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a) Có:
\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(\Rightarrow M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=\left(6x^2-5x^2\right)+\left(9xy+2xy\right)-y^2\)
\(\Rightarrow M=x^2+11xy-y^2\)
b) Có:
\(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Rightarrow N=3xy-4y^2-\left(x^2-7xy+8y^2\right)\)
\(N=3xy-4y^2-x^2+7xy-8y^2\)
\(N=\left(3xy+7xy\right)+\left(-4y^2-8y^2\right)-x^2\)
\(\Rightarrow N=10xy+\left(-12y^2\right)-x^2\)
Hay \(N=10xy-12y^2-x^2\)
Chúc bạn học tốt!
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:
\(A+B=4x^4-5xy+5y^2+3x^2+2xy-y=4x^4+3x^2-3xy+5y^2-y\)
\(A-B=4x^4-5xy+5y^2-3x^2-2xy+y=4x^4-3x^2+5y^2-7xy+y\)
\(B-A=-\left(A-B\right)=-4x^4+3x^2-5y^2+7xy-y\)
Bài 1:
\(A+B=7x^2-3xy+2y^2\)
\(A-B=x^2-7xy+4y^2\)
Bài 2:
a) \(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)
\(M=x^2+11xy-y^2\)
b) \(N=\left(3xy-4y^2\right)-\left(x^2-7xy+8y^2\right)\)
\(N=-x^2-12y^2+10xy\)
a: \(M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\)
b: \(N=3xy-4y^2-x^2+7xy-8y^2=-x^2+10xy-12y^2\)
\(A-\left(3xy-4y^2\right)=x^2-7xy+8y^2\)
\(A=x^2-7xy+8y^2+\left(3xy-4y^2\right)\)
\(A=x^2-7xy+8y^2+3xy-4y^2\)
\(A=x^2-\left(7xy-3xy\right)+\left(8y^2-4y^2\right)\)
\(A=x^2-4xy+4y^2\)
\(A=\left(x-2y\right)^2\)
\(A-\left(3xy-4y^2\right)=x^2-7xy+8y^2\)
=>A=\(x^2-4xy+4y^2\)
=>A=\(\left(x-2y\right)^2\)
Vậy A=\(\left(x-2y\right)^2\)
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:
\(M+5\cdot\left(\frac{5}{2}\right)^2-2\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)=6\cdot\left(\frac{5}{2}\right)^2+9\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(\Leftrightarrow M+\frac{455}{12}=\frac{103}{18}\)
\(\Rightarrow M=-\frac{1159}{36}\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=x^2+11xy-y^2\)
\(N=3xy-4y^2-x^2+7xy-8y^2\)
\(N=-x^2+10xy-12y^2\)
a. \(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(\Rightarrow M=6x^2+9xy-y^2-5x^2+2xy\)
b. \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Rightarrow N=3xy-4y^2-x^2+7xy-8y^2\)