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

a) \(x^3-x\)

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

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

b) \(\left(2x+y\right)^2-4y^2\)

\(=\left(2x+y-2y\right)\left(2x+y+2y\right)\)

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

c) \(x^3+y^3+x+y\)

\(=\left(x^3+y^3\right)+\left(x+y\right)\)

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

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

d) \(x^3+4x^2y+4xy^2-9x\)

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

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

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

e) \(x^2+4-y^2+4x\)

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

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

\(=\left(x+2-y\right)\left(x+2+y\right)\)

f) \(x^4+4\)

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

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

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

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

1)

a) => 16x² - 8x + 1 - 8(2x² + 3x - 4x - 6) = 15

=> 16x² - 8x + 1 - 8(2x² - x - 6) = 15

=> 16x² - 8x + 1 - 16x² + 8x + 48 = 15

=> 49 = 15 (?) (vô lí)

=> Không tìm được x thoả mãn

b) (5x - 2)(x - 2) - 4(x - 3) = x² + 3

=> 5x² - 10x - 2x + 4 - 4x + 12 = x² + 3

=> 5x² - 16x + 16 = x² + 3

=> 4x² - 16x + 16 = 3

=> (2x)² - 2.2x.4 + 4² = 3

=> (2x - 4)² = 3

=> \(\left[{}\begin{matrix}2x-4=\sqrt{3}\\2x-4=-\sqrt{3}\end{matrix}\right.\)           \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{2}\\x=\dfrac{4-\sqrt{3}}{2}\end{matrix}\right.\)

Mong bạn xem lại đề bài!

2) 

a) 5x² - 10xy + 5y² - 20z²

= 5(x² - 2xy + y² - 4z²)

= 5[(x - y)² - (2z)²]

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

b) a³ - ay - a²x + xy

= a(a² - y) - x(a² - y)

= (a - x)(a² - y)

c) 3x² - 6xy + 3y² - 12z²

= 3(x² - 2xy + y² - 4z²)

= 3[(x - y)² - (2z)²]

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

d) x² - 2xy + tx - 2ty

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

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

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

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

\(=2\cdot6x\)

\(=12x\)

_________

\(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=2x\cdot2y\)

\(=4xy\)

\(\left(x+y\right)^3+\left(x-y\right)^3\)

\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

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

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

______

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3+3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2-xy-xz-yz\right)\)

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

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

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

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

\(\Leftrightarrow\left(x-3\right)\left(4-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

___________

\(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot\dfrac{1}{2}\cdot x^2+3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

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

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{1}{2}\)

\(\Leftrightarrow x=-\dfrac{1}{4}\)

\(72^2+144\cdot16+16^2-12^2\)

\(=\left(72^2+144\cdot16+16^2\right)-12^2\)

\(=\left(72^2+2\cdot72\cdot16+16^2\right)-12^2\)

\(=\left(72+16\right)^2-12^2\)

\(=88^2-12^2\)

\(=\left(88+12\right)\left(88-12\right)\)

\(=100\cdot76\)

\(=7600\)

\(48^2-42^2+64-52^2\)

\(=\left(48^2-52^2\right)-\left(42^2-64\right)\)

\(=\left(48^2-52^2\right)-\left(42^2-8^2\right)\)

\(=\left(48-52\right)\left(48+52\right)-\left(42+8\right)\left(42-8\right)\)

\(=-4\cdot100-50\cdot34\)

\(=-8\cdot50-50\cdot34\)

\(=-50\cdot\left(8+34\right)\)

\(=-50\cdot42\)

\(=-2100\)

a: =(x+2)^3

b: =(x-1)^3

c: =(1-3x)^3

d: =(x+1/2)^3

e: =(3x-2y)^3

a: =(x-5)^2(x+5)^2-(x-5)^2

=(x-5)^2[(x+5)^2-1]

=(x-5)^2*(x+5-1)(x+5+1)

=(x+4)(x+6)(x-5)^2

b: =(2x-5)^2*(2x+5)^2-9(2x-5)^2

=(2x-5)^2[(2x+5)^2-9]

=(2x-5)^2(2x+5-3)(2x+5+3)

=4(x+4)(x+1)(2x-5)^2

c: =(2x-3)^2[4-9(2x+3)^2]

=(2x-3)^2*[2^2-(6x+9)^2]

=(2x-3)(2-6x-9)(2+6x+9)*(2x-3)

=(-6x-7)(11+6x)(2x-3)^2

d: =a^2(a^4-a^2+2a+2)

=a^2[a^2(a-1)(a+1)+2(a+1)]

=a^2(a+1)(a^2-a+2)

e: =(3x^2+3x+2-3x^2-3x+2)(3x^2+3x+2+3x^2+3x-2)

=(6x^2+6x)*4

=24x(x+1)

ab(a+b) - bc(b + c) + ac(a - c)

=a²b + ab² - b²c - bc² + a²c - ac²

=(ab² - b²c) + (a²c - ac²) + (a²b - bc²)

=b²(a - c) + ac(a - c) + b(a² - c²)

=b²(a - c) + ac(a - c) + b(a - c)(a + c)

=(a - c )[b² + ac + b(a + c)]

=(a - c)[(b² + bc) + (ab + ac)]

=(a - c)[b(b + c) + a(b + c)]

=(a - c)(a + b)(b + c)