Question

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Bài 1: Tìm x:

a, \(\left[x\right]^2-5\left[x\right]+4=0\)

b, \(\left[x\right]^2-6\left[x\right]+8=0\)

Bài 2: Tìm x: \(2\left[x\right]=x+2\left\{x\right\}\)

Bài 3: Cho a;b;c;d là các số nguyên dương ; \(A=\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\). Tìm \(\left[A\right]\)

Answer 1

[Nguyễn Việt Lâm]

Bài 1:

a/ \(\Leftrightarrow\left(\left[x\right]-1\right)\left(\left[x\right]-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[x\right]=1\\\left[x\right]=4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}1\le x< 2\\4\le x< 5\end{matrix}\right.\)

b/ \(\Leftrightarrow\left(\left[x\right]-2\right)\left(\left[x\right]-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[x\right]=2\\\left[x\right]=4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2\le x< 3\\4\le x< 5\end{matrix}\right.\)

Bài 2:

\(\Leftrightarrow2\left[x\right]=\left[x\right]+\left\{x\right\}+2\left\{x\right\}\)

\(\Leftrightarrow\left[x\right]=3\left\{x\right\}\)

\(\Rightarrow0\le\left[x\right]< 3\)

- Với \(\left[x\right]=0\Rightarrow\left\{x\right\}=0\Rightarrow x=0\)

- Với \(\left[x\right]=1\Rightarrow\left\{x\right\}=\frac{1}{3}\Rightarrow x=\frac{4}{3}\)

- Với \(\left[x\right]=2\) \(\Rightarrow\left\{x\right\}=\frac{2}{3}\Rightarrow x=\frac{8}{3}\)

Bài 3:

\(A>\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=1\)

\(A< \frac{2a}{a+b+c+d}+\frac{2b}{a+b+c+d}+\frac{2c}{a+b+c+d}+\frac{2d}{a+b+c+d}=2\)

\(\Rightarrow1< A< 2\Rightarrow\left[A\right]=1\)

Answer 2

[👁💧👄💧👁]

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