Đáp án:
\(\begin{array}{l}
d)4{x^4} - {x^3} - 5{x^2}\\
B3:\\
a)Q = 75\\
b)Q = - \dfrac{{391}}{4}\\
B4:;\\
1)x = - \dfrac{5}{{19}}\\
2)x = - \dfrac{{38}}{{69}}\\
3)x = \dfrac{3}{{14}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
d)2x.3{x^2} - 2x.x - 4{x^2}.x + 4{x^2}.{x^2} - 4{x^2} + x.x - 3{x^2}.x\\
= 6{x^3} - 2{x^2} - 4{x^3} + 4{x^4} - 4{x^2} + {x^2} - 3{x^3}\\
= 4{x^4} - {x^3} - 5{x^2}\\
B3:\\
a)Q = 5x.{x^2} - 15x + 7{x^2} - 5{x^3} - 7{x^2}\\
= - 15x\\
Thay:x = - 5\\
\to Q = - 15.\left( { - 5} \right) = 75\\
b)Q = {x^2} - xy + xy - {y^2}\\
= {x^2} - {y^2}\\
Thay:x = \dfrac{3}{2};y = 10\\
\to Q = {\left( {\dfrac{3}{2}} \right)^2} - {10^2} = - \dfrac{{391}}{4}\\
B4:;\\
1)10{x^2} - 35x + 16x - 10{x^2} = 5\\
\to - 19x = 5\\
\to x = - \dfrac{5}{{19}}\\
2)\dfrac{3}{5}{x^2} - \dfrac{3}{{10}}x - \dfrac{3}{5}{x^2} - \dfrac{{39}}{{100}}x = \dfrac{{19}}{{50}}\\
\to - \dfrac{{69}}{{100}}x = \dfrac{{19}}{{50}}\\
\to x = - \dfrac{{38}}{{69}}\\
3){x^2} - \dfrac{1}{3}x - {x^2} + \dfrac{3}{2}x = \dfrac{1}{4}\\
\to \dfrac{7}{6}x = \dfrac{1}{4}\\
\to x = \dfrac{3}{{14}}
\end{array}\)
