\(\frac{26-x}{995}\)+2+\(\frac{22-x}{997}\)+\(\frac{18-x}{999}\)+2=\(\frac{12-x}{334}\)+6

\(\frac{2016-x}{995.2}\)+\(\frac{2016-x}{997.2}\)+\(\frac{2016-x}{999.2}\)-\(\frac{2016-x}{334.6}\)=0

(2016-x)(\(\frac{1}{995.2}\)+\(\frac{1}{997.2}\)\(\frac{1}{999.2}\)-\(\frac{1}{334.6}\))=0

Dễ thấy giá trị biểu thức trong ngoặc thứ 2 khác 0

Nên 2016-x=0=>x=2016

Vậy giá trị của x là 2016

Bài 3:

\(a,\dfrac{x-1}{10}+\dfrac{x-1}{11}=\dfrac{x-1}{12}+\dfrac{x-1}{13}\)

\(\Rightarrow\dfrac{x-1}{10}+\dfrac{x-1}{11}-\dfrac{x-1}{12}-\dfrac{x-1}{13}=0\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)=0\)

Mà \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\ne0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

Vậy x = 1

b, \(\dfrac{x-2000}{10}+\dfrac{x-1999}{9}=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}\)

\(\Rightarrow\dfrac{x-2000}{10}+1+\dfrac{x-1999}{9}+1=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}+1\)

\(\Rightarrow\dfrac{x-1990}{10}+\dfrac{x-1990}{9}-\dfrac{x-1990}{8}-\dfrac{x-1990}{7}=0\)

\(\Rightarrow\left(x-1990\right)\left(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\right)=0\)

Mà \(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\ne0\)

\(\Rightarrow x-1990=0\Rightarrow x=1990\)

\(\frac{25-x}{995}+\frac{21-x}{997}+\frac{17-x}{999}=\frac{11-x}{334}\)

\(\Rightarrow\frac{25-x}{995}+2+\frac{21-x}{997}+\frac{17-x}{999}+2=\frac{11-x}{334}+6\)

\(\Rightarrow\frac{25-x}{995}+\frac{1990}{995}+\frac{21-x}{997}+\frac{1994}{997}+\frac{17-x}{999}+\frac{1998}{999}=\frac{11-x}{334}+\frac{2004}{334}\)

\(\Rightarrow\frac{2015-x}{995}+\frac{2015-x}{997}+\frac{2015-x}{999}=\frac{2015-x}{334}\)

\(\Rightarrow\frac{2015-x}{995}+\frac{2015-x}{997}+\frac{2015-x}{999}-\frac{2015-x}{334}=0\)

\(\Rightarrow\left(2015-x\right)\left(\frac{1}{995}+\frac{1}{997}+\frac{1}{999}+\frac{1}{334}\right)=0\)

\(\Rightarrow2015-x=0\left(\text{vì }\frac{1}{995}+\frac{1}{997}+\frac{1}{999}+\frac{1}{334}\ne0\right)\)

\(\Rightarrow x=2015\)

\(\left\{{}\begin{matrix}26⋮x\\x\ge13\end{matrix}\right.\Rightarrow x\in\left\{13;26\right\}\)

\(\left\{{}\begin{matrix}16⋮x\\x< 8\end{matrix}\right.\Rightarrow x\in\left\{1;2;4\right\}\)

\(\left\{{}\begin{matrix}18⋮x\\0< x< 40\end{matrix}\right.\Rightarrow x\in\left\{1;2;3;6;9;18\right\}\)

\(\left\{{}\begin{matrix}x⋮15\\30< x< 40\end{matrix}\right.\Rightarrow x\in\varnothing\)

\(\left\{{}\begin{matrix}x⋮12\\22\le5x\le50\end{matrix}\right.\Rightarrow x\in\varnothing\)

\(\left\{{}\begin{matrix}x⋮4\\16\le x\le36\end{matrix}\right.\Rightarrow x\in\left\{16;20;24;28;32;36\right\}\)

h)x² + 7x-30=6+7x

không phải trừ nha

\(197\text{≥}xx+xx< 199\)

không có x thỏa mãn ≤197;>199

⇒ x∃