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 A – B – C:
−10x2+2xy- 10{{{x}}^2} + 2{{x}}y−10x2+2xy
−2x2−10xy- 2{{{x}}^2} - 10{{x}}y−2x2−10xy
2x2+10xy2{{{x}}^2} + 10{{x}}y2x2+10xy
2x2−10xy2{{{x}}^2} - 10{{x}}y2x2−10xy
Đáp án: <p><span data-type="math-inline" data-latex="2{{{x}}^2} - 10{{x}}y">$2{{{x}}^2} - 10{{x}}y$</span></p>
Phương pháp giải:
Áp dụng quy tắc trừ các đa thức.
Lời giải chi tiết:
A−B−C=(4x2−5xy+3y2)−(3x2+2xy+y2)−(−x2+3xy+2y2)=4x2−5xy+3y2−3x2−2xy−y2+x2−3xy−2y2=(4x2−3x2+x2)+(−5xy−2xy−3xy)+(3y2−y2−2y2)=2x2−10xy\begin{array}{l}A - B - C = \left( {4{{{x}}^2} - 5{{x}}y + 3{y^2}} \right) - \left( {3{{{x}}^2} + 2{{x}}y + {y^2}} \right) - \left( { - {x^2} + 3{{x}}y + 2{y^2}} \right)\\ = 4{{{x}}^2} - 5{{x}}y + 3{y^2} - 3{{{x}}^2} - 2{{x}}y - {y^2} + {x^2} - 3{{x}}y - 2{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)\\ = 2{{{x}}^2} - 10{{x}}y\end{array}A−B−C=(4x2−5xy+3y2)−(3x2+2xy+y2)−(−x2+3xy+2y2)=4x2−5xy+3y2−3x2−2xy−y2+x2−3xy−2y2=(4x2−3x2+x2)+(−5xy−2xy−3xy)+(3y2−y2−2y2)=2x2−10xy
