Toán Lớp 8: 1. Tính nhanh: ($2x^{3}$ + $x^{2}$ – 8x + 3) : (2x – 3)
2. Thực hiện phép tính: $\frac{x + 1}{3(x – 2)}$ : $\frac{x}{3(x^{2} – 4x + 4)}$
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Toán Lớp 8: 1. Tính nhanh: ($2x^{3}$ + $x^{2}$ – 8x + 3) : (2x – 3)
2. Thực hiện phép tính: $\frac{x + 1}{3(x – 2)}$ : $\frac{x}{3(x^{2} – 4x + 4)}$
Comments ( 2 )
$\text{ @HV }$
$\textit{ CÂU 1: }$
(2x^3 + x^2 – 8x + 3) : (2x – 3)
= (2x^3 – 3x^2 + 4x^2 – 6x – 2x + 3) : (2x – 3)
= [x^2. (2x – 3) + 2x. (2x – 3) – 1. (2x – 3)] : (2x – 3)
= [(x^2 + 2x – 1). (2x – 3)] : (2x – 3)
= x^2 + 2x – 1
_______________________
$\textit{ CÂU 2: }$
\frac{x + 1}{3. (x – 2)} : \frac{x}{3. (x^2 – 4x + 4)}
= \frac{x + 1}{3. (x – 2)} * \frac{3. (x – 2)^2}{x}
= \frac{(x + 1). (x – 2)}{x}
1, (2x^3+x^2-8x+3):(2x-3)
=(2x^3-3x^2+4x^2-6x-2x+3):(2x-3)
=[x^2(2x-3)+2x(2x-3)-(2x-3)]:(2x-3)
=(2x-3)(x^2+2x-1):(2x-3)
=x^2+2x-1
___________________________________________
2, \frac{x+1}{3(x-2)}:\frac{x}{3(x^2-4x+4)} (ĐK:x \ne 0,x \ne 2)
=\frac{x+1}{3(x-2)}.\frac{3(x-2)^2}{x}->HĐT 2:(a-b)^2=a^2-2ab+b^2
=\frac{(x+1).3(x-2)^2}{3(x-2).x}
=\frac{(x+1)(x-2)}{x}
=\frac{x^2-2x+x-2}{x}
=\frac{x^2-x-2}{x}