Bài 1:
a.
[tex](x+y+z)^3=(x+y)^3+z^3+3(x+y+z)(x+y)z=x^3+y^3+3xy(x+y)+3(x+y+z)(x+y)z[/tex]
\Rightarrow [tex]A=(x+y+z)^3-x^3-y^3-z^3=3(x+y)[xy+z(x+y+z)][/tex]
\Rightarrow [tex]A=3(x+y)(xy+xz+zy+z^2)=3(x+y)[x(y+z)+z(y+z)]=3(x+y)(y+z)(z+x)[/tex]
b.
[tex]B=x^3+y^3+z^3-3xyz=(x+y)^3-3xy(x+y)+z^3-3xyz=(x+y)^3+z^3-3xy(x+y+z)[/tex]
[tex]B=(x+y+z)[(x+y)^2+z^2-z(x+y)]-3xy(x+y+z)[/tex]
[tex]B=(x+y+z)(x^2+y^2+2xy+z^2-xz-zy-3xy)[/tex]
[tex]B=(x+y+z)(x^2+y^2+z^2-xz-zy-xy)[/tex]
c.
[tex]x^4-2x^3-2x^2-2x-3[/tex]-hình như sai đề
ĐỀ đúng
[TEX]x^4+2x^3-2x^2+2x-3[/tex]
[tex] = (x^4 + 3x^3+x^2+3x)-(x^3+3x^2+x+3)[/tex]
[tex]= x(x^3+3x^2+x+3) - (x^3+3x^2+x+3) [/tex]
[tex]= (x^3+3x^2+x+3)(x-1)[/tex]
[tex]=(x-1)[(x^3+3x^2) + (x+3)][/tex]
[tex] = (x-1)[x^2(x+3) + (x+3)][/tex]
[tex]= (x-1)(x+3)(x^2+1)[/TEX]
Bài 2
[tex] 2x^2-8x-10=2(x-2)^2-18 \geq -18[/tex]
Dấu = xảy ra \Leftrightarrow x=2
[tex]9x-2x^2=-\frac{1}{8}(16x^2-72x)=\frac{-1}{8}[(4x-9)^2-81]\leq \frac{81}{8}[/tex]
Cấm chép bài dưới \forall hình thức
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