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\(\dfrac{2}{x^2-x-6}+\dfrac{x+1}{x^2+x-12}=\dfrac{x}{x^2+6x+8}\)
\(\Leftrightarrow\dfrac{2}{\left(x-3\right)\left(x+2\right)}+\dfrac{x+1}{\left(x-3\right)\left(x+4\right)}=\dfrac{x}{\left(x+2\right)\left(x+4\right)}\)
=> 2(x+4)+(x+1)(x+2)=x(x-3)
⇔2x+8+x2+2x+x+2=x2-3x
⇔x2+5x+10=x2-3x
⇔x2-x2+5x+3x=-10
⇔8x=-10
\(\Leftrightarrow\dfrac{-5}{4}\)
Vậy S={-\(\dfrac{5}{4}\)}
\(x\ne\left\{3;4;5;6\right\}\)
\(\frac{3}{x-3}-\frac{5}{x-5}=\frac{4}{x-4}-\frac{6}{x-6}\)
\(\Leftrightarrow\frac{3}{x-3}+1-\frac{5}{x-5}-1=\frac{4}{x-4}+1-\frac{6}{x-6}+1\)
\(\Leftrightarrow\frac{x}{x-3}-\frac{x}{x-5}=\frac{x}{x-4}-\frac{x}{x-6}\)
\(\Leftrightarrow x\left(\frac{1}{x-3}+\frac{1}{x-6}-\frac{1}{x-4}-\frac{1}{x-5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x-3}+\frac{1}{x-6}=\frac{1}{x-4}+\frac{1}{x-5}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{2x-9}{\left(x-3\right)\left(x-6\right)}=\frac{2x-9}{\left(x-4\right)\left(x-5\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-9=0\\\left(x-3\right)\left(x-6\right)=\left(x-4\right)\left(x-5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\18=20\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\overline{x}=\frac{\frac{9}{2}+0}{2}=\frac{9}{4}\)
\(A=\left(\frac{x+2}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
\(A=\left[\frac{\left(x+2\right)^2}{4-x^2}+\frac{4x^2}{4-x^2}-\frac{\left(2-x\right)^2}{4-x^2}\right]:\left[\frac{x\left(x-3\right)}{x^2.\left(2-x\right)}\right]\)
\(A=\left[\frac{x^2+4x+4+4x^2-4+4x-x^2}{4-x^2}\right]:\left[\frac{x-3}{x\left(2-x\right)}\right]\)
\(A=\frac{4x^2+8x}{4-x^2}:\frac{x-3}{x\left(2-x\right)}\)
\(A=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(A=\frac{4x^2}{x-3}\)
a.
$4(x+5)(x+6)(x+10)(x+12)=3x^2$
$4[(x+5)(x+12)][(x+6)(x+10)]=3x^2$
$4(x^2+17x+60)(x^2+16x+60)=3x^2$
Đặt $x^2+16x+60=a$ thì pt trở thành:
$4(a+x)a=3x^2$
$4a^2+4ax-3x^2=0$
$4a^2-2ax+6ax-3x^2=0$
$2a(2a-x)+3x(2a-x)=0$
$(2a-x)(2a+3x)=0$
Nếu $2a-x=0\Leftrightarrow 2(x^2+16x+60)-x=0$
$\Leftrightarrow 2x^2+31x+120=0\Rightarrow x=\frac{-15}{2}$ hoặc $x=-8$
Nếu $2a+3x=0\Leftrightarrow 2(x^2+16x+60)+3x=0$
$\Leftrightarrow 2x^2+35x+120=0\Rightarrow x=\frac{-35\pm \sqrt{265}}{4}$
b.
$(x+1)(x+2)(x+3)(x+6)=120x^2$
$[(x+1)(x+6)][(x+2)(x+3)]=120x^2$
$(x^2+7x+6)(x^2+5x+6)=120x^2$
Đặt $x^2+6=a$ thì pt trở thành:
$(a+7x)(a+5x)=120x^2$
$\Leftrightarrow a^2+12ax-85x^2=0$
$\Leftrightarrow a^2-5ax+17ax-85x^2=0$
$\Leftrightarrow a(a-5x)+17x(a-5x)=0$
$\Leftrightarrow (a-5x)(a+17x)=0$
Nếu $a-5x=0\Leftrightarrow x^2+6-5x=0$
$\Leftrightarrow (x-2)(x-3)=0\Rightarrow x=2$ hoặc $x=3$
Nếu $a+17x=0\Leftrightarrow x^2+17x+6=0$
$\Rightarrow x=\frac{-17\pm \sqrt{265}}{2}$
Vậy.........
bài 1:
\(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}\)
<=>\(\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}+-1\right)+\left(\dfrac{x-6}{1998}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}=0\)
<=>(x-2004)\(\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}\right)\)
vì 1/1994+1/1996+1/1998-1/2-1/4-1/6 khác 0
nên x-2004=0=>x=2004
vyaj.......
bài 2:
\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)
<=>\(\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)
<=>\(\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\)
<=>\(\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)
vì 1/15+1/13+1/11+1/9 khác 0
=>x-100=0<=>x=100
Lời giải:
Xét \(x< 0\Rightarrow \frac{x}{(x+2004)^2}< 0< \frac{1}{8016}\)
Xét \(x\geq 0\)
Ta có \((x+2004)^2-8016x=x^2+2004^2+4008x-8016x\)
\(=(x-2004)^2\geq 0\)
Suy ra \((x+2004)^2\geq 8016x\)
\(\Rightarrow \frac{x}{(x+2004)^2}\leq \frac{x}{8016x}=\frac{1}{8016}\)
Ta có đpcm