a. a^4 + a^3 +3a^2 + 2a +2
b. x^3 + y^3 + 3x^2 +3x + 1
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a) A = (a - 2b + c) - (a - 2b - c)
= a - 2b + c - a + 2b + c
= (a - a) - (2b - 2b) + (c + c)
= 2c
b) tương tự trên
c) C = 2(3a + b - 1) - 3(2a + b - 2)
= 6a + 2b - 2 - 6a - 3b + 3
= (6a - 6a) + (2b - 3b) - (2 - 3)
= 0 - b + 1
= -b + 1
d) D = 4(x - 1) - (3x + 2)
= 4x - 4 - 3x - 2
= (4x - 3x) - (4 + 2)
= x - 6
a ) \(x^3+3x^2-3x+1\)
\(=x^3-3x+3x^2-1\)
\(=\left(x-1\right)^3\)
\(\left(x+1\right)\left(x^2-x-x^2+x-1\right)=-\left(x+1\right)\)
\(\left(2a^2+1\right)^2-4a^2-\left(2a^2+1\right)^2=-4a^2\)
\(\left(a^2+b^2+c^2+a^2-b^2-c^2\right)\left(a^2+b^2+c^2-a^2+b^2+c^2\right)=2a^2\left(2b^2+2c^2\right)=4a^2b^2+4a^2c^2\)
\(\left(a-5\right)^2\left(a+5\right)^2=\left(a^2-25\right)^2\)
\(\left(3a^3+1\right)^2-9a^2-\left(3a^3+1\right)^2=-9a^2\)
1.a (3x-2y)2= (3x)2 - 2. 3x . 2y - (2y)2 = 9x2 - 12xy - 4y2
2.b (2x - 1/2)2 = (2x)2 - 2.2x.1/2 - (1/2)2= 4x2 - 2 - 1/4
3.c (x/2 - y) (x/2+y)= (x/2)2 - (y)2 = x/4 - y2
Bài 1 :
\(\left(3x-2y\right)^2=9x^2-12xy+4y^2\)
\(\left(2x-\frac{1}{2}\right)^2=4x^2-4x+\frac{1}{4}\)
\(\left(\frac{x}{2}-y\right)\left(\frac{x}{2}+y\right)=\frac{x^2}{4}-y^2\)
\(\left(x+\frac{1}{3}\right)^3=x^3+x^2+\frac{1}{3}x+\frac{1}{27}\)
\(\left(x-2\right)\left(x^2+2x+2^2\right)=x^3-8\)
(3x^3 - 2x^2 + x + 2)(5x^2)
= 15x^5 - 10x^4 + 5x^3 + 10x^2
(3x^2 + 5x - 2)(2x^2 - 4x + 3)
= 3x^4 - 12x^3 + 9x^2 + 10x^3 - 20x^2 + 15x - 4x^2 + 8x - 6
= 6x^4 - 2x^3 - 15x^2 + 23x - 6
a) \(=6a-3+15-5a=a+12\)
b) \(=25x-12x+4+35-14x=-x+39\)
d) \(=2ab+8a^2-b^2-4ab+2ab-6a^2=2a^2-b^2\)
e) \(=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4=-x^5+2x+1\)
f) \(=6y^3-3y^2+y-y+y^2-y^3-y^2+y=5y^3-3y^2+y\)
a) 3( 2a -1) +5( 3-a)
= 3. 2a -3.1 +5. 3- 5.a
= 6a -3+ 15-5a
=(6a -5a )+ (-3+ 15)
b) 25x - 4(3x - 1) +7(5 - 2x)
= 25x -4.3x + 4.1 + 7.5 - 7.2
=25x - 12x + 4 +35 - 14x
= (25x-12x-14x)+(4+35)
= -x=39
c) -12x3 -x1-2x-18x2
= -36x-x-2x-36x
= -75x
d) (2a-b)(b+4a)+2a(b-3a)
= 2ab+2a4a-bb-b4a+2ab-2a3b
= 2ab+8a2-b2-4ab+2ab-6a2
=(2ab-4ab+2ab)+(8a2-6a2)-b2
= 2a2-b2
e) (x+1)(2+x-x2+x3-x4)
= (x+1)(2-2x)
= x2-x2x+1.2-1.2x
=(2x-2x)-2x2+2
= -2x2+2
Bài 1:
a) \(\left(x+y\right)^2-y^2=x^2+2xy+y^2-y^2=x^2+2xy=x\left(x+2y\right)\)
b) Sửa đề: \(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)^2\left(x+y\right)^2\)
c) \(x\left(x-3y\right)^2+y\left(y-3x\right)^2=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)\)
\(=x^3-6x^2y+9xy^2+y^3-6xy^2+9x^2y\)
\(=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3\)
Bài 2:
a) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)\)
\(=2a\left(a^2+3b^2\right)\)
b) \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2b\left(b^2+3a^2\right)\)
b.x^3+y^3+3x^2+3x+1=(x^3+3x^2+3x+1)+y^3
=(x+1)^3+y^3
=(x+1+y).((x+1)^2-x+1.y+y^2))
Phần b trước nà