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Ta có:
\(x\left(x-a\right)\left(x+a\right)\left(x+2a\right)+a^4\)
\(\Leftrightarrow\) \(\left[x.\left(x+a\right)\right]\left[\left(x-a\right).\left(x+2a\right)\right]+a^4\)
\(\Leftrightarrow\) \(\left(x^2+ax\right)\left(x^2+ax-2a^2\right)+a^4\)
Đặt b = \(\left(x^2+ax\right)\)
Khi đó đa thức đã cho có dạng:
\(b\left(b-2a^2+a^4\right)\)
\(\Leftrightarrow\) \(b^2-2a^2b+a^4\)
\(\Leftrightarrow\) \(\left(b-a^2\right)^2\)
\(\Leftrightarrow\) \(\left(x^2+ax-a^2\right)^2\)
hay \(x\left(x-a\right)\left(x+a\right)\left(x+2a\right)+a^4\) là bình phương của 1 đa thức
Trả lời:
Bài 4:
b, B = ( x + 1 ) ( x7 - x6 + x5 - x4 + x3 - x2 + x - 1 )
= x8 - x7 + x6 - x5 + x4 - x3 + x2 - x + x7 - x6 + x5 - x4 + x3 - x2 + x - 1
= x8 - 1
Thay x = 2 vào biểu thức B, ta có:
28 - 1 = 255
c, C = ( x + 1 ) ( x6 - x5 + x4 - x3 + x2 - x + 1 )
= x7 - x6 + x5 - x4 + x3 - x2 + x + x6 - x5 + x4 - x3 + x2 - x + 1
= x7 + 1
Thay x = 2 vào biểu thức C, ta có:
27 + 1 = 129
d, D = 2x ( 10x2 - 5x - 2 ) - 5x ( 4x2 - 2x - 1 )
= 20x3 - 10x2 - 4x - 20x3 + 10x2 + 5x
= x
Thay x = - 5 vào biểu thức D, ta có:
D = - 5
Bài 5:
a, A = ( x3 - x2y + xy2 - y3 ) ( x + y )
= x4 + x3y - x3y - x2y2 + x2y2 + xy3 - xy3 - y4
= x4 - y4
Thay x = 2; y = - 1/2 vào biểu thức A, ta có:
A = 24 - ( - 1/2 )4 = 16 - 1/16 = 255/16
b, B = ( a - b ) ( a4 + a3b + a2b2 + ab3 + b4 )
= a5 + a4b + a3b2 + a2b3 + ab4 - ab4 - a3b2 - a2b3 - ab4 - b5
= a5 + a4b - ab4 - b5
Thay a = 3; b = - 2 vào biểu thức B, ta có:
B = 35 + 34.( - 2 ) - 3.( - 2 )4 - ( - 2 )5 = 243 - 162 - 48 + 32 = 65
c, ( x2 - 2xy + 2y2 ) ( x2 + y2 ) + 2x3y - 3x2y2 + 2xy3
= x4 + x2y2 - 2x3y - 2xy3 + 2x2y2 + 2y4 + 2x3y - 3x2y2 + 2xy3
= x4 + 2y4
Thay x = - 1/2; y = - 1/2 vào biểu thức trên, ta có:
( - 1/2 )4 + 2.( - 1/2 )4 = 1/16 + 2. 1/16 = 1/16 + 1/8 = 3/16
a)\(\left(x+y\right)^2:\left(x+y\right)=x+y\)
b)\(\left(x-y\right)^5:\left(y-x\right)^4=\left(x-y\right)^5:\left(x-y\right)^4=x-y\)
c)\(\left(5x^4-3x^3+x^2\right):3x^2=\frac{5}{3}x^2-x+\frac{1}{3}^{ }\)
d)\(\left(x^3y^3-\frac{1}{2}x^2y^3+x^3y^2\right):\frac{1}{2}x^2y^2=2xy-y+x\)
a: \(=-8x^5+6x^3-2\)
b: \(=-\dfrac{2}{3}x+7-x^2y\)
c: \(=\dfrac{7\left(x-y\right)^4+4\left(x-y\right)^3}{\left(x-y\right)^2}=7\left(x-y\right)^2+4\left(x-y\right)\)
d: \(=\dfrac{6\left(x-3y\right)^4}{5\left(x-3y\right)}=\dfrac{6}{5}\left(x-3y\right)^3\)
\(\left(x+2\right)\left(x^2+2x-9\right)\)
\(=x^3+2x^2-9x+2x^2+4x-18\)
\(=x^3+4x^2-5x-18\)
\(\left(x^{2y}-6\right)\left(x^2-5\right)\)
\(=x^{4y}-5x^{2y}-6x^2+30\)
\(\left(x+y\right)\left(xy-4+y\right)\)
\(=x^2y-4x+xy+xy^2-4y+y^2\)
câu còn lại tương tự nha
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
A=\(\left(x-y\right)^2+\left(x+y\right)^2=x^2-2xy+y^2+x^2+2xy+y^2=2x^2+2y^2\)
B=\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=\left(2y\right).\left(2x\right)\)
C=\(\left(2a+b\right)^2-\left(2a-b\right)^2=\left(2a+b-2a+b\right)\left(2a+b+2a-b\right)=\left(2b\right).\left(4a\right)\)
D=\(\left(2x-1\right)^2-2\left(2x-3\right)^2+4=4x^2-4x+1-4x+6+4=4x^2-8x+11\)
E=\(\left(x+3y\right)^2-\left(x-3y\right)^2=\left(x+3y-x+3y\right)\left(x+3y+x-3y\right)=\left(6y\right).\left(2x\right)\)
F=\(\left(2x+y\right)^2-\left(2x-y\right)^2=\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)=\left(2y\right).\left(4x\right)\)
G=\(\left(x-2y\right)^2+4\left(x-2y\right)y+4y^2=x^2-4xy+4y^2+4xy-8y^2+4y^2=x^2\)
H=\(\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y^{ }\right)^2=x^2-2xy+y^2-4\left(x^2+2xy-xy-2y^2\right)+4x+8y=x^2-2xy+y^2-4x^2-8xy+4xy+8y^2+4x+8y=3x^2+12xy-9y^2+4x+8y\)
Ta có:
a) A= (x-y)^2 + (x+y)^2
A= x^2 -2xy + y^2 + x^2 + 2xy + y^2
A= 2x^2+ 2y^2
b) B= (x+y)^2 -( x-y)^2
B= (x+y-x+y)(x+y+x-y)
B= 2y.2x= 4xy
c) C= (2a+b)^2 -( 2a-b)^2
C= (2a+b-2a+b)(2a+b+2a-b)
C= 2b.4a
C= 8ab
d) D= (2x-1)^2 -2(2x-3)^2+4
D= 4x^2 -4x+1 -2( 4x^2 -12x + 9) +4
D= 4x^2 -4x+1 -8x^2 + 24x -18 +4
D= -4x^2 + 20x-13
e) E= (x+3y)^2-(x-3y)^2
E= (x+3y-x+3y)(x+3y+x-3y)
E= 6y.2x= 12xy
f) F= (2x+y)^2-(2x-y)^2
F=(2x+y-2x+y)(2x+y+2x-y)
F= 2y.4x= 8xy
g) G= (x-2y)^2 + 4(x-2y)y + 4y^2
G= (x-2y)^2 + 2(x-2y)2y + (2y)^2
G= (x-2y+2y)^2
G= x^2
h) H= (x-y)^2 -4(x-y)(x+2y)+ 4(x+2y)^2
H= (x-y)^2 - 2(x-y)2(x+2y) + [2(x+2y)]^2
H= (x-y- 2x-4y)^2
H= (-x-5y)^2
Lưu ý (-A-B)^2 = ( A+ B)^2
=> H= (x+5y)^2
\(\text{a) }\left(x-1\right)\left(x^2+y\right)-\left(x^2-y\right)\left(x-2\right)-x\left(x+2y\right)+3\left(y-5\right)\)
\(=\left(x^3+xy-x^2-y\right)-\left(x^3-2x^2-xy+2y\right)-\left(x^2+2xy\right)+\left(3y-15\right)\)
\(=x^3+xy-x^2-y-x^3+2x^2+xy-2y-x^2-2xy+3y-15\)
\(=\left(x^3+x^3\right)+\left(-x^2+2x^2-x^2\right)+\left(xy+xy-2xy\right)+\left(-y-2y+3y\right)-15\)
\(=0+0+0+0-15\)
\(=-15\)
\(\text{b) }6\left(x^3y+x-3\right)-6x\left(2xy^3+1\right)-3x^2y\left(2x-4y^2\right)\)
\(=\left(6x^3y+6x-18\right)-\left(12x^2y^3+6x\right)-\left(6x^3y-12x^2y^3\right)\)
\(=6x^3y+6x-18-12x^2y^3-6x-6x^3y+12x^2y^3\)
\(=\left(6x^3y-6x^3y\right)+\left(6x-6x\right)+\left(-12x^2y^3+12x^2y^3\right)-18\)
\(=0+0+0-18\)
\(=-18\)
\(\text{c) }\left(x^2+2xy+4y^2\right)\left(x-2y\right)-6\left(\frac{1}{2}-\frac{4}{3}y^3\right)\)
\(=\left(x^3-2x^2y+2x^2y-4xy^2+4xy^2-8y^3\right)-\left(3-8y^3\right)\)
\(=\left(x^3-8y^3\right)-\left(3-8y^3\right)\)
\(=x^3-8y^3-3+8y^3\)
\(=x^3-3\)
a: \(=2\left[\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\right]-3\left(x^4+y^4\right)\)
\(=2x^4-2x^2y^2+2y^4-3x^4-3y^4\)
\(=-\left(x^4+2x^2y^2+y^4\right)\)
\(=-\left(x^2+y^2\right)^2=-1\)
b: Bạn xem lại đề chỗ 3y^ nhé bạn