\(-5x^2+x-7=-5\left(x^2-\frac{1}{5}x+\frac{7}{5}\right)=-5\left(x^2-2\cdot\frac{1}{10}\cdot x+\frac{1}{100}-\frac{1}{100}+\frac{7}{5}\right)\)

\(=-5\left(x-\frac{1}{10}\right)^2+\frac{139}{20}\)

\(-5\left(x-\frac{1}{10}\right)^2+\frac{139}{20}\le\frac{139}{20}\)

GTLN của đa thức trên là 139/20

Ta có: \(2x^2-x+9\)

\(=2\left(x^2-\frac12x+\frac92\right)\)

\(=2\left(x^2-2\cdot x\cdot\frac14+\frac{1}{16}+\frac{71}{16}\right)=2\left(x-\frac14\right)^2+\frac{71}{8}\ge\frac{71}{8}\forall x\)

=>\(\frac{41}{2x^2-x+9}\le41:\frac{71}{8}=\frac{328}{71}\forall x\)

=>\(C=\frac{41}{2x^2-x+9}+2021\le\frac{328}{71}+2021=\frac{143819}{71}\forall x\)

Dấu '=' xảy ra khi \(x-\frac14=0\)

=>\(x=\frac14\)

B=2x²+10x-1

=2(x²+5x-\(\frac{1}{2}\))

=2(x²+2x.\(\frac{5}{2}\)\(+\frac{25}{4}\)\(-\frac{27}{4}\))

=2[(x²+\(\frac{5}{2}\))²-\(\frac{27}{4}\)]

=2(x+\(\frac{5}{2}\))²-\(\frac{27}{2}\)\(\ge\frac{-27}{2}\)(vì (x+5/2)²\(\ge0\))

Dấu = xảy ra khi :

x+\(\frac{5}{2}\)=0

<=>x=\(\frac{-5}{2}\)

Vậy GTNN của B là \(\frac{-27}{2}\)khi x= \(\frac{-5}{2}\)

Tính GTNN của Biểu thức 

2x²+40x-1

a: \(x^2-2x+3\)

\(=x^2-2x+1+2=\left(x-1\right)^2+2\ge2\forall x\)

=>\(A=\frac{37}{x^2-2x+3}\le\frac{37}{2}\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(x^2-5x+10\)

\(=x^2-5x+\frac{25}{4}+\frac{15}{4}\)

\(=\left(x-\frac52\right)^2+\frac{15}{4}\ge\frac{15}{4}\forall x\)

=>\(\frac{26}{x^2-5x+10}\le26:\frac{15}{4}=26\cdot\frac{4}{15}=\frac{104}{15}\forall x\)

=>\(B=-\frac{26}{x^2-5x+10}\ge-\frac{104}{15}\forall x\)

Dấu '=' xảy ra khi \(x-\frac52=0\)

=>\(x=\frac52\)

c: \(x^2-x+6\)

\(=x^2-x+\frac14+\frac{23}{4}\)

\(=\left(x-\frac12\right)^2+\frac{23}{4}\ge\frac{23}{4}\forall x\)

=>\(\frac{2023}{x^2-x+6}\le2023:\frac{23}{4}=2023\cdot\frac{4}{23}=\frac{8092}{23}\forall x\)

=>\(C=-\frac{2023}{x^2-x+6}\ge-\frac{8092}{23}\forall x\)

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

d: \(x^2+x+5\)

\(=x^2+x+\frac14+\frac{19}{4}\)

\(=\left(x+\frac12\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)

=>\(D=\frac{0.75}{x^2+x+5}\le\frac34:\frac{19}{4}=\frac{3}{19}\forall x\)

Dấu '=' xảy ra khi \(x+\frac12=0\)

=>\(x=-\frac12\)

e: \(2x^2-x+37=2\left(x^2-\frac12x+\frac{37}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\frac14+\frac{1}{16}+\frac{295}{16}\right)=2\left(x-\frac14\right)^2+\frac{295}{8}\ge\frac{295}{8}\forall x\)

=>\(\frac{13}{2x^2-x+37}\le13:\frac{295}{8}=\frac{104}{295}\forall x\)

Dấu '=' xảy ra khi \(x-\frac14=0\)

=>\(x=\frac14\)

f: \(3x^2-x+19\)

\(=3\left(x^2-\frac13x+\frac{19}{3}\right)\)

\(=3\left(x^2-2\cdot x\cdot\frac16+\frac{1}{36}+\frac{227}{36}\right)=3\left(x-\frac16\right)^2+\frac{227}{12}\ge\frac{227}{12}\forall x\)

=>\(\frac{61}{3x^2-x+19}\le61:\frac{227}{12}=61\cdot\frac{12}{227}=\frac{732}{227}\forall x\)

=>\(-\frac{61}{3x^2-x+19}\ge-\frac{732}{227}\forall x\)

Dấu '=' xảy ra khi x-1/6=0

=>x=1/6

A =  2x^2 +10x - 1

2A = 4x^2 + 20x -1

2A = (2x+5)² - 26

A = (2x+5)²/2 - 13

A có GTNN thì (2x+5)²/2 = 0

2x+ 5 =0

x = -5/2

Sai hết rồi 

My Nguyễn ơi,bạn truy cập vào đường link này để tìm câu hỏi tương tự của câu a/Bài 1 nhé

https://vn.answers.yahoo.com/question/index?qid=20110206184834AAokV5m&sort=N

Ko biết đợi đứa khác đê

a: \(A=x^2-2x+19\)

\(=x^2-2x+1+18\)

\(=\left(x-1\right)^2+18\ge18\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(B=-x^2-5x+20\)

\(=-\left(x^2+5x-20\right)\)

\(=-\left(x^2+5x+\frac{25}{4}-\frac{25}{4}-20\right)\)

\(=-\left(x+\frac52\right)^2+\frac{105}{4}\le\frac{105}{4}\forall x\)

Dấu '=' xảy ra khi \(x+\frac52=0\)

=>\(x=-\frac52\)