Câu hỏi của Higurashi Kagome

Question

Tìm x biết$\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5$

Phùng Minh Quân · Accepted Answer

Ta có : $\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5$$\Leftrightarrow$$\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35$$\Leftrightarrow$$\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+3}{2002}+1\right)=32-35+3$$\Leftrightarrow$$\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=-3+3$$\Leftrightarrow$$\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0$Vì $\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0$Nên $x+2005=0$$\Rightarrow$$x=-2005$Vậy $x=-2005$Chúc bạn học tốt ~ Câu trả lời: Ta có : $\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5$$\Leftrightarrow$$\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35$$\Leftrightarrow$$\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+3}{2002}+1\right)=32-35+3$$\Leftrightarrow$$\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=-3+3$$\Leftrightarrow$$\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0$Vì $\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0$Nên $x+2005=0$$\Rightarrow$$x=-2005$Vậy $x=-2005$Chúc bạn học tốt ~

Dương Lam Hàng · Answer

Ta có: $\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5$$\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35$$\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=-3$$\Rightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1+\frac{x+3}{2002}+1=-3+3$$\Rightarrow\frac{x+1+2004}{2004}+\frac{x+2+2003}{2003}+\frac{x+3+2002}{2002}=0$$\Rightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=0$$\Rightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0$Vì $\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0$Nên x + 2005 = 0=> x                = -2005Vậy x = -2005Câu trả lời: Ta có: $\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5$$\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35$$\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=-3$$\Rightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1+\frac{x+3}{2002}+1=-3+3$$\Rightarrow\frac{x+1+2004}{2004}+\frac{x+2+2003}{2003}+\frac{x+3+2002}{2002}=0$$\Rightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=0$$\Rightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0$Vì $\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0$Nên x + 2005 = 0=> x                = -2005Vậy x = -2005

kudo shinichi · Answer

$\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5$$\frac{x+1}{2004}+1+\frac{x+2}{2003}+1+\frac{x+3}{2002}+1+32=32$ $\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=32-32$ $(x+2005).(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002})=0$ Mà  $\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0$$\Rightarrow x+2005=0$ $\Rightarrow x=-2005$Vậy x=-2005Câu trả lời: $\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5$$\frac{x+1}{2004}+1+\frac{x+2}{2003}+1+\frac{x+3}{2002}+1+32=32$ $\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=32-32$ $(x+2005).(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002})=0$ Mà  $\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0$$\Rightarrow x+2005=0$ $\Rightarrow x=-2005$Vậy x=-2005

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