a, \(\Leftrightarrow2x^2=72\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x=\pm6\)

Vậy ...

\(b,\Leftrightarrow\dfrac{3}{5}x-0,75=2\dfrac{4}{5}.\dfrac{3}{7}=\dfrac{6}{5}\)

\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{6}{5}+0,75=\dfrac{39}{20}\)

\(\Leftrightarrow x=\dfrac{39}{20}:\dfrac{3}{5}=\dfrac{13}{4}\)

Vậy ...

\(c,\Leftrightarrow2x=1\dfrac{5}{6}.\dfrac{6}{11}-\dfrac{3}{10}=\dfrac{7}{10}\)

\(\Leftrightarrow x=\dfrac{7}{10}:2=\dfrac{7}{20}\)

Vậy ...

\(d,\Leftrightarrow\dfrac{1}{x-7\dfrac{1}{3}}=1.5:2\dfrac{1}{4}=\dfrac{2}{3}\)

\(\Leftrightarrow x-7\dfrac{1}{3}=\dfrac{3}{2}\)

\(\Leftrightarrow x=\dfrac{3}{2}+7\dfrac{1}{3}=\dfrac{53}{6}\)

Vậy ...

a) 2x² - 72 = 0

\(\Rightarrow\) 2x² = 72

\(\Rightarrow\) x² = 36 = 6² = (- 6)²

\(\Rightarrow\) x = 6 hoặc x = - 6

Vậy x = 6 hoặc x = - 6

b) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(2\dfrac{4}{5}\)

\(\Rightarrow\)  (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(\dfrac{14}{5}\)

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

\(\Rightarrow\) \(\dfrac{3}{5}\)x = \(\dfrac{39}{20}\)

\(\Rightarrow\) x = \(\dfrac{13}{4}\)

Vậy x = \(\dfrac{13}{4}\)

c) \(2x+\dfrac{3}{10}=1\dfrac{5}{6}.\dfrac{6}{11}\)

\(\Rightarrow\) \(2x+\dfrac{3}{10}=\dfrac{11}{6}.\dfrac{6}{11}=1\)

\(\Rightarrow\) \(2x=\dfrac{7}{10}\)

\(\Rightarrow\) \(x=\dfrac{7}{20}\)

Vậy \(x=\dfrac{7}{20}\)

d) \(2\dfrac{1}{4}:\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\Rightarrow\) \(\dfrac{9}{4}:\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}\)

\(\Rightarrow\) \(x-\dfrac{22}{3}=\dfrac{3}{2}\)

\(\Rightarrow\) \(x=\dfrac{53}{6}\)

Vậy \(x=\dfrac{53}{6}\)

1) 2x – (3 – 5x) = 4( x +3)

<=>2x-3+5x=4x+12

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2

tham khảo

1) 2x – (3 – 5x) = 4( x +3)

<=>2x-3+5x=4x+12

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2

a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

\(a,\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}=-\dfrac{3}{20}\\ b,\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\\ c,\Rightarrow\dfrac{1}{4}:x=-\dfrac{2}{5}-\dfrac{3}{4}=-\dfrac{23}{20}\\ \Rightarrow x=\dfrac{1}{4}:\left(-\dfrac{23}{20}\right)=-\dfrac{5}{23}\)

\(\dfrac{3}{2}x+\dfrac{3}{7}=-\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{3}{2}x=-\dfrac{43}{35}\)

\(\Leftrightarrow x=-\dfrac{86}{105}\)

Vậy \(x=-\dfrac{86}{105}\)

\(-\dfrac{11}{12}x+0,25=\dfrac{5}{6}\)

\(\Leftrightarrow-\dfrac{11}{12}x+\dfrac{1}{4}=\dfrac{5}{6}\)

\(\Leftrightarrow-\dfrac{11}{12}x=\dfrac{7}{12}\)

\(\Leftrightarrow x=-\dfrac{7}{11}\)

Vậy \(x=-\dfrac{7}{11}\)

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

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy x = {3; 1}\(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

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

Vậy \(x=-\dfrac{1}{2}\)

       (x - 2)² = 1

<=>\(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy x = 3; 1

(2x - 1)³ = -8

<=> 2x - 1 = -2

<=> 2x = -1

<=> x = \(\dfrac{-1}{2}\)

Vậy x = \(\dfrac{-1}{2}\)

\(|\) 2,5 - x\(|\) = 1,3

1,6- \(|\)x - 0,2 \(|\)= 0

13^x = 169

\(\dfrac{-2}{x}\)=\(\dfrac{-x}{8\dfrac{ }{25}}\)

\(\dfrac{x}{-15}\)=\(\dfrac{-60}{x}\)

\(|\)x\(|\)+0,573=2

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

1) \(\frac{17}{6}-\left(x-\frac{7}{6}\right)=\frac{7}{4}\)

\(\Rightarrow x-\frac{7}{6}=\frac{17}{6}-\frac{7}{4}\)

\(\Rightarrow x=\frac{13}{12}+\frac{7}{6}=\frac{9}{4}\)

2) \(\frac{3}{35}-\left(\frac{3}{5}-x\right)=\frac{2}{7}\)

\(\Rightarrow\)\(\frac{3}{5}-x=\frac{3}{35}-\frac{2}{7}=-\frac{1}{5}\)

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

3) 4) Hjhj^_^^_^