Bài 1:
a, $2.(x+1)=3.(4x-1)$
$⇒2x+2=12x-3$
$⇒2x-12x=-3-2$
$⇒-10x=-5$
$⇒x=1/2$
$P=\frac{2x+1}{2x+5}=$ $\frac{(2x+5)-4}{2x+5}=1-$ $\frac{4}{2x+5}$
Tại $x=1/2$
$⇒P=$$1-\frac{4}{2.1/2+5}=$ $1-\frac{4}{6}=1/3$
b, Ta có: $A=(x-5)(y^2-9)=0$
⇒\(\left[ \begin{array}{l}x-5=0\\y^2-9=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=5\\y=±3\end{array} \right.\)
Vậy $x=5$ hoặc $y=±3$
Bài 2:
a, $P=(\frac{-2}{3}x^2y^3z^2).($ $\frac{-1}{2}xy)^3.(xy^2z)^2$
⇒$P=\frac{-2}{3}x^2y^3z^2.$ $\frac{-1}{8}x^3y^3.x^2y^4z^2$
⇒$P=\frac{1}{12}x^7y^{10}x^4$
b, Hệ số: $\frac{1}{12}$
Bậc: $21$
c, Ta có: $y^{10}$$\geq0;x^4$ $\geq0$
Để $P<0⇒x<0$
Vậy $P<0$ khi $x<0$
