Đáp án:
Giải thích các bước giải:
\(12)\ 5xy(4x^2-3xy+y^2)-3x^2(5xy-4y^2-1)+5y\bigg(x^3+\dfrac35x^2y+y\bigg)\\ =20x^3y-15x^2y^2+5xy^3-15x^3y+12x^2y^2+3x^2+5x^3y+3x^2y^2+5y^2\\ =(20x^3y-15x^3y+5x^3y)+(-15x^2y^2+12x^2y^2+3x^2y^2)+5xy^3+3x^2+5y^2\\ =10x^3y+5xy^3+3x^2+5y^2\\ 13)\ 3xy(x^2-2xy+2y^2)+5x(x^2y+xy^2-y^3)-8y(x^3+x^2y)\\ =3x^3y-6x^2y^2+6xy^3+5x^3y+5x^2y^2-5xy^3-8x^3y-8x^2y^2\\ =(3x^3y+5x^3y-8x^3y)+(-6x^2y^2+5x^2y^2-8x^2y^2)+(6xy^3-5xy^3)\\ =-9x^2y^2+xy^3\\ 14)\ 7x^2(x^2-5x+2)-5x(x^3-7x^2+3x)\\ =7x^4-35x^3+14x^2-5x^4+35x^3-15x^2\\ =(7x^4-5x^4)+(-35x^3+35x^3)+(14x^2-15x^2)\\ =2x^4-x^2\\ \text{Thay x = $-\dfrac12$ vào biểu thức ta có:}\\ 2.\bigg(-\dfrac12\bigg)^4-\bigg(-\dfrac12\bigg)^2=2.\dfrac{1}{16}-\dfrac{1}{4}=\dfrac18-\dfrac14=-\dfrac18\\ 15)\ \dfrac34xy^2\bigg(x^2+\dfrac23xy+\dfrac43y^2\bigg)-\dfrac12xy\bigg(-\dfrac12x^2y+xy^2+y^3)\\ =\dfrac34x^3y^2+\dfrac12x^2y^3+xy^4+\dfrac14x^3y^2-\dfrac12x^2y^3-\dfrac12xy^4\\ =\bigg(\dfrac34x^3y^2+\dfrac14x^3y^2\bigg)+\bigg(\dfrac12x^2y^3-\dfrac12x^2y^3\bigg)+\bigg(xy^4-\dfrac12xy^4\bigg)\\ =x^3y^2+\dfrac12xy^4\\ \text{Thay x = $\dfrac12$ và y = 2 vào biểu thức ta có:}\\ \bigg(\dfrac12\bigg)^3.2^2+\dfrac12.\dfrac12.2^4=\dfrac18.4+\dfrac14.16=\dfrac12+4=\dfrac92\\ 16)\ 5x(x^2+2x-1)-3x^2(x-2)=x(2x^2-1)+4x(4x-1)\\ VT=5x(x^2+2x-1)-3x^2(x-2)\\ =5x^3+10x^2-5x-3x^3+6x^2\\ =(5x^3-3x^3)+(10x^2+6x^2)-5x\\ =2x^3+16x^2-5x\ \ \ \ \ \ (1)\\ VP=x(2x^2-1)+4x(4x-1)\\ =2x^3-x+16x^2-4x\\ =2x^3+16x^2+(-x-4x)\\ =2x^3+16x^2-5x\ \ \ \ \ \ (2)\\ (1);\ (2)\Rightarrow VT=VP\ \Rightarrow Đpcm.\\ 17)\ xy(2x^3-3y^3)-x^2y^2(5x+4y)=2x^2y(x^2-xy+y^2)-3xy^2(x^2+2xy+y^2)\\ VT=xy(2x^3-3y^3)-x^2y^2(5x+4y)\\ =2x^4y-3xy^4-5x^3y^2-4x^2y^3\ \ \ \ \ \ (1)\\ VP=2x^2y(x^2-xy+y^2)-3xy^2(x^2+2xy+y^2)\\ =2x^4y-2x^3y^2+2x^2y^3-3x^3y^2-6x^2y^3-3xy^4\\ =2x^4y-3xy^4+(-2x^3y^2-3x^3y^2)+(2x^2y^3-6x^2y^3)\\ =2x^4y-3xy^4-5x^3y^2-4x^2y^3\ \ \ \ \ \ (2)\\ (1);\ (2)\Rightarrow VT=VP\Rightarrow Đpcm\\ 18)\ 2y\bigg(x^3+x^2y-\dfrac14y^3\bigg)-\dfrac12x\bigg(2x^3+4xy^2-y^3\bigg)=\dfrac12y^3(x-y)-x^3(x-2y)\\ VT=2y\bigg(x^3+x^2y-\dfrac14y^3\bigg)-\dfrac12x\bigg(2x^3+4xy^2-y^3\bigg)\\ =2x^3y+2x^2y^2-\dfrac12y^4-x^4-2x^2y^2+\dfrac12xy^3\\ =2x^3y+(2x^2y^2-2x^2y^2)-\dfrac12y^4-x^4+\dfrac12xy^3\\ =2x^3y-\dfrac12y^4-x^4+\dfrac12xy^3\ \ \ \ \ \ (1)\\ VP=\dfrac12y^3(x-y)-x^3(x-2y)\\ =\dfrac12xy^3-\dfrac12y^4-x^4+2x^3y\\ =2x^3y-\dfrac12y^4-x^4+\dfrac12xy^3\ \ \ \ \ \ (2)\\ (1);\ (2)\Rightarrow VT=VP\Rightarrow Đpcm\)
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