a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)

\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)

c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)

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

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)

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

a: Ta có: \(4x^2+12x+1\)

\(=4x^2+12x+9-8\)

\(=\left(2x+3\right)^2-8\ge-8\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

b: Ta có: \(4x^2-3x+10\)

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

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)

c: Ta có: \(2x^2+5x+10\)

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

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)

Bạn xem lại đề câu d nhé.

D=x^2+5y^2-4xy-6x+8y+12

Bạn cũng cần xem lại đề câu c nhé.

\(B=5x^2+2y^2+4xy-2x+4y+2020\)

\(=4x^2+4xy+y^2+x^2-2x+1+4y^2+4y+1+2018\)

\(=\left(2x+y\right)^2+\left(x-1\right)^2+\left(2y+1\right)^2+2018\ge2018\left(\text{với mọi x;y}\right)\)

\(\text{Dấu "=" xảy ra khi: }x-1=0;2x+1=0\Leftrightarrow x=1;y=\frac{-1}{2}\)

\(\text{Vậy GTNN của }D\text{ là }2018\text{ tại }x=1;y=\frac{-1}{2}\)

=4.x^2+x^2+y^2+y^2+4xy-2x+4y+1+4+2015

=[4.x^2+4xy+y^2]+[x^2-2x+1]+[y^2-4y+4]

=[2x+y]^2+[x-1]^2+[y-2]^2+2015>hoặc bằng2015

giá trị nhỏ nhất là 2015

\(A=5x^2+2y^2-4xy-8x-4y+2031\)

\(\Rightarrow5A=25x^2+10y^2-20xy-32x-16y+10155\)

\(=\left(25x^2-20xy+4y^2\right)+6\left(y^2-2\cdot\frac{8}{9}+\frac{64}{81}\right)+\left(10155-6\cdot\frac{64}{81}\right)\)

\(=\left(5x-2y\right)^2+6\left(y-\frac{8}{9}\right)^2+\left(10155-6\cdot\frac{64}{81}\right)\ge10155-6\cdot\frac{64}{81}\)

\(\Rightarrow A\ge2031-\frac{6}{5}\cdot\frac{64}{81}\)

Dấu "=" xảy ra tại \(y=\frac{8}{9};x=\frac{16}{45}\)

PS:Is that true ???

Gợi ý:

\(A=2\left(y-x-1\right)^2+3\left(x-2\right)^2+2017\ge2017\)

Đẳng thức xảy ra khi \(x=2;y=3\)

Vậy \(A_{min}=2017\Leftrightarrow x=2;y=3\)

zZz Cool Kid_new zZz sai từ dòng thứ 2, kiểm tra lại đi nhé.

\(N = 5x^2 + 2y^ 2 + 4xy - 2x + 4y + 2015\)

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

\(( y^2 + 4y + 4 ) + 2010\)

\(N = ( 2x + y )^2 + ( x - 1 )^2 + ( y + 2 )^2 + 2010\)

\(\ge\)\(2010\)

\(Dấu " = " xảy ra \)\(\Leftrightarrow\) \(2x + y = 0 và\)\(x - 1 = 0 và y + 2 = 0\)

\(\Rightarrow\)\(x = 1 và y = - 2\)

\(Min N = 2010\)\(\Leftrightarrow\)\(x = 1 và y = - 2\)

a, \(P=2x^2+5y^2+4xy+8x-4y+15\)

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

Dấu "="xảy ra khi:\(\hept{\begin{cases}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-4\\y=2\end{cases}}\)

Vậy...

b, \(C=2x^2+4xy+4y^2-3x-1\)

\(=\left(x+2y\right)^2+\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

sau đó giải tương tự câu a nhé