=>2ab-3a+b-9=0

=>b(2a+1)-3a-4,5-*4,5=0

=>b(2a+1)-1,5(2a+1)=4,5

=>(2a+1)(b-1,5)=4,5

=>(2a+1)(2b-3)=9

=>\(\left(2a+1;2b-3\right)\in\left\{\left(1;9\right);\left(3;3\right);\left(9;1\right)\right\}\)

=>\(\left(a,b\right)\in\left\{\left(0;6\right);\left(1;3\right);\left(4;2\right)\right\}\)

\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left[\left(x+6\right)\left(x-1\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

\(P=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-6^2.P_{min}\Leftrightarrow x^2+5xđạtGTNN\)

\(x^2+5x\ge0\Leftrightarrow x\left(x+5\right)\ge0\)

Dấu "=" xảy ra <=> \(x\in\left\{0;-5\right\}\)

Vậy: P_min=-36 <=> x E {0;-5}

CHờ tí mk lm câu b

cảm ơn bạn nhiều

\(A=\frac{x^3-3x^2-7x-15}{x^5-x^4-10x^3-38x^2-51x-45}\)

\(=\frac{x^2\left(x-5\right)+2x\left(x-5\right)+3\left(x-5\right)}{x^4\left(x-5\right)+4x^3\left(x-5\right)+10x^2\left(x-5\right)+12x\left(x-5\right)+9\left(x-5\right)}\)

\(=\frac{\left(x-5\right)\left(x^2+2x+3\right)}{\left(x-5\right)\left(x^4+4x^3+10x^2+12x+9\right)}\)

\(=\frac{x^2+2x+3}{x^4+4x^3+10x^2+12x+9}\)

\(=\frac{x^2+2x+3}{\left(x^2\right)^2+2.x^2.2x+\left(2x\right)^2+6x^2+12x+9}\)

\(=\frac{x^2+2x+3}{\left(x^2+2x\right)^2+2.\left(x^2+2x\right).3+3^2}\)

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

b, \(A=\frac{1}{x^2+2x+3}=\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)

Dấu "=" xảy ra khi: \(x+1=0\Rightarrow x=-1\)

Vậy GTLN của A là \(\frac{1}{2}\) khi x = -1

0,5x - 2/3x = 5/12

=> x(0,5 - 2/3) = 5/12

=> -1/6x = 5/12

=> x = (5/12) / (-1/6)

=> x = -5/2