Bài 1 :

Câu a : \(A=x^2-3x+5=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\)

Câu b : \(A=x^2-3x+5=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

Vậy \(GTNN\) của \(A\) là \(\dfrac{11}{4}\) . Dấu \("="\) xảy ra khi \(\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)

Bài 2 :

Câu a : \(x^2-6x+y^2-4y+13=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(y^2-4y+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y-2\right)^2=0\)

Do : \(\left(x-3\right)^2\ge0\) and \(\left(y-2\right)^2\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Vậy \(x=3\) and \(y=2\)

Câu b : \(4x^2-4x+y^2+6y+10=0\)

\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(y+3\right)^2=0\)

Because the : \(\left(2x-1\right)^2\ge0\) and \(\left(y+3\right)^2\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(2x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{2}\) và \(y=-3\)

a) Ta có:

\(x^2+2xy+y^2+1\)

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

Vì \(\left(x+y\right)^2\ge0\) với mọi x và y

\(\Rightarrow\left(x+y\right)^2+1\ge1\)

\(\Rightarrow\left(x+y\right)^2+1>0\) với mọi x

b) Ta có:

\(x^2-x+1\)

\(=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\)

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

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi x

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

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) với mọi x

chết r đăng nhầm

Bài 1:

a) \(x^2-x+1\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0;\forall x\)

b) \(25x^2+10x+2\)

\(=25x^2+10x+1+1\)

\(=\left(5x+1\right)^2+1\ge1>0;\forall x\)

c) \(3x^2+2x+14\)

\(=3x^2+2x+\dfrac{1}{3}+\dfrac{41}{3}\)

\(=\left(\sqrt{3}x+\dfrac{\sqrt{3}}{3}\right)^2+\dfrac{41}{3}\ge\dfrac{41}{3}>0;\forall x\)

d) \(2x^2+y^2-2xy-2x+2\)

\(=x^2+y^2-2xy-2x+x^2+1+1\)

\(=\left(x-y\right)^2+\left(x-1\right)^2+1\ge1>0;\forall x\)

Vậy ...

Đáp án D

  T u → M = M ' => 3x’ + ( y’ – 3) – 2 = 0   3x’ + y’ – 5 = 0

 Phương trìnhđường thẳng cần tìm: 3x + y – 5 = 0

Bài 1:

Ta có:\(x^2+xy+y^2+1\)

\(=x^2+\dfrac{1}{2}xy+\dfrac{1}{2}xy+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2+1\)

\(=\left(x^2+\dfrac{1}{2}xy\right)+\left(\dfrac{1}{2}xy+\dfrac{1}{4}y^2\right)+\dfrac{3}{4}y^2+1\)

\(=x.\left(x+\dfrac{1}{2}y\right)+\dfrac{1}{2}y.\left(x+\dfrac{1}{2}y\right)+\dfrac{3}{4}y^2+1\)

\(=\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+1\)

Với mọi giá trị của \(x\in R\) ta có:

\(\left(x+\dfrac{1}{2}y\right)^2\ge0;\dfrac{3}{4}y^2\ge0\)

\(\Rightarrow\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\ge0\Rightarrow\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+1\ge1>0\)

Hay \(x^2+xy+y^2+1>0\) (đpcm)

Chúc bạn học tốt!!!

Để thỏa mãn BPT thì:

\(\left\{{}\begin{matrix}m-1>0\\\Delta< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>1\\\left[{}\begin{matrix}m>\sqrt{2}\\m< -\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)

=> \(m>\sqrt{2}\)

a/ \(x^2+xy+y^2+1=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}+1=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1>0\)

b/ \(x^2+5y^2+2x-4xy-10y+14\)

\(=\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+1+\left(y^2-6y+9\right)+4\)

\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y-3\right)^2+4\)

\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\)