Toán Lớp 8: Đặt tính rồi tính
a) $\frac{3x+11}{x+5}$- $\frac{x+1}{x+5}$
b) ( $2x^{3}$- $3x^{2}$+x+15): (2x+3)
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: Đặt tính rồi tính
a) $\frac{3x+11}{x+5}$- $\frac{x+1}{x+5}$
b) ( $2x^{3}$- $3x^{2}$+x+15): (2x+3)
Comments ( 2 )
Giải đáp+Lời giải và giải thích chi tiết:
$\begin{array}{|c|}\hline \text{}&\text{@thienne2k7}&\text{}\\\hline\end{array}$
a)(3x+11)/(x+5)-(x+1)/(x+5) (đk:xne-5)
=(3x+11-x-1)/(x+5)
=(2x+10)/(x+5)
=(2(x+5))/(x+5)
=2
b)(2x^3 -3x^2 +x+15):(2x+3) (đk:xne-3/2)
=(2x^3 -6x^2 +10x+3x^2 -9x+15)/(2x+3)
=(2x(x^2 -3x+5)+3(x^2 -3x+5))/(2x+3)
=((x^2 -3x+5)(2x+3))/(2x+3)
=x^2 -3x+5
$a)\dfrac{3x+11}{x+5}-\dfrac{x+1}{x+5}$
$=\dfrac{(3x+11-(x+1)}{x+5}$
$=\dfrac{3x+11-x-1}{x+5}$
$=\dfrac{2x+10}{x+5}$
$=\dfrac{2(x+5)}{x+5}$
$=2$
$b)(2x^3-3x^2+x+15):(2x+3)$
$=x^2-3x+5$
$#mie1208$