Toán Lớp 8: x^2 + 4x + 4 3x – 3
——————- . ———-
x-1 x+2
(x+2)^2 x+2
———– : ——–
x-1 3x-3
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Toán Lớp 8: x^2 + 4x + 4 3x – 3
——————- . ———-
x-1 x+2
(x+2)^2 x+2
———– : ——–
x-1 3x-3
Ai chỉ với
Comments ( 1 )
Giải đáp:
$\begin{array}{l}
Dkxd:x \ne 1;x \ne – 2\\
\dfrac{{{x^2} + 4x + 4}}{{x – 1}}.\dfrac{{3x – 3}}{{x + 2}}\\
= \dfrac{{{{\left( {x + 2} \right)}^2}}}{{x – 1}}.\dfrac{{3\left( {x – 1} \right)}}{{x + 2}}\\
= \dfrac{{x + 2}}{1}.\dfrac{3}{1}\\
= 3.\left( {x + 2} \right)\\
= 3x + 6
\end{array}$