Đặt \(A=10x^2+15x+8x-12xy+4y^2\)

\(=\left(9x^2-12xy+4y^2\right)+\left(x^2+15x+8x\right)\)

\(=\left(3x-2y\right)^2+\left(x^2+23x\right)\)

\(=\left(3x-2y\right)^2+\left(x^2+2.x\dfrac{23}{2}+\dfrac{529}{4}-\dfrac{529}{4}\right)\)

\(=\left(3x-2y\right)^2+\left(x+\dfrac{23}{2}\right)^2-\dfrac{529}{4}\)

Vì \(\left(3x-2y\right)^2\ge0\) với mọi x,y

\(\left(x+\dfrac{23}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(3x-2y\right)^2+\left(x+\dfrac{23}{2}\right)^2-\dfrac{529}{4}\ge-\dfrac{529}{4}\) với mọi x,y

\(\Rightarrow Amin=-\dfrac{529}{4}\Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\x+\dfrac{23}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3x=2y\\x=-\dfrac{23}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3.\left(-\dfrac{23}{2}\right)=2y\\x=-\dfrac{23}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-\dfrac{69}{2}=2y\\x=-\dfrac{23}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=-\dfrac{69}{4}\\x=-\dfrac{23}{2}\end{matrix}\right.\)

Vậy giá trị nhỏ nhất của biểu thức là -529/4 khi y = -69/4 và x = -23/2

Ta có : \(\left|x-5\right|+12\ge12\)

\(\Rightarrow\frac{-8}{\left|x-5\right|+12}\ge-\frac{8}{12}=-\frac{2}{3}\)

\(\Rightarrow A=10+\frac{-8}{\left|x-5\right|+12}\ge10-\frac{2}{3}=\frac{28}{3}\)

Dấu ''='' xảy ra khi x = 5

Vậy GTNN của A là 28/3 tại x = 5

x= 0 

GTNN = 62/7 

mình nghĩ thế

2^x+2^x+1=48

Giúp mình nha

\(a,=x\left(x^2-10x+25\right)=x\left(x-5\right)^2\\ b,=y\left(x+y\right)-\left(x+y\right)=\left(y-1\right)\left(x+y\right)\\ c,=\left(x-5\right)^2\\ d,=\left(x-8\right)\left(x+8\right)\)

(x^2+y^2-12y-12x+36)+(5y^2-10y+5)+4=(x-y-6)^2+5(y-1)^2+4>=4

GTNN A=4

khi y=1

x=7

Ta có \(x^2+x+5=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{19}{4}=\left(x+\frac{1}{2}\right)^2+\frac{19}{4}\)

Nhận thấy \(\left(x+\frac{1}{2}\right)^2\ge0\forall x=>\left(x+\frac{1}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)

Dấu "=" xảy ra khi x+1/2=0 => x=-1/2

Vậy ...

\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)

b:

\(D=-25x^2+10x-1-10\)

\(=-\left(25x^2-10x+1\right)-10\)

\(=-\left(5x-1\right)^2-10< =-10\)

Dấu = xảy ra khi x=1/5

\(E=-9x^2-6x-1+20\)

\(=-\left(9x^2+6x+1\right)+20\)

\(=-\left(3x+1\right)^2+20< =20\)

Dấu = xảy ra khi x=-1/3

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

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

Dấu = xảy ra khi x=1

\(A=9x^2+6xy+y^2+x^2-6x+9+4=\left(3x+y\right)^2+\left(x-3\right)^2+4>=4\)

Dấu '=' xảy ra khi x=3 và y=-3x=-9