1.(x+1).(x+2).(x+3).(x+4)=120 2. 3/2x-16+3x-20/x-8+1/8=13x-102/3x-24

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1. Đáp án:

    $1) {\left[\begin{aligned}x=1\\x=-6\end{aligned}\right.}$

   $2) x=12$

   Giải thích các bước giải:

    $1)(x+1).(x+2).(x+3).(x+4)=120\\
   \Leftrightarrow \left [(x+1)(x+4)  \right ]\left [ (x+2)(x+3) \right ]=120\\
   \Leftrightarrow (x^2+4x+x+4)(x^2+3x+2x+6)=120\\
   \Leftrightarrow (x^2+5x+4)(x^2+5x+6)=120$
   Đặt $x^2+5x+4=t$
   $\Rightarrow t.(t+2)=120\\
   \Leftrightarrow t^2+2t-120=0\\
   \Leftrightarrow t^2+12t-10t-120=0\\
   \Leftrightarrow t(t+12)-10(t+12)=0\\
   \Leftrightarrow (t-10)(t+12)=0\\
   \Leftrightarrow {\left[\begin{aligned}t=10\\t=-12\end{aligned}\right.}\\
   \Rightarrow {\left[\begin{aligned}x^2+5x+4=10\\ x^2+5x+4=-12\end{aligned}\right.}\\
   \Leftrightarrow {\left[\begin{aligned}x^2+5x-6=0\\ x^2+5x+16=0\end{aligned}\right.}\\
   +) x^2+5x-6=0\\
   \Leftrightarrow x^2-x+6x-6=0\\
   \Leftrightarrow x(x-1)+6(x-1)=0\\
   \Leftrightarrow (x-1)(x+6)=0\\
   \Leftrightarrow {\left[\begin{aligned}x=1\\x=-6\end{aligned}\right.}\\
   +) x^2+5x+16=0\\
   \Leftrightarrow \left (x^2+5x+\frac{25}{4}  \right )+\frac{39}{4}=0\\
   \Leftrightarrow (x+\frac{5}{2})^2+\frac{39}{4}=0$
   Vì $(x+\frac{5}{2})^2>0\Rightarrow (x+\frac{5}{2})^2+\frac{39}{4}>0$
   Do đó phương trình vô nghiệm

   $2) \frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\\
   ĐK: x-8\neq 0\Leftrightarrow x\neq 0\\
   \Rightarrow \frac{3.12}{24(x-8)}+\frac{24(3x-20)}{24(x-8)}+\frac{3(x-8)}{24(x-8)}=\frac{8(13x-102)}{24(x-8)}\\
   \Leftrightarrow 36+72x-480+3x-24=104x-816\\
   \Leftrightarrow 75x-104x=-816+468\\
   \Leftrightarrow -29x=-348\\
   \Leftrightarrow x=12$

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