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_

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

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