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\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=x^2+11xy-y^2\)
\(N=3xy-4y^2-x^2+7xy-8y^2\)
\(N=-x^2+10xy-12y^2\)
a. \(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(\Rightarrow M=6x^2+9xy-y^2-5x^2+2xy\)
b. \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Rightarrow N=3xy-4y^2-x^2+7xy-8y^2\)
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) \(N=\left(3xy-4y^2\right)-\left(x^2-7xy+8y^2\right)\)
\(N=-x^2-12y^2+10xy\)
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:
\(A+B=4x^4-5xy+5y^2+3x^2+2xy-y=4x^4+3x^2-3xy+5y^2-y\)
\(A-B=4x^4-5xy+5y^2-3x^2-2xy+y=4x^4-3x^2+5y^2-7xy+y\)
\(B-A=-\left(A-B\right)=-4x^4+3x^2-5y^2+7xy-y\)
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!
\(a.M+(5x^2-2xy)=6x^2+9xy-y^2
\)
\(M=(6x^2+9xy-y^2)-(5x^2-2xy)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=(6x^2-5x^2)+(9xy+2xy)-y^2\)
\(M=x^2+11xy-y^2\)
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 \(M=-x^2-2xy^3\)
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\(^2\)
a)(25u2v-13uv2+u3)-M=11u2v-2u3
=>M=25u2v-13uv2+u3-11u2v-2u3
=(25u2v-11u2v)-(2u3-u3)-13uv2
=14u2v-u3-13uv2
=u(14uv-u2-13v2)
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:
\(M+5\cdot\left(\frac{5}{2}\right)^2-2\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)=6\cdot\left(\frac{5}{2}\right)^2+9\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(\Leftrightarrow M+\frac{455}{12}=\frac{103}{18}\)
\(\Rightarrow M=-\frac{1159}{36}\)
Bài 2:
\(M\left(3\right)=3^2-4\cdot3+3=0\)
=>x=3 là nghiệm của M(x)
\(M\left(-1\right)=\left(-1\right)^2-4\cdot\left(-1\right)+3=1+3+4=8\)
=>x=-1 không là nghiệm của M(x)
a/M+(5x2-2xy)=6x2+9xy-y2
=>M =(6x2+9xy-y2)-(5x2-2xy)=6x2+9xy-y2-5x2+2xy
=6x2-5x2+9xy+2xy-y2=x2+11xy-y2
b/M-(4xy-3y2)=x2-7xy+8y2
=>M=(x2-7xy+8y2)+(4xy-3y2)
=x2-7xy+8y2+4xy-3y2
=8y2-3y2-7xy+4xy+x2=5y2-3xy+x2