a) Ta có: \(A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

c) Ta có: \(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu '=' xảy ra khi x(x+5)=0

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

d) Ta có: \(x^2+5y^2-2xy+4y+3\)

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

\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

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

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

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

a: A=(x-1)(x-3)(x²-4x+5)

\(=\left(x^2-4x+3\right)\left(x^2-4x+5\right)\)

\(=\left(x^2-4x\right)^2+8\left(x^2-4x\right)+15\)

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

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

Dấu = xảy ra khi x-2=0

=>x=2

b: \(B=x^2-2xy+2y^2-2y+1\)

\(=x^2-2xy+y^2+y^2-2y+1\)

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

Dấu = xảy ra khi x-y=0 và y-1=0

=>x=y=1

c: \(C=5+\left(1-x\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=-\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)+5\)

\(=-\left(x^2+5x-6\right)\left(x^2+5x+6\right)+5\)

\(=-\left[\left(x^2+5x\right)^2-36\right]+5\)

\(=-\left(x^2+5x\right)^2+36+5\)

\(=-\left(x^2+5x\right)^2+41< =41\)

Dấu = xảy ra khi \(x^2+5x=0\)

=>x(x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Bài 2 : 

a, \(x^2-4x+4+1=\left(x-2\right)^2+1\ge1\)

Dấu ''='' xảy ra khi x = 2 

b, Ta có \(\left(x+1\right)^2+10\ge10\Rightarrow\dfrac{-100}{\left(x+1\right)^2+10}\ge-\dfrac{100}{10}=-10\)

Dấu ''='' xảy ra khi x = -1 

 Bài 1 : 

a, Ta có \(A\left(x\right)=x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)

b, \(B\left(x\right)=x^2\left(2x+1\right)+\left(2x+1\right)=\left(x^2+1>0\right)\left(2x+1\right)=0\Leftrightarrow x=-\dfrac{1}{2}\)

c, \(C\left(x\right)=\left|2x-3\right|=\dfrac{1}{3}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}+3=\dfrac{10}{3}\\2x=-\dfrac{1}{3}+3=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

Ta có:B=(x-1/x+2)+(2-5x/4-x^2)

            =[(x-1)*(x-2)/(x+2)-(2-5x)/(x-2)*(x+2)]

            =(x^2+2x)/(x-2)*(x+2)

            =x/(x-2)

=> 5B=5x/(x-2)

=>A-5B = (x^3+2/x-2)-(5x/x-2)=x^3-5x+2/x-2=(x-2)*(x^2+2x-1)/(x-2)=x^2+2x-1=(x+1)^2-2

vì (x+1)^2>= 0

=> A-5B= (x+1)^2-2>= -2

Dấu `=' xảu ra<=> (x+1)^2 =0

=>x=-1

vậy GTNN của P=-2 <=> x=-1

Chi biet phan 5 thoi @

      Vi 3a=5b=12suy ra a=4 ;b=2,4  ta co p=a.b suy ra p=4×2.4=9.6 suy ra p>[=9.6 gtln=9.6

nguyen xuan duong sr minh viet nham dau bai 3a-5b=12