a,A=(2x)²-2.2x.2+2²+11=(2x-2)²+11

Vì (2x-2)²luôn lớn hơn hoặc bằng 0

=>A>hoặc =0+11 hay a>hoặc =11

vậy GTNN của A là 11 khi x=1

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

suy ra Amin=-1

\(B=4x^2+4x+11=4\left(x^2+x+\frac{11}{4}\right)=4\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{10}{4}\right)=4\left(x+\frac{1}{2}\right)^2+10\) Suy ra Bmin = 10

Bài 1 :                                                         Bài giải

Ta có : 

\(A=7+7^2+7^3+...+7^8\)

\(A=\left(7+7^2+7^3+7^4\right)+\left(7^5+7^6+7^7+7^8\right)\)

\(A=7\left(1+7+7^2+7^3\right)+7^4\left(1+7+7^2+7^3\right)\)

\(A=7\cdot400+7^4\cdot400\)

\(A=7\cdot8\cdot50+7^4\cdot8\cdot50\)

\(A=50\left(7\cdot8+7^4\cdot8\right)\text{ }⋮\text{ }50\)

Bài 1 :                                                         Bài giải

Ta có : 

\(A=7+7^2+7^3+...+7^8\)

\(A=\left(7+7^2+7^3+7^4\right)+\left(7^5+7^6+7^7+7^8\right)\)

\(A=7\left(1+7+7^2+7^3\right)+7^4\left(1+7+7^2+7^3\right)\)

\(A=7\cdot400+7^4\cdot400\)

\(A=7\cdot8\cdot50+7^4\cdot8\cdot50\)

\(A=50\left(7\cdot8+7^4\cdot8\right)\text{ }⋮\text{ }50\)

\(a,A=\left(x+2\right)^2+37\)

\(A_{min}=37\Leftrightarrow\left(x+2\right)^2=0\Rightarrow x+2=0\Leftrightarrow x=-2\)

\(b,B=2\left(x-3\right)^2-30\)

\(B_{min}=-30\Leftrightarrow2\left(x-3\right)^2=0\Rightarrow x-3=0\Leftrightarrow x=3\)

\(e,E=-\left(x+2\right)^2+37\)

\(E_{max}=37\Leftrightarrow-\left(x+2\right)^2=0\Rightarrow x+2=0\Leftrightarrow x=-2\)

Lời giải:

$a^2+b^2+c^2+d^2+e^2=a(b+c+d+e)$

$\Leftrightarrow 4a^2+4b^2+4c^2+4d^2+4e^2-4a(b+c+d+e)=0$

$\Leftrightarrow (a^2+4b^2-4ab)+(a^2-4c^2-4ac)+(a^2+4d^2-4ad)+(a^2+4e^2-4ae)=0$

$\Leftrightarrow (a-2b)^2+(a-2c)^2+(a-2d)^2+(a-2e)^2=0$

Ta thấy: $(a-2b)^2,(a-2c)^2,(a-2d)^2,(a-2e)^2\geq 0$ với mọi $a,b,c,d,e$ thực

Do đó để tổng của chúng bằng $0$ thì:

$(a-2b)^2=(a-2c)^2=(a-2d)^2=(a-2e)^2=0$

$\Leftrightarrow 2b=2c=2d=2e=a$

$\Rightarrow b=c=d=e$

\(\left(\dfrac{a}{2}-b\right)^2\ge0\Leftrightarrow\dfrac{a^2}{4}-ab+b^2\ge0\Leftrightarrow\dfrac{a^2}{4}+b^2\ge ab\)

CMTT ta được: \(\left\{{}\begin{matrix}\dfrac{a^2}{4}+c^2\ge ac\\\dfrac{a^2}{4}+d^2\ge ad\\\dfrac{a^2}{4}+e^2\ge ae\end{matrix}\right.\)

\(\Rightarrow4.\dfrac{a^2}{4}+b^2+c^2+d^2+e^2\ge ab+ac+ad+ae\)

\(\Rightarrow a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\)

\(ĐTXR\Leftrightarrow\dfrac{a}{2}=b=c=d=e\)

a: \(A=-3\left(x^2-2x+\dfrac{2}{3}\right)\)

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

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

Dấu '=' xảy ra khi x=1

b: \(B=-\left(16x^2+8x-4\right)\)

\(=-\left(16x^2+8x+1-5\right)\)

\(=-\left(4x+1\right)^2+5< =5\)

Dấu '=' xảy ra khi x=-1/4

d: \(x^2+2x+3=\left(x+1\right)^2+2>=2\)

=>E<=1/2

Dấu '=' xảy ra khi x=-1