E = x^2 + x + 1

E = (x^2 + 2x.\(\frac{1}{2}\)+1/4 ) + 3/4 

E = (x+ 1/4 )^2 + 3/4 

Do ...... ( đến đây bn tự làm nha) 

H = ( x-1)^2 + ( x-7)^2 

H = x^2 - 2x + 1 + x^2 - 14x + 49 

H = 2x^2 - 16x + 50

H = [\(\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\frac{16}{2.\sqrt{2}}+32\)] + 18

H = ( \(\sqrt{2}x-\frac{16}{2\sqrt{2}}\))² + 18

.....

D = x^2 -20x + 101

D =( x^2 - 2.x.10 + 100) + 1

D = (x-10) ^2 + 1 

....

G = x^2 + 10x + 26 + y^2 + 2y + 2020

G = ( x^2 + 10x + 25) + (y^2+2y+1) + 2020

G = (x+5)^2 + ( y+1)^2 + 2020

....

Có gì ko hiểu hỏi mik

E=X²+2.X.1/2 + (1/2)²-(1/2)²+1
E=(X+1/2)²+3/4 >=3/4
vậy MIN E=3/4 khi x=-1/2
các câu khác phân tích tương tự

Dấu ''='' chế tự giải nha

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

\(H=\left(x-1\right)^2+\left(x-7\right)^2=x^2-2x+1+x^2-14x+49=2x^2-16x+50\)

\(=2\left(x^2-8x+16+9\right)=2\left(x-4\right)^2+18\ge18\)

\(D=x^2-20x+101=x^2-20x+100+1=\left(x-10\right)^2+1\ge1\)

\(\dfrac{2x+1}{5}-\dfrac{4x-2}{4}=10x\Leftrightarrow\dfrac{4\left(2x+1\right)}{20}-\dfrac{5\left(4x-2\right)}{20}=\dfrac{10x.20}{20}\)

\(\Leftrightarrow4\left(2x+1\right)-5\left(4x-2\right)=200x\)

\(\Leftrightarrow8x+4-20x+10=200x\)

\(\Leftrightarrow8x-20x-200x=-4-10\)

\(\Leftrightarrow-228x=-14\)\(\Leftrightarrow-\dfrac{14}{-228}=\dfrac{7}{114}\)

\(\dfrac{2x+1}{5}-\dfrac{4x-2}{4}=10x\)

\(\rightarrow\dfrac{8x+4}{20}-\dfrac{20x-10}{20}=10x\)

\(\rightarrow\dfrac{\left(8x+4\right)-\left(20x-10\right)}{20}=10x\)

\(\rightarrow\dfrac{8x+4-20x+10}{20}=10x\)

\(\rightarrow\dfrac{\left(8x-20x\right)+\left(4+10\right)}{20}=10x\)

\(\rightarrow\dfrac{-12x+14}{20}=10x\)

\(\rightarrow\left(-12x\right)+14=10x.20\)

\(\rightarrow\left(-12x\right)+14=200x\)

\(\rightarrow14=200x-\left(-12x\right)\)

\(\rightarrow14=200x+12x\)

\(\rightarrow14=\left(200+12\right)x\)

\(\rightarrow14=212x\)

\(\rightarrow14:212=x\)

\(\rightarrow\dfrac{14}{212}=x\)

\(\rightarrow\dfrac{7}{106}=x\)

Vậy ... ... ...

Ta có:

\(A=1993-x^2-3y^2+2xy-10x+14y\\ =2020-\left(x^2-2xy+y^2\right)-10\left(x-y\right)-25-\left(2y^2-4y+2\right)\\ =2020-\left(x-y-5\right)^2-2\left(y-1\right)^2\)

Với mọi x; y thì \(2020-\left(x-y-5\right)^2-2\left(y-1\right)^2\ge2020\)

Để A=2020 thì

\(\left\{{}\begin{matrix}x-y=5\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=1\end{matrix}\right.\)

Vậy...

\(\frac{100}{x^2-20x+25}=\frac{100}{\left(x^2-20x+100\right)-75}=\frac{100}{\left(x-10\right)^2-75}\le\frac{100}{-75}=-\frac{4}{3}\)