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

Tìm đa thức M biết :

a, M +5 (5x² - 2xy) = 6x² +9xy - y²

M + 5. 5x² - 5. 2xy = 6x² + 9xy - y²

M + 25x² - 10xy = 6x² + 9xy - y²

M = 6x² + 9xy - y² + 10xy - 25x²

M = ( 6x² - 25x² ) + ( 9xy + 10xy ) - y²

M = -19x² + 19xy - y²

b, M - ( 3xy - 4y² ) = x² - 7xy + 8xy

M - 3xy + 4y² = x² - 15xy

M = x² - 15xy - 4y² + 3xy

M = x² + ( 15xy + 3xy ) - 4y²

M = x² + 18xy - 4y²

c, (25 . x²y - 13xy²+ y³ ) - M = 11x²y - 2y³

25x²y - 13xy²+ y³ - M = 11x²y - 2y³

M = 25x²y - 13xy²+ y³ - 11x²y - 2y³

M = ( 25x²y - 11x²y ) + ( y³ - 2y³ ) - 13xy²

M = 14x²y - y³ - 13xy²

d, M + (5x² - 2xy )= 6x² + 9xy -y²

M + 5x² - 2xy = 6x² + 9xy -y²

M = 6x² + 9xy -y² + 2xy - 5x²

M = ( 6x² - 5x² ) + ( 9xy + 2xy ) - y²

M = x² + 11xy - y²

\(M=8x^2-2xy-y^2-5x^2+2xy+3y^2=3x^2+2y^2>=0\forall x,y\)

Cảm ơn bn

Thanks

\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\)   \(\forall x,y\)

mà \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)  (đề bài ) \(\Rightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)

\(\Rightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}}\)

Rút gọn biểu thức

\(m+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)

=> \(m=x^2+11xy-y^2\)

Thay x,y, vừa tìm được vào biểu thức đã được rút gọn ta tính được m 

Đây là bài hướng dẫn, có gì thắc mắc hãy hỏi lại!!

Có M+5x^2-2xy-y^2=6x^2+9xy-y^2

=> M= 6x^2+9xy-y^2-(5x^2-2xy-y^2)

=> M=6x^2-9xy-y^2-5x^2+2xy+y^2

=>M=(6x^2-5x^2)+(-9xy+2xy)+(y^2-y^2)

=> M= x^2-7xy

     M + 5x² - 2xy - y² = 6x² + 9xy - y²

=> M = (6x² + 9xy - y²) - ( 5x² - 2xy - y² )    (chuyển vế)

=> M = 6x² + 9xy - y²  - 5x² + 2xy + y²        ( bỏ dấu ngoặc)

=> M = ( 6x² - 5x²) + (9xy + 2xy) + ( - y² + y²)

=> M = x² + 11xy