Bạn viết thiếu đề bài nhé, phải là -x² + x - 1 nhỏ hơn hoặc bằng 0 với mọi x!! ^ . ^

Ta có: 

         -x² + x - 1 = - (x² - x + 1)

                        = - (x - 1)²   _{(hằng đẳng thức đấy bạn)}

Vì (x - 1)² \(\ge\)0 với mọi x => - (x - 1)² \(\le\)với mọi x.

 Dấu bằng xảy ra <=> x - 1 = 0 <=> x = 1.

_Kik nhé!! ^ ^

Không phải chỉ có nhỏ hơn thôi

Nhỏ hơn hoặc bằng mà..

a) Xét \(\Delta\) = b² - 4ac = (-m)² - 4(2m - 4)

= m² - 8m + 16 = ( m - 4 )²

Ta có: ( m - 4 )² \(\ge\) 0

=> Pt luôn có nghiệm

b) Vì phương trình luôn có nghiệm nên áp dụng định lí Ta- lét:

\(\left\{{}\begin{matrix}x_1+x_2=\frac{-b}{a}==m\\x_1x_2=2m-4\end{matrix}\right.\)
Xét phương trình: x₁² + x₂² - 9

= x₁² + x₂² + 2x₁x₂ - 2x₁x₂ - 9

= (x₁ + x₂)² - 2x₁x₂ - 9

= (-m)² - 2(2m - 4) - 9

= m² - 4m + 8 - 9

= m² - 4m - 1 = m² - 4m + 4 - 5

= (m - 2)² - 5

Xét (m - 2)² \(\ge\) 0

=> (m - 2)² - 5 \(\ge\) -5

Dấu " =" xảy ra khi m - 2 = 0

<=> m = 2

\(\Delta=m^2-8m+16=\left(m-4\right)^2\ge0\Rightarrow\) pt luôn có nghiệm

Khi đó theo Viet \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=2m-4\end{matrix}\right.\)

\(A=x_1^2+x_2^2-9=\left(x_1+x_2\right)^2-2x_1x_2-9\)

\(A=m^2-2\left(2m-4\right)-9\)

\(A=m^2-4m-1\)

\(A=\left(m-2\right)^2-5\ge-5\)

\(\Rightarrow A_{min}=-5\) khi \(m=-2\)

a) \(A=x^2+3x+4=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

\(minA=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(B=2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)

\(minB=\dfrac{7}{8}\Leftrightarrow x=\dfrac{1}{4}\)

c) \(C=5x^2+2x-3=5\left(x+\dfrac{1}{5}\right)^2-\dfrac{16}{5}\ge-\dfrac{16}{5}\)

\(minC=-\dfrac{16}{5}\Leftrightarrow x=-\dfrac{1}{5}\)

d) \(D=4x^2+4x-24=\left(2x+1\right)^2-25\ge-25\)

\(minD=-25\Leftrightarrow x=-\dfrac{1}{2}\)

e) \(E=x^2+6x-11=\left(x+3\right)^2-20\ge-20\)

\(minE=-20\Leftrightarrow x=-3\)

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

\(minG=-\dfrac{4}{3}\Leftrightarrow x=-2\)

a: Ta có: \(A=x^2+3x+4\)

\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

d: Ta có: \(D=4x^2+4x-24\)

\(=4x^2+4x+1-25\)

\(=\left(2x+1\right)^2-25\ge-25\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

e: ta có: \(E=x^2+6x-11\)

\(=x^2+6x+9-20\)

\(=\left(x+3\right)^2-20\ge-20\forall x\)

Dấu '=' xảy ra khi x=-3

vâng ạ

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: \(\text{Δ}=\left(m-1\right)^2-4\cdot1\cdot\left(-m\right)=\left(m+1\right)^2>=0\)

=>(5) luôn có nghiệm

b: \(x_1^2+x_2^2-2x_1x_2-\left(x_1\cdot x_2\right)^2=2m+1\)

=>\(\left(x_1+x_2\right)^2-4x_1x_2-\left(x_1\cdot x_2\right)^2=2m+1\)

=>\(\left(m-1\right)^2-4\cdot\left(-m\right)-\left(-m\right)^2=2m+1\)

=>\(m^2-2m+1+4m-m^2=2m+1\)

=>2m+1=2m+1(luôn đúng)

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 ...