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Bài 1 : Phân tích các đa thức sau thành nhân tử : ( tách một hạn tử thành nhiều hạng tử )
a, 3x2 + 9x - 30
= 3(x2 + 3x - 10)
= 3(x2 + 5x - 2x - 10)
= 3[x(x + 5) - 2(x + 5)]
= 3(x + 5)(x - 2)
b, x2 - 3x + 2
= x2 - x - 2x + 2
= x(x - 1) - 2(x - 1)
= (x - 1)(x - 2)
c, x2 - 9x + 18
= x2 - 6x - 3x + 18
= x(x - 6) - 3(x - 6)
= (x - 6)(x - 3)
d, x2 - 6x + 8
= x2 - 4x - 2x + 8
= x(x - 4) - 2(x - 4)
= (x - 4)(x - 2)
e, x2 - 5x - 14
= x2 + 2x - 7x - 14
= x(x + 2) - 7(x + 2)
= (x + 2)(x - 7)
f, x2 + 6x + 5
= x2 + x + 5x + 5
= x(x + 1) + 5(x + 1)
= (x + 1)(x + 5)
h, x2 - 7x + 12
= x2 - 3x - 4x + 12
= x(x - 3) - 4(x - 3)
= (x - 3)(x - 4)
i, x2 - 7x + 10
= x2 - 2x - 5x + 10
= x(x - 2) - 5(x - 2)
= (x - 2)(x - 5)
#Học tốt!
Bài 2 : Phân tích các đa thức sau thành nhân tử :
a, x2 + 7x + 12
= x2 + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3)(x + 4)
b, 3x2 - 8x + 5
= 3x2 - 3x - 5x + 5
= 3x(x - 1) - 5(x - 1)
= (x - 1)(3x - 5)
c, x4 + 5x2 - 6
= x4 - x2 + 6x2 - 6
= x2(x2 - 1) + 6(x2 - 1)
= (x2 - 1)(x2 + 6)
= (x - 1)(x + 1)(x2 + 6)
d, x4 - 34x2 + 225
= x4 - 9x2 - 25x2 + 225
= x2(x2 - 9) - 25(x2 - 9)
= (x2 - 9)(x2 - 25)
= (x - 3)(x + 3)(x - 5)(x + 5)
e, x2 - 5xy + 6y2
= x2 + xy - 6xy + 6y2
= x(x + y) - 6y(x + y)
= (x + y)(x - 6y)
f, 4x2 - 17xy + 13y2
= 4x2 - 4xy - 13xy + 13y2
= 4x(x - y) - 13y(x - y)
= (x - y)(4x - 13y)
a) = (x + 3)2 - y2 = (x + 3 - y)(x + 3 + y)
b) = x2(x - 3) -4(x - 3) = (x - 3)(x2 - 4) = (x - 3)(x - 2)(x + 2)
c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)
d) Nhầm đề. tui sửa lại x3 + y3 + 2x2 - 2xy + 2y2
= x3 + y3 + 2(x2 - xy + y2) = (x + y)(x2 - xy + y2) + 2(x2 - xy + y2) = (x2 - xy + y2)(x + y + 2)
e) = x4 - x3 - x3 + x2 - x2 + x + x - 1 = x3(x - 1) - x2(x - 1) - x(x - 1) + x - 1 = (x - 1)(x3 - x2 - x + 1) = (x - 1)(x - 1)(x2 - 1) = (x - 1)3(x + 1)
f) = x3 - 3x2 - x2 + 3x + 9x - 27 = x2(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x2 - x + 9)
g) chắc là 3xyz
= x2y + xy2 + y2z + yz2 + x2z + xz2 + 3xyz = x2y + xy2 + xyz + y2z + yz2 + xyz + x2z + xz2 + xyz = (x + y + z)(xy + yz + xz)
h) = 23 -(3x)3 = (2 - 3x)(4 + 6x + 9x2)
i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy
k) = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy +y2)(x + y)(x2 - xy +y2).
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) (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)
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)
Bài 4:
a) Ta có: \(a^4+a^2+1\)
\(=a^4+2a^2+1-a^2\)
\(=\left(a^2+1\right)^2-a^2\)
\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)
b) Ta có: \(a^4+a^2-2\)
\(=a^4+2a^2-a^2-2\)
\(=a^2\left(a^2+2\right)-\left(a^2+2\right)\)
\(=\left(a^2+2\right)\left(a^2-1\right)\)
\(=\left(a^2+2\right)\left(a-1\right)\left(a+1\right)\)
c) Ta có: \(x^4+4x^2-5\)
\(=x^4+5x^2-x^2-5\)
\(=x^2\left(x^2+5\right)-\left(x^2+5\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
d) Ta có: \(x^3-19x-30\)
\(=x^3-25x+6x-30\)
\(=x\left(x^2-25\right)+6\left(x-5\right)\)
\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
e) Ta có: \(x^3-7x-6\)
\(=x^3-4x-3x-6\)
\(=x\left(x^2-4\right)-3\left(x+2\right)\)
\(=x\left(x-2\right)\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x-3\right)\)
\(=\left(x+2\right)\left(x^2-3x+x-3\right)\)
\(=\left(x+2\right)\left[x\left(x-3\right)+\left(x-3\right)\right]\)
\(=\left(x+2\right)\left(x-3\right)\left(x+1\right)\)
f) Ta có: \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x^2-7x+2x-14\right)\)
\(=x\left[x\left(x-7\right)+2\left(x-7\right)\right]\)
\(=x\left(x-7\right)\left(x+2\right)\)
a/ \(3x^2-5x-2\)
\(=3x^2-3x-2x-2\)
\(=3x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(3x+2\right)\)
b/ \(2x^2+x-6\)
\(=2x^2+4x-3x-6\)
\(=2x\left(x+2\right)+3\left(x+2\right)\)
\(=\left(x+2\right)\left(2x+3\right)\)
c/ \(7x^2+50x+7\)
\(=7x^2+49x+x+7\)
\(=7x\left(x+7\right)+\left(x+7\right)\)
\(=\left(x+7\right)\left(7x+1\right)\)
d/ \(12x^2+7x-12\)
\(=12x^2-9x+16x-12\)
\(=3x\left(4x-3\right)+4\left(4x-3\right)\)
\(=\left(4x-3\right)\left(3x+4\right)\)
e/ \(15x^2+7x-2\)
\(=15x^2+10x-3x-2\)
\(=5x\left(3x+2\right)-\left(3x+2\right)\)
\(=\left(3x+2\right)\left(5x-1\right)\)
f/ \(a^2-5a-14\)
\(=a^2+2a-7a-14\)
\(=a\left(a+2\right)-7\left(a+2\right)\)
\(=\left(a+2\right)\left(a-7\right)\)
g/ \(2m^2+10m+8\)
\(=2m^2+2m+8m+8\)
\(=2m\left(m+1\right)+8\left(m+1\right)\)
\(=\left(m+1\right)\left(2m+8\right)\)
h/ \(4p^2-36p+56\)
\(=4p^2-28p-8p+56\)
\(=4p\left(p-7\right)-8\left(p-7\right)\)
\(=\left(p-7\right)\left(4p-8\right)\)
câu i) tách 5x sao vậy ạ ?