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Câu hỏi của Bangtan Sonyeondan - Toán lớp 8 - Học toán với OnlineMath
c) \(E=\left(x+a\right)\left(x+2a\right)\left(a+3a\right)\left(x+4a\right)+a^4\)
\(=\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(a+3a\right)+a^4\)
\(=\left(x^2+5ax+4a^2\right)\left(a^2+5ax+6a^2\right)+a^4\)(1)
Đặt \(x^2+5ax+4a^2=t\)
\(\Rightarrow\left(1\right)=t\left(t+2a^2\right)+a^4\)
\(=t^2+2a^2t+a^4=\left(t+a^2\right)^2\)(2)
Mà \(x^2+5ax+4a^2=t\)
Nên \(\left(2\right)=\left(x^2+5ax+5a^2\right)^2\)
a)x7+x5+1=x7+x6-x6+2x5-x5+x4-x4+x3-x3+x2-x2+1
=x7-x6+x5-x3+x2+x6-x5+x4-x2+x+x5-x4+x3-x+1
=x2(x5-x4+x3-x+1)+x(x5-x4+x3-x+1)+1(x5-x4+x3-x+1)
=(x2+x+1)(x5-x4+x3-x+1)
b)4x4-32x2+1=4x4+12x3+2x2-12x3-36x2-6x+2x2+6x+1
=2x2(2x2+6x+1)-6x(2x2+6x+1)+1(2x2+6x+1)
=(2x2-6x+1)(2x2+6x+1)
c)x6+27=(x2+3)(x2-3x+3)(x2+3x+3)
d)3(x4+x2+1)-(x2+x+1)
=3x4-3x3+2x2+3x3-3x2+2x+3x2-3x+2
=x2(3x2-3x+2)+x(3x2-3x+2)+1(3x2-3x+2)
=(x2+x+1)(3x2-3x+2)
e)bạn tự làm nhé
Phân tích đa thức thành nhân tử
a. 3ab ( x+ y) - 6ab ( y+ x)
=( x + y) ( 3ab - 6ab )
= ( x +y ) ( - 3ab)
b.7a (x - 3)+a2(x2 - 9)
=7a( x- 3) + a2 ( x2 - 32)
=7a ( x - 3 ) + a2 ( x- 3 ) ( x+3 )
= ( x- 3) . 7a + a2 ( x + 3)
= ( x- 3) ( 7a +a2x + 3a2)
c. 34 (x + y) -x -y
= 34 ( x+ y) - ( x+y)
=(x +y ) ( 34 - 1) = 33 ( x+ y)
d. 25 x4 - 942
=( 5x2 )2 - 942
=( 5x2 - 94 ) ( 5x2+94)
e.( 5a - b )2 - ( 2a +3b)2
=( 5a -b -2a - 3b) (5a -b + 2a + 3b)
=(3a - 4b) (7a+ 2b)
k. 22 -3a - b2 +3b
=( 22 - b2 ) + ( -3a +3b)
=( 2-b) (2+b) + 3( -a +b)
a) \(\left(2x+5\right)^2\)\(-\left(x-9\right)^2\)
=\(\left(2x+5+x-9\right).\left(2x+5-x+9\right)\)
=\(\left(3x-4\right).\left(x+14\right)\)
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2
a: \(x^2-9-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(1-x^2\right)\)
\(=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\)
b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
d: \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
e: \(3x^2-4x-4\)
\(=3x^2-6x+2x-4\)
\(=3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x+2\right)\)
g: \(x^4+64y^4\)
\(=x^4+16x^2y^2+64y^4-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
h: \(a^2+b^2+2a-2b-2ab\)
\(=a^2-2ab+b^2+2a-2b\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)
i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)
\(=\left(x+1-y+3\right)^2\)
\(=\left(x-y+4\right)^2\)
k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
Phân tích đa thức thành nhân tử:
a, \(36a^2-\left(a^2+9\right)^2\)
\(=\left(6a\right)^2-\left(a^2+9\right)^2\)
\(=\left(6a-a^2-9\right)\left(6a+a^2+9\right)\)
b, \(\left(a+3b\right)^2-\left(a^2+9\right)^2\)
\(=\left(a+3b-a^2-9\right)\left(a+3b+a^2+9\right)\)
c, \(9\left(2a-x\right)^2-4\left(3a-x\right)^2\)
\(=\left[3\left(2a-x\right)\right]^2-\left[2\left(3a-x\right)\right]^2\)
\(=\left(6a-3x\right)^2-\left(6a-2x\right)^2\)
\(=\left(6a-3x-6a+2x\right)\left(6a-3x+6a-2x\right)\)
\(=\left(-x\right)\left(12a-5x\right)\)
e, \(x^4+x^3+x+1\)
\(=\left(x^4+x^3\right)+\left(x+1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x^3+1\right)\left(x+1\right)\)
thank bạn nhiều nha !!!