Cho đa thức
A=4x2−5xy+3y2B=3x2+2xy+y2C=−x2+3xy+2y2\begin{array}{l}A = 4{{{x}}^2} - 5{{x}}y + 3{y^2}\\B = 3{{{x}}^2} + 2{{x}}y + {y^2}\\C = - {x^2} + 3{{x}}y + 2{y^2}\end{array}A=4x2−5xy+3y2B=3x2+2xy+y2C=−x2+3xy+2y2
Tính C – A – B:
8x2+6xy+2y28{{{x}}^2} + 6{{x}}y + 2{y^2}8x2+6xy+2y2
−8x2+6xy−2y2- 8{{{x}}^2}{{ + 6x}}y - 2{y^2}−8x2+6xy−2y2
8x2−6xy−2y28{{{x}}^2}{{ - 6x}}y - 2{y^2}8x2−6xy−2y2
8x2−6xy+2y28{{{x}}^2} - 6{{x}}y + 2{y^2}8x2−6xy+2y2
Đáp án: <p><span data-type="math-inline" data-latex="- 8{{{x}}^2}{{ + 6x}}y - 2{y^2}">$- 8{{{x}}^2}{{ + 6x}}y - 2{y^2}$</span></p>
Phương pháp giải:
Áp dụng quy tắc trừ các đa thức
Lời giải chi tiết:
C−A−B=(−x2+3xy+2y2)−(4x2−5xy+3y2)−(3x2+2xy+y2)=−x2+3xy+2y2−4x2+5xy−3y2−3x2−2xy−y2=(−4x2−3x2−x2)+(5xy−2xy+3xy)+(−3y2−y2+2y2)=−8x2+6xy−2y2\begin{array}{l}C - A - B = \left( { - {x^2} + 3{{x}}y + 2{y^2}} \right) - \left( {4{{{x}}^2} - 5{{x}}y + 3{y^2}} \right) - \left( {3{{{x}}^2} + 2{{x}}y + {y^2}} \right) \\ = - {x^2} + 3{{x}}y + 2{y^2} - 4{{{x}}^2} + 5{{x}}y - 3{y^2} - 3{{{x}}^2} - 2{{x}}y - {y^2} \\ = \left( { - 4{{{x}}^2} - 3{{{x}}^2} - {x^2}} \right) + \left( {5{{x}}y - 2{{x}}y + 3{{x}}y} \right) + \left( { - 3{y^2} - {y^2} + 2{y^2}} \right)\\ = - 8{{{x}}^2}{{ + 6x}}y - 2{y^2}\end{array}C−A−B=(−x2+3xy+2y2)−(4x2−5xy+3y2)−(3x2+2xy+y2)=−x2+3xy+2y2−4x2+5xy−3y2−3x2−2xy−y2=(−4x2−3x2−x2)+(5xy−2xy+3xy)+(−3y2−y2+2y2)=−8x2+6xy−2y2
