Đáp án:
Giải thích các bước giải:
$A(x) + B(x)=3x^4-\dfrac{3}{4}x^3+2x^2-3+8x^4+\dfrac{1}{5}x^3-9x+\dfrac{1}{2}$
$=3x^4-\dfrac{3}{4}x^3+2x^2-3+8x^4+\dfrac{1}{5}x^3-9x+\frac{1}{2}$
$=11x^4-\frac{11}{20}x^3+2x^2-9x+\dfrac{1}{2}-3$
$=-\dfrac{5}{2}-\dfrac{11x^3}{20}+11x^4+2x^2-9x$
$=11x^4-d\frac{11}{20}x^3+2x^2-9x-\dfrac{5}{2}$
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$A(x)-B(x)=3x^4-\dfrac{3}{4}x^3+2x^2-3-8x^4+\dfrac{1}{5}x^3-9x+\dfrac{1}{2}$
$=3x^4-8x^4-\dfrac{3}{4}x^3+\dfrac{1}{5}x^3+2x^2-9x+\dfrac{1}{2}-3$
$=-5x^4-\dfrac{11}{20}x^3+2x^2-9x+\frac{1}{2}-3$
$=-\dfrac{5}{2}-\dfrac{11}{20}x^3-5x^4+2x^2-9x$
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$B(x)-A(x)=8x^4+\dfrac{1}{5}x^3-9x+\dfrac{1}{2}-3x^4-\dfrac{3}{4}x^3+2x^2-3$
$=8x^4-3x^4-\dfrac{11}{20}x^3+2x^2-9x+\dfrac{1}{2}-3$
$=5x^4-\dfrac{11}{20}x^3+2x^2-9x+\dfrac{1}{2}-3$
$=5x^4-\frac{11}{20}x^3+2x^2-9x-\dfrac{5}{2}$
