1, B=\(\frac{3}{4x^2-4x+5}\)

=\(\frac{3}{\left(4x^2-2.2x+4\right)+5-4}\)

=\(\frac{3}{\left(2x-2\right)^2+1}\le\frac{3}{1}=3\)

Để B=3 thì : (2x-2)²=0

\(\Leftrightarrow2x-2=0\)

\(\Leftrightarrow x=1\)

Vậy Max B =3 \(\Leftrightarrow x=1\)

\(A=-2x^2+5x-8\)

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

\(A=-2\left(x^2-2\cdot x\cdot\frac{5}{4}+\frac{25}{16}+\frac{39}{16}\right)\)

\(A=-2\left[\left(x-\frac{5}{4}\right)^2+\frac{39}{16}\right]\)

\(A=-2\left(x-\frac{5}{4}\right)^2-\frac{39}{6}\le\frac{-39}{6}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{5}{4}\)

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

\(2B=-2x^2-2y^2+2xy-4x-4y\)

\(2B=-\left(2x^2+2y^2-2xy+4x+4y\right)\)

\(2B=-\left(x^2-2xy+y^2+x^2+4x+4+y^2+4y+4-8\right)\)

\(2B=-\left[\left(x-y\right)^2+\left(x+2\right)^2+\left(y+2\right)^2-8\right]\)

\(B=-\frac{\left(x-y\right)^2+\left(x+2\right)^2+\left(y+2\right)^2}{2}+4\le4\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=-2\)

\(C=\frac{3}{4x^2-4x+5}=\frac{3}{\left(2x-1\right)^2+4}\le\frac{3}{4}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)

\(D=\frac{x^2-6x+14}{x^2-6x+12}=\frac{x^2-6x+12+2}{x^2-6x+12}\)

\(=1+\frac{2}{\left(x-3\right)^2+3}\le1+\frac{2}{3}=\frac{5}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=3\)

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

a) Theo mình thì chỉ min thôi nhé!

\(A=\frac{8x^2-1}{4x^2+1}+1+11=\frac{12x^2}{4x^2+1}+11\ge11\)

b)Bạn rút gọn lại giùm mìn, lười quy đồng lắm:(

a.\(\frac{3x-1}{3x+1}+\frac{x-3}{x+3}=2\)

\(\frac{\left(3x-1\right)\left(x+3\right)+\left(3x+1\right)\left(x-3\right)}{\left(3x+1\right)\left(x+3\right)}=\frac{3x^2+8x-3+3x^2-8x-3}{\left(3x+1\right)\left(x+3\right)}=\frac{6x^2-6}{\left(3x+1\right)\left(x+3\right)}=2\)

\(6x^2-6=2\left(3x^2+10x+3\right)\)

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

-20x-12=0

x=\(\frac{-3}{5}\)

Có: \(C=\frac{1}{\sqrt{x^2-4x+5}}\)

\(\Leftrightarrow C=\frac{1}{\sqrt{\left(x-2\right)^2+1}}\)\(\le1\)

Vậy C_min=1 \(\Leftrightarrow x=2\)

Có: \(B=5-\sqrt{x^2-6x+14}\)

\(\Leftrightarrow B=5-\sqrt{\left(x-3\right)^2+5}\) \(\le5-\sqrt{5}\)

Vậy \(B_{min}=5-\sqrt{5}\Leftrightarrow x=3\)

a) \(M=-\left(x^2+6x+9\right)+23=23-\left(x-3\right)^2\le23\Rightarrow MaxM=23\Leftrightarrow x=3\)

b) \(N=\left(9x^2+12x+4\right)+16=\left(3x+2\right)^2+16\ge16\Leftrightarrow MinN=16\Leftrightarrow x=-\frac{2}{3}\)

c) \(P=-\left(x^2-4x+4\right)-\left(4y^2+4y+1\right)+8=8-\left(x-2\right)^2-\left(2y+1\right)^2\le8\Rightarrow MaxP=8\Leftrightarrow x=2;y=-\frac{1}{2}\)

câu b k tìm đc GTLN chỉ tìm được GTNN thôi nha

câu b mình ghi đề sai,phải là N=-9x^2+12x+20 nha bạn

\(N=-9x^2+12x+20\)

\(=-\left[\left(3x\right)^2-2.3x.2+4+16\right]\)

\(=-\left[\left(3x-2\right)^2+16\right]\)

\(=-\left(3x-2\right)^2-16\le-16\)

Vậy \(N_{max}=-16\Leftrightarrow3x-2=0\Leftrightarrow x=\frac{2}{3}\)

Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI

1/

a, \(A=4x^2-4x+5=4x^2-4x+1+4=\left(2x-1\right)^2+4\ge4\)

Dấu "=" xảy ra khi x=1/2

Vậy Amin=4 khi x=1/2

b, \(B=3x^2+6x-1=3\left(x^2+2x+1\right)-4=3\left(x+1\right)^2-4\ge-4\)

Dấu "=" xảy ra khi x=-1

Vậy Bmin = -4 khi x=-1

2/

a, \(A=10+6x-x^2=-\left(x^2-6x+9\right)+19=-\left(x-3\right)^2+19\le19\)

Dấu "=" xảy ra khi x=3

Vậy Amax = 19 khi x=3

b, \(B=7-5x-2x^2=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{31}{8}=-2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}\le\frac{31}{8}\)

Dấu "=" xảy ra khi x=5/4

Vậy Bmax = 31/8 khi x=5/4