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Bài 1:
\(M=6x^2+xyz+2xy+3-y^2+3xyz-5x^2+7xy-9\)
\(=x^2+4xyz+9xy-y^2-6\)
Bài 2:
a: Sửa đề: \(x^2+2x+3\)
Đặt \(x^2+2x+3=0\)
\(\Delta=2^2-4\cdot1\cdot3=4-12=-8< 0\)
Do đó: Phương trình vô nghiệm
b: Đặt \(x^2+4x+6=0\)
\(\Leftrightarrow x^2+4x+4+2=0\)
\(\Leftrightarrow\left(x+2\right)^2+2=0\)(vô lý)
a) Ta có: \(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy\)
\(\Leftrightarrow M=x^2+11xy-y^2\)
Vậy: \(M=x^2+11xy-y^2\)
b) Ta có: \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2\)
\(\Leftrightarrow N=-x^2+10xy-12y^2\)
Vậy: \(N=-x^2+10xy-12y^2\)
a, (6x2+9xy-y2) - ( 5x2-2xy)=M
=> M= (6x2+9xy-y2) - ( 5x2-2xy)
=> M= 6x2+9xy-y2 - 5x2+2xy
=> M=(6x2- 5x2)+(9xy+2xy)-y2
=>M= 1x2 + 11xy - y2
Vậy M= 1x2 + 11xy - y2
b, N= (3xy-4y2) - (x2-7xy+8y2)
=> N= 3xy-4y2 - x2+7xy-8y2
=> N= (3xy+7xy)-(4y2+8y2)-x2
=> N= 10xy - 12y2 -x2
Vậy N= 10xy - 12y2 -x2
a: Ta có: \(M+5x^2-2xy=6x^2+9xy-y^2\)
\(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy\)
\(\Leftrightarrow M=x^2+11xy-y^2\)
b: Ta có: \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2\)
\(\Leftrightarrow N=-x^2+10xy-12y^2\)
Ta có : \(\left(2x-5\right)^{2012}\ge0\forall x\)
\(\left(3y+4\right)^{2014}\ge0\forall y\)
\(\rightarrow\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}\ge0\forall x,y\)
Theo bài : \(\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}\le0\)
\(\rightarrow\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}=0\)
\(\rightarrow\left(2x-5\right)^{2012}=0,\left(3y+4\right)^{2014}=0\)
\(\rightarrow2x-5=0,3y+4=0\)
\(\rightarrow x=\frac{5}{2};y=\frac{-4}{3}\)
Tự tìm M nhé bạn
1, 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
Ta có:
M + 5 x 2 − 2 x y = 6 x 2 + 10 x y − y 2 ⇒ M = 6 x 2 + 10 x y − y 2 − 5 x 2 − 2 x y ⇒ M = 6 x 2 + 10 x y − y 2 − 5 x 2 + 2 x y ⇒ M = 6 x 2 − 5 x 2 + ( 10 x y + 2 x y ) − y 2 ⇒ M = x 2 + 12 x y − y 2
Chọn đáp án A
\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\\ \Leftrightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\\ \Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\\ \Leftrightarrow M=\dfrac{25}{4}-11\cdot\dfrac{4}{3}\cdot\dfrac{5}{2}-\dfrac{16}{9}=\dfrac{25}{4}-\dfrac{110}{3}-\dfrac{16}{9}=-\dfrac{1159}{36}\)
Vũ Minh Tuấn,Băng Băng 2k6
1)
\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(M+5x^2-2xy=6x^2+9xy-y^2\)
\(M=\left(6x^2+9xy-y^2\right)-\left(5x^2+2xy\right)\)
\(M=6x^2+9xy-y^2-5x^2-2xy\)
\(M=\left(6x^2-5x^2\right)+\left(9xy-2xy\right)-y^2\)
\(M=x^2+7xy-y^2.\)
Chúc em học tốt!