Bài 9: Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp

a) 2x³ + x² - 4x - 12

=( x³ - 2³) +( x³ - 2³) + (x² - 2.2x + 4)

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

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

= ( x- 2)( 2x² + 5x + 6)

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

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

= ( x - 2)( x - 2) - 1

= ( x - 2)² - 1

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

= ( x - 3)( x - 1)

c) 6x³ + x² - 2x

= 6x³ + 4x² - 3x² - 2x

= 3x²( 2x - 1) + 2x( 2x -1 )

= ( 2x - 1)( 3x²+ 2x)

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

A = x⁴( y -z) + y⁴( z - x) + z⁴( x - y)

A= x⁴y - x⁴z + y⁴z - y⁴x + z⁴( x - y)

A = xy( x³ - y³) - xz( x³ - y³) + z⁴( x - y)

Áp dụng hằng đẳng thức nha ( Dài tớ lười)

no ra hang dang thuc dau tien nha ban .

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

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

\(=4x^2\)

(3x+y)${}^{}$-2(x+y)(3x+y)+(x+y)${}^{}$

= [(3x-y)-(x+y)]²

=(3x-y-x-y)²

= (2x-2y)²

đề bài

(5x${}^{}$y${}^{}$+2x${}^{}$y${}^{}$-10x${}^{}$y${}^{}$):5x${}^{}$y${}^{}$

-2y²

(5x${}^{}$y${}^{}$+2x${}^{}$y${}^{}$-10x${}^{}$y${}^{}$):5x${}^{}$y${}^{}$

= x+ \(\dfrac{2}{5}\) -2y²

4x(x + y) - (2x + y)(2x - y)

= 4x² + 4xy - (4x² - y²)

= 4x² + 4xy - 4x² + y²

= 4xy + y²

đề

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

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

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

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

\(=y\left(4x+y\right)\)

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

b) Đặt \(x^2+2x=a\)

\(\Rightarrow B=a\left(a+4\right)+3\) ( Đặt biểu thức trên là B)

= \(a^2+4a+3\)

=\(a^2+a+3a+3\)

=\(\left(a^2+a\right)+\left(3a+3\right)\)

=\(a\left(a+1\right)+3\left(a+1\right)\)

=\(\left(a+1\right)\left(a+3\right)\)

Thay \(a=x^2+2x\)

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

a) x⁴ + 2x² - 3

= x⁴ + 2x² + 1 - 4

= (x² + 1)² - 2²

= (x² + 1 - 2)(x² + 1 + 2)

= (x² - 1)(x² + 3)

= (x - 1)(x + 1)(x² + 3)

c) 2xy² + 4xy + 2x - 2xz² + 4xzt - 2xt²

= 2x(y² + 2y + 1 - z² + 2zt - t²)

= \(2x\left [ (y + 1)^{2} - (z -t)^{2} \right ]\)

= 2x(y - 1 - z + t)(y + 1 + z - t)

= 2x(y - z + t - 1)(y + z - t + 1)

câu b mk đang suy nghĩ

Ta có:\(\left(a+b\right)^2=a^2+2ab+b^2=2a^2+2b^2\)

\(\Rightarrow a^2-2ab+b^2=0\) hay\(\left(a-b\right)^2=0\Leftrightarrow a=b\left(đpcm\right)\)

\(A=2x^2+10x-1=2\left(x^2+5x-\dfrac{1}{5}\right)=2\left(x^2+5x+\dfrac{25}{4}-\dfrac{129}{20}\right)=2\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{129}{10}=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{129}{10}\ge-\dfrac{129}{10}\)

Dấu "=" xảy ra khi: \(x=-\dfrac{5}{2}\)

Xem lại đề.

a) x³ - 4x² + 4x - xy²

= x(x² - 4x + 4 - y²)

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

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

c) 3x² - 7x - 10

= 3x² + 3x - 10x - 10

= 3x(x + 1) - 10(x + 1)

= (x + 1)(3x - 10)

d) 5x³ - 5x²y - 10x² + 10xy

= 5x(x² - xy - 2x + 2y)

= \(5x\left [ x(x - y) - 2(x - y) \right ]\)

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

b) sửa đề: x³ - 4x² + 12x - 27

= x³ - 27 - 4x² + 12x

= (x - 3)(x² + 3x + 9) - 4x(x - 3)

= (x - 3)(x² + 3x - 4x + 9)

= (x - 3)(x² - x + 9)