题目
3.已知 (x)=(x)^2-2x+3 , (x)=3(x)^2-x+1, 求下列-|||-(1) (x)+2g(x);-|||-(2) (x)-g(x);-|||-(3)f(x)·g(x );-|||-(4) dfrac (2f(x))(3g(x))

题目解答
答案
本题考查了整式的加减,熟练掌握运算法则是解本题的关键.
(1)把f(x)与g(x)代入f(x)+2g(x)中,去括号合并即可得到结果;
(2)把f(x)与g(x)代入2f(x)-g(x)中,去括号合并即可得到结果;
(3)把f(x)与g(x)代入f(x)•g(x)中,利用多项式乘以多项式法则计算即可得到结果;
(4)把f(x)与g(x)代入f(x)÷g(x)中,利用多项式除以多项式法则计算即可得到结果.
(1)$f\left ( {x} \right )+2g\left ( {x} \right )={x}^{2}-2x+3+2\left ( {3{x}^{2}-x+1} \right )$
$={x}^{2}-2x+3+6{x}^{2}-2x+2$
$=7{x}^{2}-4x+5$
(2)$2f\left ( {x} \right )-g\left ( {x} \right )=2\left ( {{x}^{2}-2x+3} \right )-\left ( {3{x}^{2}-x+1} \right )$
$=2{x}^{2}-4x+6-3{x}^{2}+x-1$
$=-{x}^{2}-3x+5$
(3)$f\left ( {x} \right )\cdot g\left ( {x} \right )=\left ( {{x}^{2}-2x+3} \right )\left ( {3{x}^{2}-x+1} \right )$
$=3{x}^{4}-{x}^{3}+{x}^{2}-6{x}^{3}+2{x}^{2}-2x+9{x}^{2}-3x+3$
$=3{x}^{4}-7{x}^{3}+12{x}^{2}-5x+3$
(4)$\dfrac {2f\left ( {x} \right )} {3g\left ( {x} \right )}=\dfrac {2\left ( {{x}^{2}-2x+3} \right )} {3\left ( {3{x}^{2}-x+1} \right )}$
$=\dfrac {2{x}^{2}-4x+6} {9{x}^{2}-3x+3}$
(1)把f(x)与g(x)代入f(x)+2g(x)中,去括号合并即可得到结果;
(2)把f(x)与g(x)代入2f(x)-g(x)中,去括号合并即可得到结果;
(3)把f(x)与g(x)代入f(x)•g(x)中,利用多项式乘以多项式法则计算即可得到结果;
(4)把f(x)与g(x)代入f(x)÷g(x)中,利用多项式除以多项式法则计算即可得到结果.
(1)$f\left ( {x} \right )+2g\left ( {x} \right )={x}^{2}-2x+3+2\left ( {3{x}^{2}-x+1} \right )$
$={x}^{2}-2x+3+6{x}^{2}-2x+2$
$=7{x}^{2}-4x+5$
(2)$2f\left ( {x} \right )-g\left ( {x} \right )=2\left ( {{x}^{2}-2x+3} \right )-\left ( {3{x}^{2}-x+1} \right )$
$=2{x}^{2}-4x+6-3{x}^{2}+x-1$
$=-{x}^{2}-3x+5$
(3)$f\left ( {x} \right )\cdot g\left ( {x} \right )=\left ( {{x}^{2}-2x+3} \right )\left ( {3{x}^{2}-x+1} \right )$
$=3{x}^{4}-{x}^{3}+{x}^{2}-6{x}^{3}+2{x}^{2}-2x+9{x}^{2}-3x+3$
$=3{x}^{4}-7{x}^{3}+12{x}^{2}-5x+3$
(4)$\dfrac {2f\left ( {x} \right )} {3g\left ( {x} \right )}=\dfrac {2\left ( {{x}^{2}-2x+3} \right )} {3\left ( {3{x}^{2}-x+1} \right )}$
$=\dfrac {2{x}^{2}-4x+6} {9{x}^{2}-3x+3}$