\(\frac{3x^2+10x+11}{x^2+2x+3}\)

\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)

b:

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

\(=-\left(25x^2-10x+1\right)-10\)

\(=-\left(5x-1\right)^2-10< =-10\)

Dấu = xảy ra khi x=1/5

\(E=-9x^2-6x-1+20\)

\(=-\left(9x^2+6x+1\right)+20\)

\(=-\left(3x+1\right)^2+20< =20\)

Dấu = xảy ra khi x=-1/3

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

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

Dấu = xảy ra khi x=1

\(1.\)

\(-17-\left(x-3\right)^2\)

Ta có: \(\left(x-3\right)^2\ge0\)với \(\forall x\)

\(\Leftrightarrow-\left(x-3\right)^2\le0\)với \(\forall x\)

\(\Leftrightarrow17-\left(x-3\right)^2\le17\)với \(\forall x\)

Dấu '' = '' xảy ra khi: 

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Vậy \(Max=-17\)khi \(x=3\)

\(2.\)

\(A=x\left(x+1\right)+\frac{3}{2}\)

\(A=x^2+x+\frac{3}{2}\)

\(A=\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\)

\(\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\ge\frac{5}{4}\)với \(\forall x\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\ge\frac{5}{4}\)với \(\forall x\)

Vậy \(Max=\frac{5}{4}\)khi \(x=\frac{-1}{2}\)

\(5.\)

\(x^2-48x+65\)

\(=\left(x-24\right)^2\ge0\)với \(\forall x\)

\(\left(x-24\right)^2\ge0\)với \(\forall x\)

\(\Leftrightarrow\left(x-24\right)^2-511\ge-511\)với \(\forall x\)

Vậy \(Max=-511\)khi \(x=24\)

B=\(x^2+3x+7\)

=>B= \(x^2+2\times\frac{3}{2}x+\frac{9}{4}+\frac{19}{4}\)

=>B=\(\left(x+\frac{3}{2}\right)^2+\frac{19}{4}\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0\)   (Với mọi x)

=>\(\left(x+\frac{3}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\)   (Với mọi x )

Dấu "='' xảy ra  <=> \(x+\frac{3}{2}=0=>x=-\frac{3}{2}\)

Vậy min B bằng 19/4 <=>x=-3/2

Phần b thì mk làm đc n phần a hình như sai đề pn ạ !!!
 

\(A=x^2+10x-37=x^2+2x.5+25-62\)\(=\left(x+5\right)^2-62\)

Vì \(\left(x+5\right)^2\ge0\Rightarrow A\ge-62\)

Dấu ''='' xảy ra \(\Leftrightarrow x=-5\)

Vậy ...

\(C=6x-x^2+3=-\left(x^2-2x.3+9\right)+12=\)\(-\left(x-3\right)^2+12\)

Vì \(-\left(x-3\right)^2\le0\Rightarrow C\le12\)

Dấu ''='' xảy ra khi x =3

Vậy ....

Câu b,d làm tương tự nhé :)

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

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{7}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{7}{16}>=\dfrac{7}{16}\)

Dấu '=' xảy ra khi x=3/8

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

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

\(=-\left(x-1\right)^2-2< =-2\)

Dấu '=' xảy ra khi x=1

1. a, => -12x+60+21-7x = 5

=> 81 - 19x = 5

=> 19x = 81 - 5 = 76

=> x = 76 : 19 = 4

Tk mk nha