Đáp án:
C5:
b) \(R = 4{x^3} + 4{x^2} - 2x - \dfrac{9}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
C4)a)\left( { - 4 + 4} \right){x^5} + 2{x^4} + \left( {1 - 3} \right){x^2} + \dfrac{3}{2}\\
= 2{x^4} - 2{x^2} + \dfrac{3}{2}\\
Thay:x = - \dfrac{1}{2}\\
\to 2.{\left( {\dfrac{1}{2}} \right)^4} - 2.{\left( {\dfrac{1}{2}} \right)^2} + \dfrac{3}{2} = \dfrac{9}{8}\\
b)7x{y^2} + \left( { - 4 - 3 - 4} \right){x^2}y + 6{x^3} - 1\\
= 7x{y^2} - 11{x^2}y + 6{x^3} - 1\\
Thay:x = 1;y = - 1\\
\to 7.1.{\left( { - 1} \right)^2} - 11.1.\left( { - 1} \right) + 6.1 - 1 = 23\\
c)\left( {5 + \dfrac{1}{2}} \right){x^2}y + xy + \dfrac{1}{3}x + \dfrac{1}{4}\\
= \dfrac{{11}}{2}{x^2}y + xy + \dfrac{1}{3}x + \dfrac{1}{4}\\
Thay:x = 1;y = - 1\\
\to \dfrac{{11}}{2}{.1^2}.\left( { - 1} \right) + 1.\left( { - 1} \right) + \dfrac{1}{3}.1 + \dfrac{1}{4} = - \dfrac{{71}}{{12}}\\
C5:\\
Q - P = 4{x^2} - {x^3} + x - \dfrac{1}{5}\\
\to Q = 4{x^2} - {x^3} + x - \dfrac{1}{5} + 2{x^3} - 4{x^2} + x - \dfrac{1}{2}\\
= {x^3} + 2x - \dfrac{7}{{10}}\\
R + P = 6{x^3} - x - 5\\
\to R = 6{x^3} - x - 5 - 2{x^3} + 4{x^2} - x + \dfrac{1}{2}\\
= 4{x^3} + 4{x^2} - 2x - \dfrac{9}{2}
\end{array}\)
