\(a.\dfrac{3}{2}+\dfrac{-1}{3}< \dfrac{x}{6}< \dfrac{1}{9}+\dfrac{31}{18}\)

\(\Leftrightarrow\dfrac{7}{6}< \dfrac{x}{6}< \dfrac{11}{6}\)

\(\Leftrightarrow7< x< 11\)

\(\Leftrightarrow x\in\left\{8;9;10\right\}\)

\(b.\dfrac{-5}{12}+\dfrac{7}{12}+\dfrac{-1}{12}< \dfrac{x}{12}< \dfrac{2}{15}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{4}{12}\)

\(\Leftrightarrow1< x< 4\)

\(\Leftrightarrow x\in\left\{2;3\right\}\)

Bn cộng 2 phân số đó lại

a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)

\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)

hay x=0

Vậy: x=0

b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)

\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)

\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)

\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)

hay \(x=\dfrac{-1}{9}\)

Vậy: \(x=\dfrac{-1}{9}\)

\(a.x=\dfrac{1}{3}-\dfrac{1}{3}\)

\(x=0\)

\(b.x-\dfrac{1}{4}=\dfrac{-1}{2}\)

\(x=\dfrac{-1}{2}+\dfrac{1}{4}\)

\(x=\dfrac{-1}{4}\)

c. \(\dfrac{-1}{6}=\dfrac{3}{2x}\)

\(-2x=18\)

\(x=-9\)

d. \(\dfrac{4}{5}=\dfrac{-12}{9-x}\)

\(4.\left(9-x\right)=-60\)

\(9-x=-15\)

\(x=24\)

\(e.\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

\(\left(x+1\right)^2=9\)

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

\(\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

f.\(\dfrac{x-1}{-4}=\dfrac{-4}{x-1}\)

\(\left(x-1\right)^2=16\)

\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}+\dfrac{3}{\left(x+3\right)\left(x+6\right)}+...+\dfrac{3}{\left(x+9\right)\left(x+12\right)}=\dfrac{3}{16}\)

=>\(\dfrac{1}{x}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+6}+...+\dfrac{1}{x+9}-\dfrac{1}{x+12}=\dfrac{3}{16}\)=>\(\dfrac{1}{x}-\dfrac{1}{x+12}=\dfrac{3}{16}\)

=>\(\dfrac{x+12-x}{x\left(x+12\right)}=\dfrac{3}{16}\)

=>12/x(x+12)=3/16

=>4/x(x+12)=1/16

=>x(x+12)=64

=>x^2+12x-64=0

=>x^2+16x-4x-64=0

=>(x+16)(x-4)=0

=>x=4 hoặc x=-16

Tự học giúp bạn có được một gia tài
Jim Rohn – Triết lý cuộc đời

a) \(x+1\dfrac{4}{7}=2\dfrac{2}{3}\)

\(x+\dfrac{11}{7}=\dfrac{8}{3}\)

\(x=\dfrac{8}{3}-\dfrac{11}{7}\)

\(x=\dfrac{23}{21}\)

b) \(3\dfrac{9}{12}:x=4\dfrac{1}{6}\)

\(\dfrac{15}{4}:x=\dfrac{25}{6}\)

\(x=\dfrac{15}{4}:\dfrac{25}{6}\)

\(x=\dfrac{9}{10}\)

a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)

b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)

c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)

\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)

1) |x + 2| = 4

\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)

2) 3 – |2x + 1| = (-5)

\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)

3) 12 + |3 – x| = 9

\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)

=>\(x=\varnothing\) 

1) I x+2 I=4

\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)

2) \(3-|2x+1|=-5\)

\(\Leftrightarrow|2x+1|=8\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)

3) \(12+|3-x|=9\)

\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)

1; |\(x\) + 2| = 4

    \(\left[{}\begin{matrix}x+2=-4\\x+2=4\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-4-2\\x=4-2\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-6\\x=2\end{matrix}\right.\)

Vậy \(x\) \(\in\) {- 6; 2}

1. \(3-|2x+1|=-5\)

\(\Rightarrow|2x+1|=8\)

\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)

Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)

2.\(12+|3-x|=9\)

\(\Rightarrow|3-x|=-3\)

Mà \(|3-x|\ge0\forall x\)

\(\Rightarrow\)Vô lí

Vậy không có x

3.\(|x+9|=12+\left(-9\right)+2\)

\(\Rightarrow|x+9|=5\)

\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)

Vậy \(x\in\left\{-4;-14\right\}\)

4.\(5x-16=40+x\)

\(\Rightarrow5x-x=40+16\)

\(\Rightarrow4x=56\)

\(\Rightarrow x=14\)

Vậy \(x=14\)

5.\(5x-7=-21-2x\)

\(\Rightarrow5x+2x=-21+7\)

\(\Rightarrow7x=-14\)

\(\Rightarrow x=-2\)

Vậy \(x=-2\)

6.\(\left(2x-1\right)\left(y-2\right)=12\)

Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)

\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

Ta có bảng : (em tự xét bảng nhé)