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 \(y=-3\)

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 ...

3x² - 6x + 4

= 3( x² - 2x + 1) + 1

= 3( x - 1)² + 1

Do : 3( x - 1)² lớn hơn hoặc bằng 0 với mọi x thuộc R

=> 3( x - 1)² + 1 > 0 với mọi x thuộc R

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

Sửa đề: \(A=3x^2-6x+4=3\left(x^2-2x+\dfrac{4}{3}\right)\)

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

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

\(A=3\left(x-1\right)^2+1>0\left(đpcm\right)\)

(8x-3)(3x+2)-(4x+7)(x+4) = (2x+1)(5x-1)-33

(24x²-9x+16x-6)-(4x²+7x+16x+28) = (10x²+5x-2x-1)-33

24x²+7x-6-4x²-23x-28 = 10x²+3x-1-33

20x²-16x-34 = 10x²+3x-34

<=> 20x²-16x = 10x²+3x

2x²-19x=0

2x(x-19)=0

=>\(\left[{}\begin{matrix}2x=0\Rightarrow x=0\\x-19=0\Rightarrow x=19\end{matrix}\right.\)

Không chắc lắm :)