A) \(A=-3x^2+x+1\)

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

\(A=-3\left(x^2-2\cdot\dfrac{1}{6}\cdot x+\dfrac{1}{36}-\dfrac{13}{36}\right)\)

\(A=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{13}{12}\)

Mà: \(-3\left(x-\dfrac{1}{6}\right)^2\le0\forall x\)

\(\Rightarrow A=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{13}{12}\le\dfrac{13}{12}\forall x\)

Dấu "=" xảy ra khi:

\(x-\dfrac{1}{6}=0\Rightarrow x=\dfrac{1}{6}\)

Vậy: \(A_{max}=\dfrac{13}{12}.khi.x=\dfrac{1}{6}\)

B) \(B=2x^2-8x+1\)

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

\(B=2\left(x^2-4x+4-\dfrac{7}{2}\right)\)

\(B=2\left(x-2\right)^2-7\)

Mà: \(2\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow B=2\left(x-2\right)^2-7\ge-7\forall x\)

Dấu "=" xảy ra khi:

\(x-2=0\Rightarrow x=2\)

Vậy: \(B_{min}=2.khi.x=2\)

câu a) bạn viết sai đề rồi

đề....

(41/9+x-133/18):47/3=2009-2008

(41/9+x-133/18):47/3=1

(41/9+x-133/18)=1*47/3=47/3

41/9+x=47/3+133/18

41/9+x=415/18

x=415/18-41/9

x=37/2

Cảm ơn nguyen anh hieu nhé

-12(x-5)+7(3-x)=5

=> -12x+60+21-7x=5

=> -19x+81=5

=>-19x=5-81

=>-19x=-76

=>x=4

Vậy x=4

\(2^{x}=46-3^{x}\)

=>\(2^{x}+3^{x}=46\)

=>\(2^{x}<46\)

mà x nguyên và \(2^{x}>0\)

nên x∈{0;1;2;3;4;5}

TH1: x=0

=>\(2^{x}+3^{x}=2^0+3^0=1+1=2<>46\)

=>LOại

TH2: x=1

\(2^{x}+3^{x}=2+3=5<>46\)

=>Loại

TH3: x=2

\(2^{x}+3^{x}=2^2+3^2=4+9=13<>46\)

=>Loại

TH4: x=3

\(2^{x}+3^{x}=2^3+3^3=8+27=35<>46\)

=>Loại

TH5: x=4

\(2^{x}+3^{x}=2^4+3^4=16+81=97<>46\)

=>Loại

TH6: x=5

\(2^{x}+3^{x}=2^5+3^5=32+243=275<>46\)

=>Loại

vậy: x∈∅