a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)

Ý là đề vầy chứ gì:

\(5^x.5^{x+1}.5^{x+2}=10^{18}:2^{18}\)

⇔\(5^{3x+3}=5^{18}\)

⇔\(3x+3=18\)

⇔\(x=5\)

Vậy x=5

\(5^{x+1}.5^x.5^{x+2}=\dfrac{1000...0}{2^{18}}\\\Rightarrow5^x.5.5^x.5^x.5^2=\dfrac{10^{18}}{2^{18}}\\\Rightarrow(5^x.5^x.5^x).(5.5^2)=5^{18}\\\Rightarrow5^{3x}.5^3=5^{18}\\\Rightarrow5^{3x}=5^{15}\\\Rightarrow3x=15\\\Rightarrow x=5\)

Vậy \(x=5\).

a: (x-1)(x+2)(-x-3)=0

=>(x-1)(x+2)(x+3)=0

=>\(\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)

b: (x-7)(x+3)<0

TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)

=>-3<x<7

mà x nguyên

nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)

câu hỏi hay......nhưng tui xin nhường cho các bn khác

Hãy tích đúng cho tui nha

THANKS

\(x\cdot\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

\(x\cdot\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x=3\end{cases}}\)

Phần cuối thay dấu ngoặc nhọn bằng dấu ngoặc vuông nha .

a,

\(x\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

vậy_

b,

\(\left(x+1\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

vậy_

c, x(x - 2)(x - 3) = 0

=> x = 0 hoặc x - 2 = 0 hoặc x - 3 = 0

=> x = 0 hoặc x = 2 hoặc x = 3

vậy_

\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

     Bài 1:

\(x\) \(\times\) \(\dfrac{15}{16}\) - \(x\) \(\times\) \(\dfrac{4}{16}\) = 2

\(x\) \(\times\) (\(\dfrac{15}{16}\) - \(\dfrac{4}{16}\)) = 2

\(x\) \(\times\) \(\dfrac{11}{16}\) = 2

\(x\) = 2 : \(\dfrac{11}{16}\)

\(x\) = 2 x \(\dfrac{16}{11}\)

\(x\) = \(\dfrac{32}{11}\)

Bài 2: 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{1}{x\left(x+1\right):2}\) = 1 : \(\dfrac{2011}{2012}\)

1 + 2\(\times\) ( \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{2\times6}\) + \(\dfrac{1}{2\times10}\) + ... + \(\dfrac{2}{2\times x\times\left(x+1\right)}\)) = \(\dfrac{2012}{2011}\)

1 + 2 \(\times\)(\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+...+ \(\dfrac{1}{x\times\left(x+1\right)}\)) = \(\dfrac{2012}{2011}\)

1 + 2 \(\times\) (\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\) + ... + \(\dfrac{1}{x\times\left(x+1\right)}\)) = 1 + \(\dfrac{1}{2011}\)

1 + 2\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\)) = 1 + \(\dfrac{1}{2011}\)

 1 + 2 \(\times\) (\(\dfrac{1}{2}\) - \(\dfrac{1}{x+1}\)) = 1 + \(\dfrac{1}{2011}\)

  1 + 1 - \(\dfrac{2}{x+1}\) = 1 + \(\dfrac{1}{2011}\)

    \(\dfrac{2}{x+1}\) = 1 + 1 - 1  - \(\dfrac{1}{2011}\)

    \(\dfrac{2}{x+1}\) =  2 - 1 - \(\dfrac{1}{2011}\)

  \(\dfrac{2}{x+1}\)  = 1 - \(\dfrac{1}{2011}\)

     \(\dfrac{2}{x+1}\) = \(\dfrac{2010}{2011}\)

       \(x\) + 1 = 2 : \(\dfrac{2010}{2011}\)

        \(x\) + 1 = \(\dfrac{2011}{1005}\)

         \(x\) = \(\dfrac{2011}{1005}\) - 1  = \(\dfrac{1006}{1005}\)(loại vì \(\dfrac{1006}{1005}\) không phải là số tự nhiên)

Vậy không có giá trị nào của \(x\) là số tự nhiên thỏa mãn đề bài. 

Bài 3:

\(\dfrac{x}{16}\) \(\times\) (2017 - 1) = 2

\(\dfrac{x}{16}\) \(\times\) 2016 = 2

 \(\dfrac{x}{16}\)             = 2 : 2016

 \(\dfrac{x}{16}\)            = \(\dfrac{1}{1008}\)

   \(x\)             = \(\dfrac{1}{1008}\) x 16

  \(x\) = \(\dfrac{1}{63}\)

Có sai không bạn

\(\frac{1}{3}+x\times\frac{2}{7}=\frac{11}{12}\)

\(x\times\frac{2}{7}=\frac{11}{12}-\frac{1}{3}\)

\(x\times\frac{2}{7}=\frac{7}{12}\)

\(x=\frac{7}{12}:\frac{2}{7}\)

\(x=\frac{49}{24}\)

~Moon~

\(\frac{1}{3}+x.\frac{2}{7}=\frac{11}{12}\)

\(x.\frac{2}{7}=\frac{11}{12}-\frac{1}{3}\)

\(x.\frac{2}{7}=\frac{7}{12}\)

\(x=\frac{7}{12}\div\frac{2}{7}\)

\(x=\frac{49}{24}\)

bạn làm thế này nhé

 1/3 + x . 2/7 = 11/12

<=> 2x/7=11/12 - 1/3

<=> 2x/7= 7/12

<=> 2x.12=7.7

<=> 24x = 49

<=> x = 49/24

    Vậy x = 49/24