x\right)^3\)

h) \(64x^3+1=\left(4x+1\right)\left(16x^2-4x+1\right)\)

i) \(216x^3-125=\left(6x-5\right)\left(36x^2+30x+25\right)\)

k) \(x^2+10x+25=\left(x+5\right)^2\)

l) \(49-36y^2=\left(7-6y\right)\left(7+6y\right)\)

m) \(9x^2-x+\dfrac{1}{36}=\left(3x-\dfrac{1}{6}\right)^2\)

e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)

\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)

\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)

\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)

13, \(9x^2-16y^2=\left(3x-4y\right)\left(3x+4y\right)\)

14, \(\dfrac{1}{4}x^2-9y^2=\left(\dfrac{1}{2}x-3y\right)\left(\dfrac{1}{2}x+3y\right)\)

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

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

17, \(\left(3x+5\right)^2-4=\left(3x+5-2\right)\left(3x+5+2\right)=\left(3x+3\right)\left(3x+7\right)\)

18, \(\left(2x+3\right)^2-16x^2=\left(2x+3-4x\right)\left(2x+3+4x\right)=\left(-2x+3\right)\left(6x+3\right)\)

13) \(9x^2-16y^2=\left(3x-4y\right)\left(3x+4y\right)\)

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

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

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

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

18) \(\left(2x+3\right)^2-16x^2=\left(2x+3-4x\right)\left(2x+3+4x\right)=\left(-2x+3\right)\cdot3\cdot\left(2x+1\right)\)

2) 9x²+ 12x+ 4

<=>(3x)²+ 2.3x.2+ 2² <=>(3x+ 2)²

3) 4x⁴+ 20x²+ 25

<=>(2x²)²+ 2.2x².5+ 5² <=>(2x²+5)²

4) 25x²- 20xy+ 4y²

<=> (5x)²- 2.5x.2y+ (2y)²<=> (5x-2y)²

5) 9x⁴- 12x²y+ 4y²

<=> (3x²)²- 2.3x².2.y+ (2y)²<=> (3x²- 2y)²

6) 4x⁴- 16x²y³+ 16y⁶

<=> (2x²)²- 2.2x².4y³+ (4y³)²<=> (2x²- 4y³)²

7) 9x⁴- 12x⁵+ 4x⁶

<=> (3x²)²- 2.3x².2x³+ (2x³)²<=> (3x²- 2x³)²

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

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

\(x^2+6x+9=\left(x+3\right)^2\)

\(x^2+10x+25=\left(x+5\right)^2\)

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

\(9x^2-6x+1=\left(3x-1\right)^2\)

\(9x^2+12x+4=\left(3x+2\right)^2\)

\(4x^2-12x+9=\left(2x-3\right)^2\)

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

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

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

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

\(9x^2-16y^2=\left(3x-4y\right)\left(3x+4y\right)\)

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

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

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

\(\left(2x+3\right)^2-16x^2=\left(2x+3\right)^2-\left(4x\right)^2=\left(2x+3-4x\right)\left(2x+3+4x\right)=\left(3-x\right)\left(6x+3\right)\)

\(4\left(3x+1\right)^2-9=\left(6x+2\right)^2-3^2=\left(6x+2-3\right)\left(6x+2+3\right)=\left(6x-1\right)\left(6x+5\right)\)

\(\left(2x-5\right)^2-9x^2=\left(2x-5\right)^2-\left(3x\right)^2=\left(2x-5-3x\right)\left(2x-5+3x\right)=\left(-x-5\right)\left(5x-5\right)\)

\(=-5\left(x+5\right)\left(x+1\right)\)

\(\left(3x-2\right)^2-16=\left(3x-2\right)^2-4^2=\left(3x-2-4\right)\left(3x-2+4\right)=\left(3x-6\right)\left(3x+2\right)=3\left(x-2\right)\left(3x+2\right)\)

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

\(\left(2x-3\right)^2-\left(x-5\right)^2=\left(2x-3-x+5\right)\left(2x-3+x-5\right)=\left(x+2\right)\left(3x-8\right)\)

2) \(x^2-4x+4=\left(x-2\right)^2\)

4) \(x^2+10x+25=\left(x+5\right)^2\)

23) \(\left(2x-3\right)^2-\left(x-5\right)^2=\left(2x-3-x+5\right)\left(2x-3+x-5\right)=\left(x+2\right)\left(3x-8\right)\)

29) \(4x^2-\left(x^2+1\right)=\left(2x-x^2-1\right)\left(2x+x^2+1\right)=-\left(x-1\right)^2\cdot\left(x+1\right)^2\)

giúp e hết bài vs ạ

từ 1=>29 e cảm ơn