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a, (3x2-2xy+y2) + (x2-xy+2y2) - (4x2-y2)
= 3x2-2xy+y2+x2-xy+2y2-4x2+y2
= 4y2-3xy
b, = x2-y2+2xy-x2-xy-2y2+4xy-1
= -3y2+5xy
c, M=5xy+x2-7y2+(2xy-4y)2 = 5xy+x2-7y2+4x2y2-16xy2+16y2 = 5xy+x2+9y2+4x2y2-16xy2
Bài 1 Tớ giải từng bài nhé ! Ko có ý đồ câu điểm.
\(A=4x^2-5xy+xy^2\)
\(B=3x^2+2xy-xy^2\)
Ta có : \(A+B=4x^2-5xy+xy^2+3x^2+2xy-xy^2\)
\(=7x^2-3xy\)
\(A-B=4x^2-5xy+xy^2-3x^2-2xy+xy^2\)
\(=x^2-7xy+2xy^2\)
Bài 2 : N ở đâu ?
Ta có : \(M+\left(5x^2-2xy\right)=xy^2+xy^3-y^2\)
\(M=xy^2+xy^3-y^2-5x^2+2xy\)
Bài 3 :
\(A=x^2y-xy^2+xy^2=x^2y\)
\(B=xy+4xy^2-2x-1\)
mk chỉ làm đc bà 1 thôi nha
M+x2+32y-5xy2-7xy-2
=M+(x2-5xy2-7xy)+(32y-2)
Để đa thức tổng ko chứa biến x thì:
M+(x2-5xy2-7xy)=0
=> M=0-(x2-5xy2-7xy)
M=-x2-5xy2-7xy
Lời giải:
a)
$M=2x^3-4xy+6y^2+x^2+3xy-y^2=2x^3-xy+5y^2+x^2$
b)
$M=-(2x^2-4xy+y^2)=-2x^2+4xy-y^2$
c)
$2M=(2x^2-7xy+3y^2)-(4x^2-5xy+9y^2)=-2x^2-2xy-6y^2$
$\Rightarrow M=-x^2-xy-3y^2$
Bài 1 :
A + B = 4x2 - 5xy + 3y2 + 3x2 + 2xy - y2
= ( 4x2 + 3x2 ) - ( 5xy - 2xy ) + ( 3y2 - y2 )
= 7x2 - 3xy + 2y2
A - B = 4x2 - 5xy + 3y2 - ( 3x2 + 2xy - y2 )
= 4x2 - 5xy + 3y2 - 3x2 - 2xy + y2
= ( 4x2 - 3x2 ) - ( 5xy + 2xy ) + ( 3y2 + y2 )
= x2 - 7xy + 4y2
Bài 2 :
a) M + (5x2 - 2xy) = 6x2 + 9xy - y2
M = 6x2 + 9xy - y2 - (5x2 - 2xy)
M = 6x2 + 9xy - y2 - 5x2 + 2xy
M = ( 6x2 - 5x2 ) + ( 9xy + 2xy ) - y2
M = x2 + 11xy - y2
Vậy M = x2 + 11xy - y2
b) (3xy - 4y2) - N = x2 - 7xy + 8y2
N = 3xy - 4y2 - x2 - 7xy + 8y2
N = ( 3xy - 7xy ) - ( 4y2 - 8y2 ) - x2
N = -4xy + 4y2 - x2
Vậy N = -4xy + 4y2 - x2
3, Cho đa thức
A(x)+B(x) = (3x4-\(\dfrac{3}{4}\)x3+2x2-3)+(8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\))
= 3x4-\(\dfrac{3}{4}\)x3+2x2-3+8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\)
= (3x4+8x4)+(-3/4x3+1/5x3)+(-3+2/5)+2x2-9x
= 11x4 -0.55x3-2.6+2x2-9x
A(x)-B(x)=(3x4-\(\dfrac{3}{4}\)x3+2x2-3)-(8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\))
= 3x4-\(\dfrac{3}{4}\)x3+2x2-3-8x4-\(\dfrac{1}{5}\)x3+9x-\(\dfrac{2}{5}\)
= (3x4-8x4)+(-3/4x3-1/5x3)+(-3-2/5)+2x2+9x
= -5x4-0.95x3-3.4+2x2+9x
B(x)-A(x)=(8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\))-(3x4-\(\dfrac{3}{4}\)x3+2x2-3)
=8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\)-3x4+\(\dfrac{3}{4}\)x3-2x2+3
=(8x4-3x4)+(1/5x3+3/4x3)+(2/5+3)-9x-2x2
= 5x4+0.95x3+2.6-9x-2x2
a) M = ( -2x^3 + x^2y + 1 ) + ( 2x^2y - 1 )
= -2x^3 + x^2y + 1 + 2x^2y - 1
= -2x^3 + ( x^2y + 2x^2y ) + ( 1 - 1 )
= -2x^3 + 3x^2y
b) M = ( 3x^2 + 3xy - x^3 ) - ( 3x^2 + 2xy -4y^2 )
= 3x^2 + 3xy - x^3 - 3x^2 - 2xy + 4y^2
= ( 3x^2 - 3x^2 ) + ( 3xy - 2xy ) - x^3 + 4y^2
= xy - x^3 + 4y^2
a, \(M-\left(3xy-4y^2-2xy\right)=\left(x^2-7xy+8y^2\right)\)
\(\Rightarrow M=\left(x^2-7xy+8y^2\right)+\left(3xy-4y^2-2xy\right)\)
\(\Rightarrow M=x^2-7xy+8y^2+3xy-4y^2-2xy\)
\(\Rightarrow M=x^2+\left[3xy-7xy-2xy\right]+\left[8y^2-4y^2\right]\)
\(\Rightarrow M=x^2-6xy+4y^2\)
b, \(N+\left(x^3-xyz+3x^2y\right)=2x^3+3xy-xy^2\)
\(\Rightarrow N=\left(2x^3+3xy-xy^2\right)-\left(x^3-xyz+3x^2y\right)\)
\(\Rightarrow N=2x^3+3xy-xy^2-x^3+xyz-3x^2y\)
\(\Rightarrow N=\left[2x^3-x^3\right]+3xy-xy^2+xyz-3x^2y\)
\(\Rightarrow N=x^3+3xy-xy^2+xyz-3x^2y\)
Tích mình nha!!!
\(1,M+N\\ =\left(2x^2-4xy+6y^2\right)+\left(2x^2+2xy-4y^2\right)\\ =2x^2-4xy+6y^2+2x^2+2xy-4y^2\\ =\left(2x^2+2x^2\right)+\left(-4xy+2xy\right)+\left(6y^2-4y^2\right)\\ =4x^2-2xy+2y^2\\ 2,M+\left(x^3-2xy^2+y^3\right)=x^3+5xy^2-y^3\\ =>M=\left(x^3+5xy^2-y^3\right)-\left(x^3-2xy^2+y^3\right)\\ =>M=x^3+5xy^2-y^3-x^3+2xy^2-y^3\\ =>M=\left(x^3-x^3\right)+\left(5xy^2+2xy^2\right)+\left(-y^3-y^3\right)\\ =>M=7xy^2-2y^3\)
1)
M + N = (2x² - 4xy + 6y²) + (2x² + 2xy - 4y²)
= 2x² - 4xy + 6y² + 2x² + 2xy - 4y²
= (2x² + 2x²) + (-4xy + 2xy) + (6y² - 4y²)
= 4x² - 2xy + 2y²
2)
M + (x³ - 2xy² + y³) = x³ + 5xy² - y³
M = x³ + 5xy² - y³ - (x³ - 2xy² + y³)
= x³ + 5xy² - y³ - x³ + 2xy² - y³
= (x³ - x³) + (5xy² + 2xy²) + (-y³ - y³)
= 7xy² - 2y³