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

Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

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

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)

c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

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

\(=-\left(x^2+2x+1-9\right)\)

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

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

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

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

\(=9x^2-6xy+y^2+y^2+1\)

\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

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

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

Có link câu này bạn tham khảo xem có được không nhé

https://h.vn/hoi-dap/question/535151.html

Học tốt nhé!

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

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

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

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

\(A=2\left(x^2+2.x\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{16}-\dfrac{1}{2}\right)\)

\(A=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\)

Vì \(2\left(x+\dfrac{1}{4}\right)^2\ge0\) với mọi x

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

\(\Rightarrow Amin=-\dfrac{9}{8}\Leftrightarrow x=-\dfrac{1}{4}\)

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

\(B=\left(2x\right)^2-2.2x.y+y^2+y^2+1\)

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

Vì \(\left(2x-y\right)^2\ge0\) với mọi x và y

\(y^2\ge0\) với mọi y

\(\Rightarrow\left(2x-y\right)^2+y^2+1\ge1\)

\(\Rightarrow Bmin=1\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

\(C=5x-3x^2+2\)

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

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

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

\(C=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{49}{12}\)

Vì \(-3\left(x-\dfrac{5}{6}\right)^2\le0\) với mọi x

\(\Rightarrow-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{49}{12}\le\dfrac{49}{12}\)

\(\Rightarrow Cmax=\dfrac{49}{12}\Leftrightarrow x=\dfrac{5}{6}\)

\(D=-8x^2+4xy-y^2+3\)

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

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

Vì \(-\left(2x-y\right)^2\le0\) với mọi x và y

\(-4x^2\le0\) với mọi x

\(\Rightarrow-\left(2x-y\right)^2-4x^2+3\le3\) với mọi x và y

\(\Rightarrow Dmax=3\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

\(E=x^2-8x+38\)

\(E=x^2-2.x.4+16+22\)

\(E=\left(x-4\right)^2+22\)

Vì \(\left(x-4\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x-4\right)^2+22\ge22\) với mọi x

\(\Rightarrow Emin=22\Leftrightarrow x=4\)

\(F=6x-x^2+1\)

\(F=-\left(x^2-6x-1\right)\)

\(F=-\left(x^2-2.x.3+9-9-1\right)\)

\(F=-\left(x-3\right)^2+10\)

Vì \(-\left(x-3\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-3\right)^2+10\le10\)

\(\Rightarrow Fmax=10\Leftrightarrow x=3\)

a: \(A=-x^2-4x-2\)

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

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

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

Dấu '=' xảy ra khi x+2=0

=>x=-2

b: \(B=-2x^2-3x+5\)

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

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}< =\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x+\dfrac{3}{4}=0\)

=>\(x=-\dfrac{3}{4}\)

c: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

\(=-x^2-2x-1+9\)

\(=-\left(x^2+2x+1\right)+9\)

\(=-\left(x+1\right)^2+9< =9\forall x\)

Dấu '=' xảy ra khi x+1=0

=>x=-1

d: \(D=-8x^2+4xy-y^2+3\)

\(=-8\left(x^2-\dfrac{1}{2}xy\right)-y^2+3\)

\(=-8\left(x^2-2\cdot x\cdot\dfrac{1}{4}y+\dfrac{1}{16}y^2\right)+\dfrac{1}{2}y^2-y^2+3\)

\(=-8\left(x-\dfrac{1}{4}y\right)^2-y^2+3< =3\forall x,y\)

Dấu '=' xảy ra khi y=0 và x-1/4y=0

=>y=0 và x=0