Bài 1:
1)ĐKXĐ:[tex]x\neq 0;x\neq \pm 3[/tex]
Rút gọn đc $A=\dfrac{-3x-9}{x}$
2)
[tex]A<-1\Leftrightarrow \dfrac{-3x-9}{x}<-1\\\Leftrightarrow \dfrac{-3x-9}{x}+1<0\\\Leftrightarrow \dfrac{-2x-9}{x}<0[/tex]
LBXD:
Vậy [tex]A<-1\iff \left[\begin{matrix}x<\dfrac{-9}{2}\\ x>0\end{matrix}\right.[/tex]
Bài 2:
[tex]1)x^4+4=x^4+4x^2+4-4x^2\\=(x^2+2)^2-(2x)^2\\=(x^2+2x+2)(x^2-2x+2)\\2)(x+2)(x+3)(x+4)(x+5)-24(1)\\=[(x+2)(x+5)][(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\\Đặt \ x^2+7x+11=y\\\Rightarrow (1)=(y-1)(y+1)-24\\=y^2-1-24\\=y^2-25=(y+5)(y-5)\\\Rightarrow (1)=(x^2+7x+11+5)(x^2+7x+11-5)\\=(x^2+7x+16)(x^2+7x+6)\\=(x+1)(x+6)(x^2+7x+16)[/tex]
Bài 3:
[tex]1)\dfrac{x-2}{2010}+\dfrac{x-3}{2009}+\dfrac{x-4}{2008}=\dfrac{x-2010}{2}+\dfrac{x-2009}{3}+\dfrac{x-2008}{4}\\\Leftrightarrow \dfrac{x-2}{2010}-1+\dfrac{x-3}{2009}-1+\dfrac{x-4}{2008}-1=\dfrac{x-2010}{2}-1+\dfrac{x-2009}{3}-1+\dfrac{x-2008}{4}-1\\\Leftrightarrow \dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}+\dfrac{x-2012}{2008}-\dfrac{x-2012}{2}-\dfrac{x-2012}{3}-\dfrac{x-2012}{4}=0\\\Leftrightarrow (x-2012)\left ( \dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4} \right )=0\\mà \ \dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}\neq 0\\\Rightarrow x-2012=0\\\Leftrightarrow x=2012\\Vậy...[/tex]
[tex]2)\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\\Leftrightarrow \dfrac{yz+xz+xy}{xyz}=0\\\Rightarrow yz+xz+xy=0\\Ta \ có:x^2+2yz=x^2+yz+yz=x^2+yz-zx-xy\\=x(x-z)-y(x-z)=(x-y)(x-z)\\Tương \ tự:y^2+2zx=(x-y)(z-y);z^2+2xy=(x-z)(y-z)\\\Rightarrow \dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2zx}+\dfrac{1}{z^2+2xy}\\=\dfrac{1}{(x-y)(x-z)}+\dfrac{1}{(x-y)(z-y)}+\dfrac{1}{(x-z)(y-z)}\\=\dfrac{y-z}{(x-y)(x-z)(y-z)}-\dfrac{x-z}{(x-y)(x-z)(y-z)}+\dfrac{x-y}{(x-y)(x-z)(y-z)}\\=\dfrac{y-z-x+z+x-y}{(x-y)(x-z)(y-z)}=0[/tex]
Bài 6:
$a)ĐKXĐ:x\neq -3\\A=\dfrac{2x+6}{x^3+27}=\dfrac{2(x+3)}{(x+3)(x^2-3x+9)}=\dfrac{2}{x^2-3x+9}\\x^2-3x+9=x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{4}\\=\left ( x-\dfrac{3}{2} \right )^2+\dfrac{27}{4}\geq \dfrac{27}{4}\\\Rightarrow A\leq \dfrac{2}{\dfrac{27}{4}}=\dfrac{8}{27}$
Dấu "=" xảy ra $\iff x=\dfrac{3}{2}$
Vậy $Max \ A=\dfrac{8}{27}\iff x=\dfrac{3}{2}$
[tex]b)\dfrac{3x^2-8x+6}{x^2-2x+1}=\dfrac{x^2-4x+4+2(x^2-2x+1)}{x^2-2x+1}\\=\dfrac{(x-2)^2}{(x-1)^2}+2\geq 2\Leftrightarrow x=2\\Vậy...[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
