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Bài 7: Phân tích đa thức thành nhân tử
a) Ta có: \(a^2-b^2-2a+2b\)
\(=\left(a-b\right)\left(a+b\right)-2\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b-2\right)\)
b) Ta có: \(3x-3y-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(3+5x\right)\)
c) Ta có: \(16-x^2+4xy-4y^2\)
\(=16-\left(x^2-4xy+4y^2\right)\)
\(=16-\left(x-2y\right)^2\)
\(=\left(4-x+2y\right)\left(4+x-2y\right)\)
d) Ta có: \(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4-2x-3y+1\right)\left(x-y+4+2x+3y-1\right)\)
\(=\left(5-x-4y\right)\left(3x+2y+3\right)\)
e) Ta có: \(x^4+x^3+2x^2+x+1\)
\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+1+x\right)\)
f) Ta có: \(\left(x+3\right)^3+\left(x-3\right)^3\)
\(=\left(x+3+x-3\right)\left[\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)+\left(x-3\right)^2\right]\)
\(=2x\cdot\left[x^2+6x+9-\left(x^2-9\right)+x^2-6x+9\right]\)
\(=2x\cdot\left(2x^2+18-x^2+9\right)\)
\(=2x\cdot\left(x^2+27\right)\)
g) Ta có: \(9x^2-3xy+y-6x+1\)
\(=\left(9x^2-6x+1\right)-y\left(3x-1\right)\)
\(=\left(3x-1\right)^2-y\left(3x-1\right)\)
\(=\left(3x-1\right)\left(3x-1-y\right)\)
h) Ta có: \(x^3-4x^2+12x-27\)
\(=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
a) \(=\left[a^2+\left(a-1\right)\right]\left[a^2-\left(a-1\right)\right]=a^4-\left(a-1\right)^2=a^4-a^2+2a-1\)
b)\(=\left(a+2\right)\left(a^2-2a+4\right)\left(a-2\right)\left(a^2+2a+4\right)=\left(a^3+8\right)\left(a^3-8\right)=a^6-64\)
c)\(=x^2-4x+4-x^2+x+3x-3=1\)
d)\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6=-8\)
e) \(=4+12y+9y^2-4x^2+12xy-9y^2-12xy=-4x^2+12y+4\)
g)\(=x^3+3x^2+3x+1-x^3+3x^2-3x+1-x^3+1-x^3+1=-2x^3+6x^2+3\)
h) \(=\left(x^3+1\right)-\left(x^3-1\right)=2\)
giải
A=(3x-5)(2x+11)-(2x+3)(3x+7)
=6x^2+33x-10x-55-(6x^2+14x+9x+21)
=6x^2+33x-10x-55-6x^2-14x-9x-21
= -76
vậy biểu thức không phụ thuộc vào biến x,y
B=(2x+3)(4x^2-6x+9)-2(4x^3-1)
=8x^3-12x^2+18x+12x^2-18x+27-8x^3+2
=29
vậy biểu thức không phụ thuộc vào biến x
Trả lời :
1) x2+8x+21
= x^2 + 8x + 16 +5
= (x + 4 )^2 +5 lớn hơn hoặc bằng 5
Vậy giá tri nhỏ nhất của biểu thức bằng 5 khi x +4 =0 hay x=-4
2) f(x) = x^3 +x ^2 +x +1 =0
= (x^3 +x ^2) +(x +1) =0
= x^2 (x + 1 ) + (x +1 ) =0
= (x ^2 +1 )(x +1) =0
Xảy ra hai trường hợp :
x^2 +1=0 hoặc x + 1 =0
mà x^2 +1 >0 nên chỉ x + 1 =0 hay x= -1
Câu 3 gợi ý thôi bạn khai triển ra rồi thu gọn lại .
Học tốt 
\(\left(5x+3y\right)^2-\left(3y-1\right)\left(3y+1\right)-\left(4-5x\right)^2-10x\left(3y+4\right)\\ =25x^2+9y^2+30xy-\left(9y^2-1\right)-\left(16-40x+25x^2\right)-\left(30xy+40x\right)\\ =25x^2+9y^2+30xy-9y^2+1-16+40x-25x^2-30xy-40x\\ =\left(25x^2-25x^2\right)+\left(9y^2-9y^2\right)+\left(30xy-30xy\right)+\left(40x-40x\right)+\left(1-16\right)\\ =-15\)
a) \(x^2-10x+9\)
\(=x^2-9x-x+9\)
\(=x\left(x-9\right)-\left(x-9\right)\)
\(=\left(x-1\right)\left(x-9\right)\)
b) \(3x^2-10xy+3y^2\)
\(=3x^2-9xy-xy+3y^2\)
\(=3x\left(x-3y\right)-y\left(x-3y\right)\)
\(=\left(3x-y\right)\left(x-3y\right)\)
a/ \(=3y^2-6y-2x+1\)
b/ \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
c/ \(=\left(2-x\right)^3\)
d/ \(=xy^2+x^2y+3xy+x^2y+x^3+3x^2-3xy-3x^2-9x\)
\(=xy\left(y+x+3\right)+x^2\left(y+x+3\right)-3x\left(y+x+3\right)\)
\(=\left(xy+x^2-3x\right)\left(y+x+3\right)=x\left(y+x-3\right)\left(y+x+3\right)\)
e/ \(=xy-x^2+2x-y^2+xy-2y\)
\(=x\left(y-x+2\right)-y\left(y-x+2\right)=\left(x-y\right)\left(y-x+2\right)\)
a) =(2x+3y-1)2
b)=-(x-1)3
c)=-(x3-6x2+12x-8)=-(x-2)3
d)x3 + 2x2y + xy2 – 9x
= x(x2 + 2xy + y2 -9)
= x[(x2 + 2xy + y2) - 32]
= x[(x + y)2 - 32]
= x (x + y – 3)(x + y + 3)
e) 2x-2y-x2+2xy-y2=2(x-y)-(x-y)2=(x-y)(2-x+y)
c, ( 2x - 3 )2 - a2 - 2a - 1
= ( 2x - 3 )\(^2\) - ( a\(^2\) + 2a + 1 )
= ( 2x - 3 )\(^2\) - ( a + 1 )\(^2\)
= ( 2x - 3 - a - 1 ) ( 2x - 3 + a + 1 )
= ( 2x - a - 4 ) ( 2x + a - 2 )
d, 8x2 + 4xy - 2ax - ay
= 4 x ( 2x + y ) - a ( 2x + y )
= ( 4x - a ) ( 2x + y )
e, x2 - 2x - 3
= x\(^2\)+ x - 3x - 3
= x ( x + 1 ) - 3 ( x+1 )
= ( x - 3 ) ( x + 1 )