\(\Leftrightarrow x\left(x-1\right)+7⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{2;0;8;-6\right\}\)

`[2-x]/x >= 1`

`<=>[2-x-x]/x >= 0`

`<=>[2-2x]/x >= 0`

`<=>0 < x <= 1`

     `->\bb B`

B

Ta có: \(\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{9}\right)>\dfrac{1}{9}.6=\dfrac{6}{9}>\dfrac{1}{2}\)  (1)

\(\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}\right)>\dfrac{1}{19}.10=\dfrac{10}{19}>\dfrac{1}{2}\)  (2)

\(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{19}>\left(1\right)+\left(2\right)\)

\(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{19}>1\left(đpcm\right)\)

1/4+1/5+1/6+...+1/19=(1/4+1/5+...+1/11)+(1/12+1/13+...+1/19)>(1/11+1/11+...+1/11)+(1/19+1/19+...+1/19)=8/11+8/19=240/209>209/209=1

⇒B>1

\(\Leftrightarrow n-1\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{2;0;4;-2\right\}\)

17,8 x 101 - 1

=(17,8 x 101) - 1

= 1797, 8 - 1

=1796,8 

\(17,8\times101-1=1797,8-1=1796,8\)

\(\left|x-2\right|+\left(x^2-2x\right)^{2014}=0\)

Ta có \(\hept{\begin{cases}\left|x+2\right|\ge0\\\left(x^2-2x\right)^{2014}\ge0\end{cases}\forall x}\)

\(\Rightarrow\left|x-2\right|+\left(x^2-2x\right)^{2014}\ge0\forall x\)

Do đó để \(\left|x-2\right|+\left(x^2-2x\right)^{2014}=0\) \(\Leftrightarrow\hept{\begin{cases}\left|x+2\right|=0\\\left(x^2-2x\right)^{2014}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-2=0\\x^2-2x=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2\\2^2-2.2=0\end{cases}}\)

\(\Leftrightarrow x=2\)

Vậy x = 2

@@ Học tốt

Chiyuki Fujito

1/51+1/52+1/53+....+1/100>1/100+1/100+1/100+...+1/100(50 so 0)=50/100=1/2

sai rồi

B

B. y=-3x-6