Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\left(\dfrac{1+2\left(x-y\right)\left(2x-2y+1\right)-2x+2y-1}{2x-2y+1}\right):\dfrac{\left(2x-2y\right)\left(2x-2y+1\right)-4x^2+8xy-4y^2}{2x-2y+1}\)
\(=\dfrac{1+\left(2x-2y\right)^2+2x-2y-2x+2y-1}{2x-2y+1}\cdot\dfrac{2x-2y+1}{\left(2x-2y\right)^2+2x-2y-4x^2+8xy-4y^2}\)
\(=\dfrac{\left(2x-2y\right)^2}{4x^2-8xy+4y^2+2x-2y-4x^2+8xy-4y^2}=2x-2y\)
=2(x-y) luôn là số chẵn
Bài 2:
a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)
\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)
\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)
b: \(B=-x^2-y^2+xy+2x+2y\)
\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)
\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)
\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)
\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)
Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0
=>y=2 và 2x=y+2=2+2=4
=>x=2 và y=2
Bài 1:
d: \(D=2x^2+3y^2+4xy-8x-2y\)
\(=2x^2+4xy+2y^2-8x-8y+y^2+6y\)
\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)
\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)
f: \(F=2x^2+8xy+11y^2-4x-2y+6\)
\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)
\(=2\left(x+2y\right)^2-4\left(x+2y\right)+2+3y^2+6y+3+1\)
\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi y+1=0 và x+2y-1=0
=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3
h: \(H=x^2+y^2-xy-x+y+1\)
\(=\frac14\left(4x^2+4y^2-4xy-4x+4y+4\right)\)
\(=\frac14\left(4x^2-4xy+y^2-4x+2y+3y^2+2y+4\right)\)
\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)
\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)
Bài 2:
a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)
\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)
\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)
b: \(B=-x^2-y^2+xy+2x+2y\)
\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)
\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)
\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)
\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)
Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0
=>y=2 và 2x=y+2=2+2=4
=>x=2 và y=2
Bài 1:
d: \(D=2x^2+3y^2+4xy-8x-2y\)
\(=2x^2+4xy+2y^2-8x-8y+y^2+6y\)
\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)
\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)
f: \(F=2x^2+8xy+11y^2-4x-2y+6\)
\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)
\(=2\left(x+2y\right)^2-4\left(x+2y\right)+2+3y^2+6y+3+1\)
\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi y+1=0 và x+2y-1=0
=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3
h: \(H=x^2+y^2-xy-x+y+1\)
\(=\frac14\left(4x^2+4y^2-4xy-4x+4y+4\right)\)
\(=\frac14\left(4x^2-4xy+y^2-4x+2y+3y^2+2y+4\right)\)
\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)
\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)
Ta có:
D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18
D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18
D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1
D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1
Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3
Hay x = 5 , y = -3
Đc chx bạn
a: \(=\left(\dfrac{x}{y\left(x-y\right)}-\dfrac{2x-y}{x\left(x-y\right)}\right):\dfrac{x+y}{xy}\)
\(=\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)}\cdot\dfrac{xy}{x+y}\)
\(=\dfrac{\left(x-y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x-y}{x+y}\)
b: \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x-y}{2y}\)
\(=\dfrac{4xy+4y^2}{2\left(x+y\right)}\cdot\dfrac{1}{2y}=\dfrac{4y\left(x+y\right)}{4y\left(x+y\right)}=1\)
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)