Lời giải:

a) 

Áp dụng BĐT Bunhiacopxky:

\((y-2x)^2\leq (16y^2+36x^2)(\frac{1}{16}+\frac{1}{9})=9.\frac{25}{144}\)

\(\Rightarrow \frac{-5}{4}\leq y-2x\leq \frac{5}{4}\Rightarrow \frac{15}{4}\leq y-2x+5\leq \frac{25}{4}\)

Vậy $A_{\min}=\frac{15}{4}$ và $A_{\max}=\frac{25}{4}$

b) 

Áp dụng BĐT Bunhiacopxky:

\((2x-y)^2\leq (\frac{x^2}{4}+\frac{y^2}{9})(16+9)=25\)

\(\Rightarrow -5\leq 2x-y\leq 5\Leftrightarrow -7\leq 2x-y-2\leq 3\)

Vậy $B_{min}=-7; B_{\max}=3$

Ai giúp mik vs

Huhu ai giúp vs

a) A=5+16-(x^2+8x+16)=21-(x+4)^2

Amax=21 khi x=-4

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

Bmin=2 khi x=1; y=2

c)C=(x-1)(x+2)(x+3)(x+6)=(x^2+5x-6)(x^2+5x+6)=(x^2+5x)^2-36

Cmin =-36 khi x=0 

a)

gồm bình phường (a^2+2ab+b^2)=(a+b)^2 (*)

5-8x-x^2=-(x^2+8x-5) đây đâu trừ ra ngoài

(....) biến đổi cho giống biểu thức trên (*)

-(x^2+2.4.x+4^2) ....(ở đây a=x; b=4)

xong như vậy ta đã thêm 4^2=16  vào biểu thức mang dấu(-)

vậy ta công trả lại 16

-(x^2+2.4.x+4^2)+16+5 { còn 5 nguyên ban đầu )

=21-(x+4)^2

{x+4}^2 luôn dương=> -(x+4)^2 luon am

=> 21-(x+4)^2 \(\ge\)21

GTNN=21

câu c áp dụng (a^2-b^2)=(a-b)(a+b) (**)

\(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)t/c giao hoán phép nhân

\(\left[x^2+6x-x-6\right]\left[x^2+3x+2x+6\right]\)nhận phân phối 

\(\left[x^2+5x-6\right]\left[x^2+5x+6\right]\)rút gọn binhf thường

   \(\left[\left(x^2+5x\right)-6\right]\left[\left(x^2+5x\right)+6\right]\)    ap dung ct (**) {a=(x^2+5x); b=6 

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

ok !!!!

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

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)

\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)

\(minC=-8\Leftrightarrow x=-1\)

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

\(maxD=-4\Leftrightarrow x=1\)

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

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

\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

a) A= 2x²-8x+10 = 2(x-2)²+2\(\ge\)2\(\Leftrightarrow\)x=2

Vậy MinA=2 \(\Leftrightarrow\)x=2

b) B= -(x-1)²-(2y+1)²+7 \(\le\)7

Dấu = xảy ra khi x=1 và y=\(\frac{-1}{2}\)

Vậy MaxB=7 ....

cảm ơn bạn nha

I don't know

...................

Sorry !

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

\(\left(x-\frac{1}{2}\right)^2\ge0\)

\(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

\(-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\le\frac{1}{4}\)

Vậy Max A = \(\frac{1}{4}\) khi x = \(\frac{1}{2}\)

***

\(B=5-8x-x^2=-\left(x^2+2\times x\times4+4^2-4^2-5\right)=-\left[\left(x+4\right)^2-21\right]\)

\(\left(x+4\right)^2\ge0\)

\(\left(x+4\right)^2-21\ge-21\)

\(-\left[\left(x+4\right)^2-21\right]\le21\)

Vậy Max B = 21 khi x = - 4 

***

\(C=5-x^2+2x-4y^2-4y=-\left(x^2-2\times x\times1+1^2-1^2+\left(2y\right)^2-2\times2y\times1+1^2-1^2-5\right)=-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\)

\(\left(x-1\right)^2\ge0\)

\(\left(2y-1\right)^2\ge0\)

\(\left(x-1\right)^2+\left(2y-1\right)^2-7\ge-7\)

\(-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\le7\)

Vậy Max C = 7 khi x = 1 và y = \(\frac{1}{2}\)