a.

\(A=\left(x^4+y^2+1-2x^2y+2x^2-2y\right)+2\left(y^2-2y+1\right)+2026\)

\(A=\left(x^2-y+1\right)^2+2\left(y-1\right)^2+2026\ge2026\)

\(A_{min}=2026\) khi \(\left(x;y\right)=\left(0;1\right)\)

b.

Đặt \(x-1=t\Rightarrow x=t+1\)

\(\Rightarrow A=\dfrac{3\left(t+1\right)^2-8\left(t+1\right)+6}{t^2}=\dfrac{3t^2-2t+1}{t^2}=\dfrac{1}{t^2}-\dfrac{2}{t}+3=\left(\dfrac{1}{t}-1\right)^2+2\ge2\)

\(A_{min}=2\) khi \(t=1\Rightarrow x=2\)

\(A=\dfrac{3x^2-8x+6}{x^2-2x+1}=\dfrac{3x^2-8x+6}{\left(x-1\right)^2}=\dfrac{2\left(x-1\right)^2+\left(x-2\right)^2}{\left(x-1\right)^2}=2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)

Dấu \("="\Leftrightarrow x=2\)

A)\(5xyz.4x^2y^2.\left(-2x^3y\right)=\left(5.4.\left(-2\right)\right).\left(xx^2x^3\right).\left(yy^2y\right)=\left(-40\right)x^6y^4\)

- BẬC : 10

- HỆ SỐ: -40 

B) \(-xy.\left(\frac{1}{2}x^3y^4\right).\left(\frac{-4}{7}x^2y^5\right)=\left(\frac{1}{2}.\frac{-4}{7}.\left(-1\right)\right).\left(xx^3x^2\right).\left(y^4y^5y\right)=\frac{2}{7}x^6y^{10}\)

- BẬC : 16

- HỆ SỐ: 2/7

C) \(\frac{5}{3}x^2y^4.\left(\frac{-6}{5}xy^3\right).\left(-xy\right)=\left(\frac{5}{3}.\frac{-6}{5}.\left(-1\right)\right).\left(x^2xx\right).\left(y^4y^3y\right)=2x^4y^8\)

- BẬC : 12

- HỆ SỐ : 2

D) \(\left(\frac{-1}{3}x^2y^5\right).\left(\frac{3}{4}xy\right).5x=\left(\frac{-1}{3}.\frac{3}{4}.5\right).\left(x^2xx\right).\left(y^5y\right)=\frac{-5}{4}x^4y^6\)

- BẬC : 10

- HỆ SỐ : -5 /4

CHÚC BN HỌC TỐT!!

cảm on

Bài 1

a) \(A=\left(x+1\right)\left(2x-1\right)=2x^2+x-1=2\left(x^2+\frac{x}{2}-\frac{1}{2}\right)=2\left(x^2+2.\frac{1}{4}.x+\frac{1}{16}-\frac{9}{16}\right)\)\(=2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]=2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\)

Vì \(\left(x+\frac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\frac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)

Dấu "=" xảy ra khi \(\left(x+\frac{1}{4}\right)^2=0\Leftrightarrow x+\frac{1}{4}=0\Leftrightarrow x=-\frac{1}{4}\)

Vậy minA=-9/8 khi x=-1/4

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

Vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\y^2\ge0\end{cases}}\)=>\(\left(2x-y\right)^2+y^2\ge0\Rightarrow B=\left(2x-y\right)^2+y^2+1\ge1\)

Dấu "=" xảy ra khi (2x-y)²=y²=0 <=> 2x-y=y=0 <=> x=y=0

Vậy minB=1 khi x=y=0

lý luận tương tự bài 1, bài này mình làm tắt

Bài 2:

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

\(=-3\left(x^2-2.\frac{5}{6}.x+\frac{25}{35}-\frac{49}{36}\right)=-3\left[\left(x-\frac{5}{6}\right)^2-\frac{49}{36}\right]=\frac{49}{12}-3\left(x-\frac{5}{6}\right)^2\le\frac{49}{12}\)

Dấu "=" xảy ra khi x=5/6

b)\(D=-8x^2+4xy-y^2+3=3-\left(8x^2-4xy+y^2\right)=3-\left[\left(4x^2-4xy+y^2\right)+4x^2\right]\)

\(=3-\left[\left(2x-y\right)^2+4x^2\right]\le3\)

Dấu "=" xảy ra khi x=y=0

\(a,A=3x^2-5x+1\)

\(=3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)-\dfrac{13}{12}\)

\(=3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\)

Với mọi giá trị của x ta có:

\(\left(x-\dfrac{5}{6}\right)^2\ge0\)

\(\Rightarrow3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\ge-\dfrac{13}{12}\)

Vậy Min \(A=-\dfrac{13}{12}\)

Để \(A=-\dfrac{13}{12}\) thì \(x-\dfrac{5}{6}=0\Rightarrow x=\dfrac{5}{6}\)

\(b,B=2x^2+5y^2-4x+2y+4xy+2017\)

\(=\left(2x^2-4x+4xy\right)+5y^2+2y+2017\)

\(=2\left(x^2-2x+2xy\right)+5y^2+2y+2017\)

\(=2\left[x^2-2x\left(1-y\right)+\left(1-y\right)^2\right]+5y^2+2y+2017+2\left(1-y\right)^2\)\(=2\left(x-1+y\right)^2+5y^2+2y+2017-2\left(1-y\right)^2\)

\(=2\left(x+y-1\right)^2+5y^2+2y+2017-2+4y-2y^2\)\(=2\left(x+y-1\right)^2+3y^2+6y+2015\)

\(=2\left(x+y-1\right)^2+3\left(y^2+2y+1\right)+2012\)

\(=2\left(x+y-1\right)^2+3\left(y+1\right)^2+2012\)

Với mọi giá trị của x ta có:

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

\(\Rightarrow2\left(x+y-1\right)^2+3\left(y+1\right)^2+2012\ge2012\) Vậy : Min B = 2012

Để B = 2012 thì \(\left\{{}\begin{matrix}x+y-1=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

a)  ... = (x^2 -2xy + y^2)+(x^2 -2x+1)+2014=(x-y)^2 + (x-1)^2 +2014 >= 2014 

Đăngt thức xay ra khi x=y=1

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

\(A=x^2-2\cdot\frac{1}{2}x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1\)

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

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

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

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

\(\Rightarrow GTNNx^2-x-1=\frac{3}{4}\)

với \(\left(x-\frac{1}{2}\right)^2=0;x=\frac{1}{2}\)

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

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

\(B=3\left(x^2-2\cdot\frac{1}{3}x+\left(\frac{1}{3}\right)^2-\frac{1}{9}+\frac{1}{3}\right)\)

\(B=3\left[\left(x-\frac{1}{3}\right)^2+\frac{2}{9}\right]\)

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

có \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow3\left(x-\frac{1}{2}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)

\(\Rightarrow GTNN3x^2-2x+1=\frac{2}{3}\)

với\(\left(x-\frac{1}{2}\right)^2=0;x=\frac{1}{2}\)

các bạn ơi trả lời câu c cho mình với