Bài 1Tìm x biết a, (x+2)^2-(x-2)(x+2)=0 b, x(x-3)-5(x-3)=0 Bài 2 rút gọn phân thức a, 12x^3y^2/18xy^5 b, 12xy/4xy c, 3x^2-12x+12/x^4-8x d, 7x^2+14x+7/

Đáp án:

 B2:

d) \(\dfrac{{7\left( {x + 1} \right)}}{3}\)

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
a,{(x + 2)^2} – (x – 2)(x + 2) = 0\\
 \to \left( {x + 2} \right)\left( {x + 2 – x + 2} \right) = 0\\
 \to 4\left( {x + 2} \right) = 0\\
 \to x =  – 2\\
b)x\left( {x – 3} \right) – 5\left( {x – 3} \right) = 0\\
 \to \left( {x – 3} \right)\left( {x – 5} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 3\\
x = 5
\end{array} \right.\\
B2:\\
a)\dfrac{{12{x^3}{y^2}}}{{18x{y^5}}} = \dfrac{{2{x^2}}}{{3{y^3}}}\\
b)\dfrac{{12xy}}{{4xy}} = 3\\
c)\dfrac{{3{x^2} – 12x + 12}}{{{x^4} – 8x}}\\
 = \dfrac{{3{{\left( {x – 2} \right)}^2}}}{{x\left( {x – 2} \right)\left( {x + 2x + 4} \right)}}\\
 = \dfrac{{3\left( {x – 2} \right)}}{{x\left( {x + 2x + 4} \right)}}\\
d)\dfrac{{7{x^2} + 14x + 7}}{{3x + 3}} = \dfrac{{7{{\left( {x + 1} \right)}^2}}}{{3\left( {x + 1} \right)}}\\
 = \dfrac{{7\left( {x + 1} \right)}}{3}
\end{array}\)