p(x)=-2x ²3×4+5x ³+x ²_1/4x_2
q(x)=3×4+x ²_1/4_3x ³ _x²
a,sắp xếp theo lũy thừa giảm dần của biến
b,p(x)+q(x) và p(x)_q(x)
c,tính 2p(x)+5q(x) và 4p(x)_3Q(x)
p(x)=-2x ²3×4+5x ³+x ²_1/4x_2
q(x)=3×4+x ²_1/4_3x ³ _x²
a,sắp xếp theo lũy thừa giảm dần của biến
b,p(x)+q(x) và p(x)_q(x)
c,tính 2p(x)+5q(x) và 4p(x)_3Q(x)
Đáp án:
$a)
P(x)=3x^4+5x^3-x^2-\dfrac{1}{4}x-2\\
Q(x)=3x^4-3x^3-\dfrac{1}{4}\\
b) P(x)+Q(x)=6x^4+2x^3-x^2-\dfrac{1}{4}x-\dfrac{9}{4}\\
P(x)-Q(x)=8x^3-x^2-\dfrac{1}{4}-\dfrac{7}{4}\\
c) 2P(x)+5Q(x)=21x^4-5x^3-2x^2-\dfrac{1}{2}x-\dfrac{21}{4}\\
4P(x)-3Q(x)=3x^4+29x^3-4x^2-x-\dfrac{29}{4}\\$
Giải thích các bước giải:
$a)
P(x)=-2x^2+3x^4+5x^3+x^2-\dfrac{1}{4}x-2\\
=3x^4+5x^3+(-2x^2+x^2)-\dfrac{1}{4}x-2\\
=3x^4+5x^3-x^2-\dfrac{1}{4}x-2\\
Q(x)=3x^4+x^2-\dfrac{1}{4}-3x^3-x^2\\
=3x^4-3x^3+(x^2-x^2)-\dfrac{1}{4}\\
=3x^4-3x^3-\dfrac{1}{4}\\
b) P(x)+Q(x)=3x^4+5x^3-x^2-\dfrac{1}{4}x-2+3x^4-3x^3-\dfrac{1}{4}\\
=(3x^4+3x^4)+(5x^3-3x^3)-x^2-\dfrac{1}{4}x+\left ( -2-\dfrac{1}{4} \right )\\
=6x^4+2x^3-x^2-\dfrac{1}{4}x-\dfrac{9}{4}\\
P(x)-Q(x)=3x^4+5x^3-x^2-\dfrac{1}{4}x-2-(3x^4-3x^3-\dfrac{1}{4})\\
=3x^4+5x^3-x^2-\dfrac{1}{4}x-2-3x^4+3x^3+\dfrac{1}{4}\\
=(3x^4-3x^4)+(5x^3+3x^3)-x^2-\dfrac{1}{4}x+\left ( -2+\dfrac{1}{4} \right )\\
=8x^3-x^2-\dfrac{1}{4}-\dfrac{7}{4}\\
c) 2P(x)+5Q(x)=2\left (3x^4+5x^3-x^2-\dfrac{1}{4}x-2 \right )+5\left ( 3x^4-3x^3-\dfrac{1}{4} \right )\\
=6x^4+10x^3-2x^2-2.\dfrac{1}{4}x-4 +15x^4-15x^3-5.\dfrac{1}{4} \\
=(6x^4+15x^4)+(10x^3-15x^3)-2x^2-\dfrac{1}{2}x+\left ( -4-\dfrac{5}{4} \right )\\
=21x^4-5x^3-2x^2-\dfrac{1}{2}x-\dfrac{21}{4}\\
4P(x)-3Q(x)=4\left (3x^4+5x^3-x^2-\dfrac{1}{4}x-2 \right )-3\left ( 3x^4-3x^3-\dfrac{1}{4} \right )\\
=12x^4+20x^3-4x^2-4.\dfrac{1}{4}x-8 -9x^4+9x^3+3.\dfrac{1}{4} \\
=(12x^4-9x^4)+(20x^3+9x^3)-4x^2-x+\left ( -8+\dfrac{3}{4} \right )\\
=3x^4+29x^3-4x^2-x-\dfrac{29}{4}\\$