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.

222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Rút gọn : E= $\frac{x+2}{x+3}$ +$\frac{5}{x^{2}+x-6}$ + $\frac{1}{2-x}$

Tháng Ba 31, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: Rút gọn : E= $\frac{x+2}{x+3}$ +$\frac{5}{x^{2}+x-6}$ + $\frac{1}{2-x}$

Comments ( 1 )

  1. tuyetnhungthu
    tuyetnhungthu
    31 Tháng Ba, 2022 at 12:23 sáng
    Reply
    Giải đáp:
    $E=\dfrac{x+1}{x+3}.$
    Lời giải và giải thích chi tiết:
    $E=\dfrac{x+2}{x+3}+\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\\ \text{ĐKXĐ: }\left\{\begin{array}{l} x +3 \ne 0 \\ x^2+x-6 \ne 0 \\ 2-x \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x  \ne -3 \\ x^2-2x+3x-6 \ne 0 \\ x \ne 2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x  \ne -3 \\ x(x-2)+3(x-2) \ne 0 \\ x \ne 2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x  \ne -3 \\ (x+3)(x-2)\ne 0 \\ x \ne 2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x  \ne -3 \\ x \ne 2\end{array} \right.\\ E=\dfrac{x+2}{x+3}+\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\\ =\dfrac{x+2}{x+3}+\dfrac{5}{(x+3)(x-2)}-\dfrac{1}{x-2}\\ =\dfrac{(x+2)(x-2)}{(x+3)(x-2)}+\dfrac{5}{(x+3)(x-2)}-\dfrac{x+3}{(x-2)(x+3)}\\ =\dfrac{(x+2)(x-2)+5-(x+3)}{(x-2)(x+3)}\\ =\dfrac{x^2-4+5-(x+3)}{(x-2)(x+3)}\\ =\dfrac{x^2-4+5-x-3}{(x-2)(x+3)}\\ =\dfrac{x^2-x-2}{(x-2)(x+3)}\\ =\dfrac{x^2+x-2x-2}{(x-2)(x+3)}\\ =\dfrac{x(x+1)-2(x+1)}{(x-2)(x+3)}\\ =\dfrac{(x-2)(x+1)}{(x-2)(x+3)}\\ =\dfrac{x+1}{x+3}.$

Leave a reply

222-9+11+12:2*14+14 = ? ( )

Thúy Mai

About Thúy Mai