chiu roi

ban oi

tk nhe@@@@@@@@@

ai tk minh minh tk lai!!

chiu roi

bAN oi

tk nhe!!!!!!!!!!

ai tk minh minh tk lai!!

a) \(A=x^2+x+1\)

\(A=x^2+x+\frac{1}{4}+\frac{3}{4}\)

\(A=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Có: \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu = xảy ra khi: \(\left(x+\frac{1}{2}\right)^2=0\Rightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)

Vậy: \(Min_A=\frac{3}{4}\) tại \(x=-\frac{1}{2}\)

b) \(B=2+x-x^2\)

\(B=\frac{9}{4}-x^2+x-\frac{1}{4}\)

\(B=\frac{9}{4}-\left(x-\frac{1}{2}\right)^2\)

Có: \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\frac{9}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{9}{4}\)

Dấu = xảy ra khi: \(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

Vậy: \(Max_B=\frac{9}{4}\) tại \(x=\frac{1}{2}\)

c) \(C=x^2-4x+1\)

\(C=x^2-4x+4-3\)

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

Có: \(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2-3\ge-3\)

Dấu = xảy ra khi: \(\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

Vậy: \(Min_C=-3\) tại \(x=2\)

Mấy bài kia tương tự, riêng bài g

g) \(G=h\left(h+1\right)\left(h+2\right)\left(h+3\right)\)

\(G=\left(h^2+3h\right)\left(h^2+3h+2\right)\)

Đặt: \(t=h^2+3h+1\)

\(\Leftrightarrow\hept{\begin{cases}h^2+3h=t-1\\h^2+3h+2=t+1\end{cases}}\)

\(\Leftrightarrow\left(h^2+3h\right)\left(h^2+3h+2\right)=\left(t-1\right)\left(t+1\right)=t^2-1=\left(h^2+3h+1\right)^2-1\)

Có: \(\left(h^2+3h+1\right)^2\ge0\Rightarrow\left(h^2+3h+1\right)^2-1\ge-1\)

Dấu = xảy ra khi: \(\left(h^2+3h+1\right)^2=0\Rightarrow h^2+3h+1=0\Rightarrow\left(h+\frac{3}{2}\right)^2-\frac{5}{4}=0\Rightarrow\orbr{\begin{cases}h=-\frac{\sqrt{5}}{2}-\frac{3}{2}\\h=\frac{\sqrt{5}}{2}-\frac{3}{2}\end{cases}}\)

Vậy: \(Min_G=-1\) tại \(\orbr{\begin{cases}h=-\frac{\sqrt{5}}{2}-\frac{3}{2}\\h=\frac{\sqrt{5}}{2}-\frac{3}{2}\end{cases}}\) 

a) \(x^2-4x+5\)

= \(\left(x^2-2.2x+4\right)+1\)

= \(\left(x-2\right)^2+1\)

Ta co: \(\left(x-2\right)^2>=0\)

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

b) \(x^2-4xy+5y^2\)

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

= \(\left(x-2y\right)^2+y^2\)

Ta co: \(\left(x-2y\right)^2>=0\)

            \(y^2>=0\)

=> \(\left(x-2y\right)^2+y^2>=0\)

c) \(3-2x-x^2\)

= \(-\left(x^2+2x\right)+3\)

= \(-\left(x^2+2.1x+1-1\right)+3\)

= \(-\left(x+1\right)^2+4\)

= 

Hình như câu này sai đề ...

a) \(x^2-4x+5\)

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

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

b) \(x^2-4xy+5y^2\)

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

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

Dấu = xảy ra khi: \(x=y=0\)

c) \(-3-2x-x^2\)

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

\(=-2-\left(x+1\right)^2=-\left[2+\left(x+1\right)^2\right]< 0\)