Đáp án:
Bài 7: $x^2-5x+1$
Bài 8: $ N=(x^2-5xy+5y^2)^2$
Bài 9: $A\le 12$
Bài 10: $M=\dfrac13$
Bài 11: $P=-1$
Giải thích các bước giải:
Bài 7:
Ta có:
$(2x^4-10x^3-x^2+15x-3):(2x^2-3)$
$=((2x^4-3x^2)-(10x^3-15x)+(2x^2-3)):(2x^2-3)$
$=(x^2(2x^2-3)-5x(2x^2-3)+(2x^2-3)):(2x^2-3)$
$=(x^2-5x+1)(2x^2-3):(2x^2-3)$
$=x^2-5x+1$
Bài 8:
Ta có:
$N=(x-y)(x-2y)(x-3y)(x-4y)+y^4$
$\to N=(x-y)(x-4y)\cdot (x-2y)(x-3y)+y^4$
$\to N=(x^2-5xy+4y^2)\cdot (x^2-5xy+6y^2)+y^4$
$\to N=(x^2-5xy+4y^2)\cdot (x^2-5xy+4y^2+2y^2)+y^4$
$\to N=(x^2-5xy+4y^2)^2+2(x^2-5xy+4y^2)\cdot y^2+y^4$
$\to N=(x^2-5xy+4y^2+y^2)^2$
$\to N=(x^2-5xy+5y^2)^2$
Bài 9:
Ta có:
$A=2+2xy+14y-x^2-5y^2-2x$
$\to A=-x^2+2x(y-1)-5y^2+14y+2$
$\to A=-x^2+2x(y-1)-(y-1)^2-4y^2+12y+3$
$\to A=-(x^2-2x(y-1)+(y-1)^2)-(4y^2-4y\cdot 3+3^2)+12$
$\to A=-(x-(y-1))^2-(2y-3)^2+12$
$\to A=-(x-y+1)^2-(2y-3)^2+12$
$\to A\le -0-0+12=12$
Dấu = xảy ra khi $x-y+1=2y-3=0\to y=\dfrac32, x=\dfrac12$
Bài 10:
Ta có:
$4a^2+b^2=5ab$
$\to (4a^2-4ab)+(b^2-ab)=0$
$\to 4a(a-b)+b(b-a)=0$
$\to 4a(a-b)-b(a-b)=0$
$\to (4a-b)(a-b)=0$
Mà $2a>b>0$
$\to 4a-b=2a+(2a-b)>0$
$\to a-b=0$
$\to a=b$
$\to M=\dfrac{a^2}{4a^2-a^2}=\dfrac13$
Bài 11:
Ta có:
$x^2+y^2+z^2=9$
$\to x^2+y^2+z^2=3^2$
$\to x^2+y^2+z^2=(x+y+z)^2$
$\to x^2+y^2+z^2=x^2+y^2+z^2+2(xy+yz+zx)$
$\to xy+yz+zx=0$
Ta có:
$\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}$
$=\dfrac{-xy-zx}{x^2}+\dfrac{-xy-yz}{y^2}+\dfrac{-yz-zx}{z^2}$ vì $xy+yz+zx=0$
$=\dfrac{-y-z}{x}+\dfrac{-x-z}{y}+\dfrac{-y-x}{z}$
$=\dfrac{x-3}{x}+\dfrac{y-3}{y}+\dfrac{z-3}{z}$
$=1-\dfrac3x+1-\dfrac3y+1-\dfrac3z$
$=3-3(\dfrac1x+\dfrac1y+\dfrac1z)$
$=3-3\cdot \dfrac{xy+yz+zx}{xyz}$
$=3-3\cdot \dfrac{0}{xyz}$
$=3$
$\to \dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}-4=3-4=-1$
$\to P=(\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}-4)^{2019}=-1$
