a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)

b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)

\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)

c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)

\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)

e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)

f) \(\left|12-x\right|-7=5\)

th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)

th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)

i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)

k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)

a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)

b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)

Làm nốt nha

a) 2.(x-5) - 3.(x+7) = 14

2.x - 10 - 3.x - 21 = 14

2.x - 3.x - (10+21) = 14

-x - 31 = 14

-x = 45

x = -45

phần b bn cx lm tương tự như phần a nha!

c) 5- (x+7) = 4

5 - x - 7 = 4

5-7 -x = 4

-2 -x = 4

x = -6

câu d lm giống câu c

a) 2.(x-5) - 3.(x+7) = 14

2.x - 10 - 3.x - 21 = 14

2.x - 3.x - (10+21) = 14

-x - 31 = 14

-x = 45

x = -45

\(a)x+30\%x=-1,31\)

\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)

\(\Leftrightarrow10x+3x=-13,1\)

\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)

\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)

\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)

\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)

\(\Leftrightarrow\frac{6x-3}{2}=9\)

\(\Leftrightarrow6x-3=18\)

\(\Leftrightarrow x=\frac{7}{2}\)

a) (x + 1/2) . (2/3 − 2x) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)

\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)

\(\Rightarrow x=-\frac{2}{11}\)

c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)

\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)

\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)

\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)

\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)

\(\Rightarrow x=1\)

d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)

\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)

\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)

\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)

\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)

\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)

\(\Rightarrow x\approx28,7\) (số hơi lẻ)

e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)

a) \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x=-\dfrac{5}{6}+\dfrac{7}{3}=\dfrac{3}{2}\\ =>x=\dfrac{3}{2}:\dfrac{-1}{4}=-6\)

b) \(\left|x-\dfrac{1}{6}\right|+-\dfrac{5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\\ < =>\left|x-\dfrac{1}{6}\right|=\left(\dfrac{4}{7}.\dfrac{14}{48}\right)-\left(-\dfrac{5}{12}\right)=\dfrac{1}{6}+\dfrac{5}{12}=\dfrac{7}{12}\\ \)

Xảy ra 2 trường hợp:

+) TH1: \(x-\dfrac{1}{6}=\dfrac{7}{12}\\ =>x=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{3}{4}->\left(a\right)\)

+) TH2" \(-\left(x-\dfrac{1}{6}\right)=\dfrac{7}{12}\\ < =>-x+\dfrac{1}{6}=\dfrac{7}{12}\\ < =>-x=\dfrac{7}{12}-\dfrac{1}{6}=\dfrac{5}{12}\\ =>x=-\dfrac{5}{12}->\left(b\right)\)

Từ (a) và (b) => \(x\in\left\{-\dfrac{5}{12};\dfrac{3}{4}\right\}\)

a, \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=\dfrac{-5}{6}\)

\(\Rightarrow\dfrac{-1}{4}x=\dfrac{-5}{6}+\dfrac{7}{3}\)

\(\Rightarrow\dfrac{-1}{4}x=\dfrac{3}{2}\Rightarrow x=-6\)

Vậy \(x=-6\)

b, \(\left|x-\dfrac{1}{6}\right|+\dfrac{-5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\)

\(\Rightarrow\left|x-\dfrac{1}{6}\right|-\dfrac{5}{12}=\dfrac{1}{6}\)

\(\Rightarrow\left|x-\dfrac{1}{6}\right|=\dfrac{7}{12}\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{-7}{12}\\x-\dfrac{1}{6}=\dfrac{7}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{12}\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{-5}{12};\dfrac{3}{4}\right\}\)

Chúc bạn học tốt!!!