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a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))
a/ (5x^2 -10xy+5y^2)-20z^2=5[(x-y)^2-(2z)^2]=5(x-y-2x)(x-y+2z)
b/16x-5x^2-3=-[5x^2-16x+3]=-[(5x^2-x)-(15x+3)]=-[x(5x-1)-3(5x-1)]=(3-x)(5x-1)
c/x^2+4x+3=(x^2+x)+(3x+3)=x(x+1)+3(x+1)=(x+1)(x+3)
2a/ 5x(X^2-9)=0=>x=0 hoặc x^2=9=>x=0 hoặc x=+-3
b/x^2-7x+10=0=>(x^2-2x)-(5x-10)=0=>x(x-2)-5(x-2)=0=>x-2=0 hoặc x-5 =0 => tự tính nhé!
Answer:
Bài 1:
\(5x² - 10xy + 5y² - 20z²\)
\(= 5( x² - 2xy + y² - 4z²)\)
\(= 5 [(x² - 2xy + y²) - (2z)²]\)
\(= 5 [(x - y)² - (2z)²]\)
\(= 5 (x - y - 2z) ( x - y + 2z)\)
\(16x - 5x² - 3 \)
\(= -( 5x² - 16x + 3)\)
\(= -( 5x² - 15x - 1x + 3)\)
\(= - [ (5x² -x) - (15x -3)]\)
\(= - [ x(5x -1) -3(5x -1)]\)
\(= - (5x-1)(x-3)\)
\(x² + 4x + 3\)
\(= x² + x + 3x + 3\)
\(= (x² + x) + (3x + 3)\)
\(= x( x + 1) +3 (x+1)\)
\(= (x+1) (x+3)\)
Bài 2:
\(5x\left(x^2-9\right)=0\)
\(\Rightarrow5x\left(x-3\right)\left(x+3\right)=0\)
Trường hợp 1: \(5x=0\Leftrightarrow x=0\)
Trường hợp 2: \(x-3=0\Leftrightarrow x=3\)
Trường hợp 3: \(x+3=0\Leftrightarrow x=-3\)
\(x^2-7x+10=0\)
\(\Rightarrow x^2-5x-2x+10=0\)
\(\Rightarrow x\left(x-5\right)-2\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}}\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a,2x2-7x+6=(2x2-4x)-(3x-6)
=2x(x-3)-3(x-2)=(x-2)(2x-3)
b,x2+x-6=(x2+3x)-(2x+6)
=x(x-3)-2(x-3)=(x-3)(x-2)
c,x3+3x2+6x+4=x3+x2+2x2+2x+4x+4
=(x+1)(x2+2x+4)
d,x10+x5+1=(x10-x)+(x5-x2)+(x2+x+1)
=x((x3)3-1)+x2(x3-1)+(x2+x+1)
=x(x3-1)(x6+x3+1)+x2(x-1)(x2+x+1)+(x2+x+1)
=x(x-1)(x2+x+1)+x2(x-1)(x2+x+1)+(x2+x+1)
(x2+x+1)(x2-x+x3-x2+1)
e,(12x2-12xy+3y2)-10x(2x-y)=3(4x2-4xy+y2)-10x(2x-y)
=3(2x-y)2-10x(2x-y)=(2x-y)(6x-3y-10x)=(2x-y)(-4x-3y)
phân tích đa thức thành nhân tử
a,2x^2-7x+6
b,x^2+x-6
c,x^3+3x^2+6x+4
d,x^10+x^5+1
e,(12x^2-12xy+3y^2)-10x(2x-y)
a) \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
a) \(^{x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)}\)
b)\(a^3-a^2x-ay=a\left(a^2-a.x-y\right)\)
c)\(5x^2-10xy+5y-20z^2=-20z^2+\left(5-10x\right)y+5x^2 \)
\(=-5\left(4z^2+2xy-y-x^2\right)\)
d)\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3xy^2+3x^2y+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
a) (x^2+2xy+y^2)-9=(x+y)^2-9=(x+y-3)(x+y+3)
b) 5(x^2-2xy+y^2-4z^2)=5[(x-y)^2-4z^2]=5[(x-y-2z)(x-y+2z)
c)x^2-2x-5x+10=x(x-2)-5(x-2)=(x-5)(x-2)
d)2x^2-4x-3x+6=2x(x-2)-3(x-2)=(2x-3)(x-2)