a: C=-2x^4+3x^2y-2xy+y^2+7

Bậc là 4

b: B=5x^4-3x^2y+2xy+y^2

D=-2x^4+3x^2y-2xy+y^2+7+5x^4-3x^2y+2xy+y^2

=3x^4+2y^2

E=-2x^4+3x^2y-2xy+y^2+7-5x^4+3x^2y-2xy-y^2

=-7x^4+6x^2y-4xy+7

\(A=4\cdot3\left(-2\right)-2\left(3+2\right)=-24-10=-34\\ B=\left(x+y\right)^2-3\left(x+y\right)=\left(x+y\right)\left(x+y-3\right)=\left(x+y\right)\left(2+1-3\right)=0\)

\(a,=3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)

Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)

\(b,=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(c,=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)

Bài 1:

Ta thấy: $(x+\frac{1}{2})^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow (x+\frac{1}{2})^2+\frac{5}{4}\geq \frac{5}{4}$

Vậy gtnn của biểu thức là $\frac{5}{4}$

Giá trị này đạt tại $x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}$

Bài 2:

$x+y-3=0\Rightarrow x+y=3$
\(M=x^2(x+y)-(x+y)x^2-y(x+y)+4y+x+2019\)

\(=-3y+4y+x+2019=x+y+2019=3+2019=2022\)

\(M=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)

\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-xy-y^2+2y+y+x-2+2019\)

\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(y+x-2\right)+2019\)

\(\Rightarrow M=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)+2019\)

\(\Rightarrow M=\left(x^2-y+1\right)\left(x+y-2\right)+2019\)

\(\Rightarrow M=\left(x^2-y+1\right).0+2019\)

\(\Rightarrow M=0+2019\)

\(\Rightarrow M=2019\)

giúp mik với các bạn

M = x³ + x²y - 2x² - xy - y² + 3y + x + 2017

M = (x³ + x²y - 2x²) - (xy + y² - 2y) + (x + y - 2) + 2019

M = x². (x + y - 2) - y(x + y - 2) + (x + y - 2) + 2019 = 2019

\(M = x^3 + x^2y - 2x^2 - xy - y^2 + 3y + x + 2017.\)

\(M=(x^3+x^2y-2x^2)-(xy-y^2+2y)+(x+y-2)+2019\)

\(M=x^2.(x+y-2)-y.(x-y+2)+(x+y-2)+2019\)

\(M=x^2.0-y.0+0+2019\)

\(M=0-0+0+2019\)

\(M=2019\)

$M={}^{}+{}^{}y-2{}^{}-xy-{}^{}+3y+x+2017.$

$M=({}^{}+{}^{}y-2{}^{})-(xy-{}^{}+2y)+(x+y-2)+2019$

$M={}^{}.(x+y-2)-y.(x-y+2)+(x+y-2)+2019$

$M={}^{}.0-y.0+0+2019$

\(1,\\ a,\dfrac{x^2}{x+1}+\dfrac{x}{x+1}=\dfrac{x^2+x}{x+1}=\dfrac{x\left(x+1\right)}{x+1}=x\)

\(b,\left(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}\right):\dfrac{x+y}{2x}=\left(\dfrac{4xy}{2\left(x-y\right)\left(x+y\right)}+\dfrac{\left(x-y\right)^2}{2\left(x-y\right)\left(x+y\right)}\right).\dfrac{2x}{x+y}=\dfrac{4xy+x^2-2xy+y^2}{2\left(x-y\right)\left(x+y\right)}.\dfrac{2x}{x+y}=\dfrac{2x\left(x^2+2xy+y^2\right)}{2\left(x-y\right)\left(x+y\right)^2}=\dfrac{2x\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)^2}=\dfrac{x}{x-y}\)

Câu 2: 

a: ĐKXĐ: \(x\ne1\)

\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)

\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)

\(-18x+13=0\)

\(x=\dfrac{13}{18}\)

Vậy \(S=\left\{\dfrac{13}{18}\right\}\)

\(b.\left(x-1\right)^3-125=0\)

\(\left(x-1\right)^3=125\)

\(x-1=5\)

\(x=6\)

Vậy \(S=\left\{6\right\}\)

\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)

\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy \(S=\left\{1;-2\right\}\)

\(d.x^2-4x+4+x^2-2xy+y^2=0\)

\(\left(x-2\right)^2+\left(x-y\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

Vậy \(S=\left\{2;2\right\}\)