ví dụ 1 bài 2x-x²

bài giải: -(x²+2x)

=-(x²-2x)

=-(x²-2x.5/2 +25/4 - 25/4)

=-(x-5/2)-25/4<25/4

=25/4+(x-5/2)<25

=>x=5/2

Tìm GTLN của biểu thức: a,A= (21|4x+6| + 33)/(3|4x+6|+5)

Ta có:

 \(A=\frac{4x+5}{x^2+2x+6}=\frac{x^2+2x+6-x^2-2x-6+4x+5}{x^2+2x+6}\)

\(=\frac{\left(x^2+2x+6\right)-x^2+2x-1}{x^2+2x+6}=1-\frac{\left(x-1\right)^2}{x^2+2x+6}\le1\)

=> max A = 1 tại x = 1

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

\(=-\frac{4}{5}+\frac{4x^2+28x+49}{5\left(x^2+2x+6\right)}=-\frac{4}{5}+\frac{\left(2x+7\right)^2}{5\left(x^2+2x+6\right)}\ge-\frac{4}{5}\)

=> min A = -4/5 <=> 2x + 7 = 0 <=> x = -7/2

Vậy...

a: \(A=-4x^2+4x-1\)

\(=-\left(4x^2-4x+1\right)\)

\(=-\left(2x-1\right)^2\le0\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

b: \(B=-x^2+5x\)

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

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

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

\(=-\left(2x-1\right)^2\le0\)

\(maxA=0\Leftrightarrow x=\dfrac{1}{2}\)

b) \(B=-x^2+5x=-\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{25}{4}\)

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

\(maxB=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)

c) \(C=-3x^2-9x+6=-3\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{51}{4}\)

\(=-3\left(x+\dfrac{3}{2}\right)^2+\dfrac{51}{4}\le\dfrac{51}{4}\)

\(maxC=\dfrac{51}{4}\Leftrightarrow x=-\dfrac{3}{2}\)

Ta có : C = 4x² + 25y² - 4x + 30y 

=> C = 4x² - 4x + 25y² + 30y

=> C = (4x² - 4x + 1) + (25y² + 30y + 9) - 10

=> C = (2x - 1)² + (5y + 3)² - 10 

Mà \(\left(2x-1\right)^2;\left(5y+3\right)^2\ge0\forall x\)

Nên C =  (2x - 1)² + (5y + 3)² - 10 \(\ge-10\forall x\)

Vậy giá trị nhỏ nhất của C là -10 tại x = \(\frac{1}{2}\) và y = \(-\frac{3}{5}\)

Ta có:

4x^2+25y^2-4x+30y

=(4x^2-4x+1)+(25y^2+30y+9)-10

=(2x-1)^2+(5y+3)^2-10

Vì (2x-1)^2>=0 với mọi x; (5y+3)^2>=0 với mọi y

=>(2x-1)^2+(5y+3)^2>=0 với mọi x,y

=>(2x-1)^2+(5y+3)^2-10>=-10 với mọi x,y

Dấu "=" xảy ra <=>2x-1=0 và 5y+3=0

<=>x=1/2 và y=-3/5

\(C=4x^2+25y^2-4x+30y\)

\(=4x^2-4x+1-1+25y^2+30y+9-9\)

\(=\left(2x-1\right)^2+\left(5y+3\right)^2-10\)

Dễ thấy: \(\hept{\begin{cases}\left(2x-1\right)^2\ge0\\\left(5y+3\right)^2\ge0\end{cases}}\)

\(\Rightarrow\left(2x-1\right)^2+\left(5y+3\right)^2\ge0\)

\(\Rightarrow\left(2x-1\right)^2+\left(5y+3\right)^2-10\ge-10\)

Xảy ra khi \(\hept{\begin{cases}\left(2x-1\right)^2=0\\\left(5y+3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{5}\end{cases}}\)

sai  đề hay sao ý phải là GTNN hay sao ý

a:Ta có: \(A=-4x^2+x-1\)

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

\(=-4\left(x^2-2\cdot x\cdot\dfrac{1}{8}+\dfrac{1}{64}+\dfrac{63}{64}\right)\)

\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{63}{16}\le-\dfrac{63}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{8}\)

b: Ta có: \(B=-3x^2+5x+6\)

\(=-3\left(x^2-\dfrac{5}{3}x-2\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{97}{36}\right)\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{6}\)

c: Ta có: \(C=-x^2+3x+4\)

\(=-\left(x^2-3x-4\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)