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

\(M=x^2+11xy-y^2\)

\(N=3xy-4y^2-x^2+7xy-8y^2\)

\(N=-x^2+10xy-12y^2\)

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

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

b.     \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)

\(\Rightarrow N=3xy-4y^2-x^2+7xy-8y^2\)

a, \(A=-x^2+4xy^2-2xz+3y^2\)

b, \(B=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\)

c, \(A=3xy-4y^2-x^2+7xy-8y^2=-x^2+10xy-12y^2\)

\(a.M+(5x^2-2xy)=6x^2+9xy-y^2 \)
\(M=(6x^2+9xy-y^2)-(5x^2-2xy)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=(6x^2-5x^2)+(9xy+2xy)-y^2\)
\(M=x^2+11xy-y^2\)
Vậy \(M=x^2+11xy-y^2\)
\(b.M+(3x^2y-2xy^3)=2x^2y-4xy^3\)
\(M=(2x^2y-4xy^3)-(3x^2-2xy^3)\)
\(M= \) \(2x^2-4xy^3-3x^2+2xy^3\)
\(M=(2x^2-3x^2)+(-4xy^3+2xy^3)\)
\(M=-x^2-2xy^3\)
Vậy \(M=-x^2-2xy^3\)

Mình sẽ giúp bạn bài này a. M+(5x^2-2xy) = 6x^2+9xy-y^2 M = (6x^2+9xy-y^2) - (5x^2-2xy) = 6x^2+9xy-y^2-5x^2+2xy = (6x^2-5x^2)+(9xy+2xy)+(-y^2) = x^2+11xy-y^2 b. M + (3x^2y−2xy^3)=2x^2y−4xy^3 M = (2x^2y−4xy^3)-(3x^2y−2xy^3) = 2x^2y−4xy^3-3x^2y+2xy^3 = (2x^2y-3x^2y)+(−4xy^3+2xy^3) = -x^2y+(-2xy^3) c.(1/2xy^2+x^2−x^2y)−M=−xy^2+x^2y+1 M =(1/2xy^2+x^2−x^2y)-(−xy^2+x^2y+1) =1/2xy^2+x^2−x^2y-xy^2-x^2y-1 = (1/2xy^2-xy^2)+(x^2y-x^2y)+x^2-1 = -1/2xy^2+x^2-1 d. M−(x^3.y^2−x^2.y+xy)=2x^3.y^2−3/2xy M= (2x^3.y^2−3/2xy)+(x^3.y^2−x^2.y+xy) = 2x^3.y^2−3/2xy+x^3.y^2−x^2.y+xy = (2x^3.y^2+x^3.y^2)+(3/2xy+xy)-x^2.y = 3x^3.y^2+5/2xy-x^2.y

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

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

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

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

M = 1x\(^2\) + 11xy - y\(^2\)

Ta có: \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\left(\forall x\right)\\\left(3y+4\right)^{2020}\ge0\left(\forall y\right)\end{cases}}\Rightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\left(\forall x,y\right)\)

Mà \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\left(\forall x,y\right)\)

\(\Rightarrow\hept{\begin{cases}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)

Khi đó thay vào ta được: 

\(M+5\cdot\left(\frac{5}{2}\right)^2-2\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)=6\cdot\left(\frac{5}{2}\right)^2+9\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(\Leftrightarrow M+\frac{455}{12}=\frac{103}{18}\)

\(\Rightarrow M=-\frac{1159}{36}\)

a/

 \(M+5x^2-2xy-6x^2-9xy+y^2=0\)

\(M-x^2-11xy+y^2=0\)

\(M-x^2+y^2-11xy=0\)

\(M=x^2-y^2+11xy\)

Vậy:..

Câu b tương tự

M + (5x² - 2xy) = 6x² + 9xy - y^{2   ta có} M = ( 6x² + 9xy - y^{2) -(}5x² - 2xy)= x2 +11xy -y2

a: \(M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\)

b: \(N=3xy-4y^2-x^2+7xy-8y^2=-x^2+10xy-12y^2\)

a) M + (5x² - 2xy) = 6x² + 9xy - y²

=> M = (6x² + 9xy - y²) - (5x² - 2xy)

=> M = 6x² + 9xy - y² - 5x² + 2xy = (6x² - 5x²) + (9xy + 2xy) - y² = x² + 11xy - y²

b) Sửa đề lại đi nhé

c) (25x²y - 13x²y + y³) - M = 11x²y - 2y²

=> M = (25x²y - 13x²y + y³) - (11x²y - 2y²)

=> M = 25x²y - 13x²y + y³ - 11x²y + 2y²

=> M = x²y + y³ + 2y²

d) M = 0 - (12x⁴ - 15x²y + 2xy² + 7) = -12x⁴ + 15x²y - 2xy² - 7

a) Ta có : M = 6x² + 9xy - y² - (5x² - 2xy)

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

                    = x² + 11xy - y²

b) Ta có M = x² - 7xy + 8y² - (3xy - 24y²)

                 = x² - 7xy + 8y² - 3xy + 24y²

                  = x² - 10xy + 32y²

c) Ta có M = 25x².y- 13x²y + y³ - (11x²y - 2y²)

                  = 25x².y- 13x²y + y³ - 11x²y + 2y²

                 = x²y + y³ + 2y²

d) Ta có M = -(12x⁴ - 15x²y + 2xy² + 7)

                 =  -12x⁴ + 15x²y - 2xy² - 7

Bài 3: 

a: \(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\)

b: \(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2=-x^2+10xy-12y^2\)

Bài 2: 

\(A+B=4x^4-5xy+5y^2+3x^2+2xy-y=4x^4+3x^2-3xy+5y^2-y\)

\(A-B=4x^4-5xy+5y^2-3x^2-2xy+y=4x^4-3x^2+5y^2-7xy+y\)

\(B-A=-\left(A-B\right)=-4x^4+3x^2-5y^2+7xy-y\)