a) \(\frac{-x}{2}+\frac{2x}{3}+x+\frac{1}{4}+2x+\frac{1}{6}=\frac{3}{8}.\)

\(\frac{-x}{2}+\frac{2x}{3}+3x+\frac{5}{12}=\frac{3}{8}\)

\(x.\left(-\frac{1}{2}+\frac{2}{3}+3\right)+\frac{5}{12}=\frac{3}{8}\)

\(x\cdot\frac{19}{6}=-\frac{1}{24}\)

x = -1/76

b) \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

\(\frac{3}{2x+1}+\frac{2.5}{2.\left(2x+1\right)}-\frac{2.3}{3.\left(2x+1\right)}=\frac{6}{13}\)

\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)

\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)

\(\frac{6}{2x+1}=\frac{6}{13}\)

=> 2x + 1 = 13

2x = 12

x = 6

\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)

a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)

ĐKXĐ: x ≠ -1

⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)

⇔ 65 + 52 = -3(x + 1)

⇔ 117 = -3x - 3

⇔ 117 + 3 = -3x

⇔ 120 = -3x 

⇔ x = \(\dfrac{120}{-3}=-40\) (TM)

b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)

⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)

⇔ 4x = -2,75

⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)

c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)

⇔  \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)

⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x² + 96x + 144x + 48

⇔ 936x + 1560x - 624x - 96x - 144x - 288x² = 48 - 312 - 520 + 312

⇔ 1632x - 288x² = -472

⇔ -288x² + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)

⇔ x = 5,942459684 \(\approx\) 6

c: Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Leftrightarrow2x+1=13\)

hay x=6

a)\(|4+5x|+5x=-4\)
\(|4+5x|=-4-5x\)
\(\Rightarrow\left[{}\begin{matrix}4+5x=-4-5x\\4+5x=-\left(-4-5x\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x+5x=-4-4\\4+5x=4+5x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}10x=-8\\5x-5x=4-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{8}{10}\Rightarrow x=\frac{4}{5}\\\\0=0\left(loai\right)\end{matrix}\right.\)
Vậy \(x=\frac{4}{5}\)

\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)

\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)

\(\Rightarrow27x+15=96\)

\(\Rightarrow27x=81\)

\(\Rightarrow x=3\left(tm\right)\)

\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\left(tm\right)\)

#Toru

a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\) 

\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)

\(\Rightarrow-6x+8x+3x+3+4x+2=32\)

\(\Rightarrow9x+5=32\)

\(\Rightarrow9x=32-5\)

\(\Rightarrow9x=27\)

\(\Rightarrow x=\dfrac{27}{9}\)

\(\Rightarrow x=3\)

b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\)) 

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=\dfrac{12}{2}\)

\(\Rightarrow x=6\left(tm\right)\)

Tìm x ∈ N

a) 2x chia hết cho 12 ⇒ 2x ∈ B(12) 

2x chia hết cho 30 ⇒ 2x ∈ B(30) 

Mà x có hai chữ số ⇒ 10 ≤ x ≤ 99 

\(\Rightarrow2x\in BC\left(12;30\right)\)

Mà: \(B\left(12\right)=\left\{0;12;24;36;48;60;72;84;96;108;...\right\}\)

\(B\left(30\right)=\left\{0;30;60;90;120;...\right\}\)

\(\Rightarrow BC\left(12;30\right)=\left\{0;60;...\right\}\)

\(\Rightarrow2x=60\)

\(\Rightarrow x=\dfrac{60}{2}\\ \Rightarrow x=30\)

b) \(9^{x+2}-9^{x+1}+9^x=657\)

\(\Rightarrow9^x\cdot\left(9^2-9+1\right)=957\)

\(\Rightarrow9^x\cdot\left(81-8\right)=657\)

\(\Rightarrow9^x\cdot73=657\)

\(\Rightarrow9^x=9\)

\(\Rightarrow9^x=9^1\)

\(\Rightarrow x=1\)