Câu a:

2.(3\(x\) - \(\frac12\)) - 2\(x\) = \(\frac12\).(2\(x\) - 3)

6\(x\) - 1 - 2\(x\) = \(x\) - \(\frac32\)

6\(x\) - 2\(x\) - \(x\) = 1 - \(\frac32\)

4\(x\) - \(x\) = - \(\frac12\)

3\(x\) = - \(\frac12\)

\(x\) = - \(\frac12\) : 3

\(x=-\frac16\)

Vậy \(x=-\frac16\)

Câu b:

(2\(x\) - \(\frac35\))\(^2\) = \(\frac{4}{25}\)

(2\(x-\frac35\))\(^2\) = \(\left(\frac{2}{25}\right)\)\(^2\)

2\(x\) - \(\frac35\) = \(\frac25\) hoặc 2\(x\) - \(\frac35\) = - \(\frac25\)

TH: 2\(x\) - \(\frac35\) = \(\frac25\)

2\(x\) = \(\frac25+\frac35\)

2\(x\) = 1

\(x=\frac12\)

2\(x\) - \(\frac35\) = - \(\frac25\)

2\(x\) = - \(\frac25\) + \(\frac35\)

2\(x\) = \(\frac15\)

\(x\) = \(\frac{13}{25}\) : 2

\(x\) = \(\frac15\)

Vậy \(x\) ∈ {1/5; 1/2}

Bài 1:

a) \(x^2-3=1\)

\(\Rightarrow x^2=1+3=4\)

\(\Rightarrow x=\pm2\)

b)\(2x^3+12=-4\)

\(\Rightarrow2x^3=-4-12=-16\)

\(\Rightarrow x^3=-8\)

\(\Rightarrow x=-2\)

c)\(\left(2x-3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{7}{2}\\-\dfrac{1}{2}\end{matrix}\right.\)

a) \(x^2-3=1\Rightarrow x^2=4\Rightarrow x=\pm2\)

b) \(2x^3+12=-4\Rightarrow2x^3=-16\)

\(\Rightarrow x^3=-\dfrac{16}{2}=-8=-2^3\)

\(\Rightarrow x=-2\)

c) \(\left(2x-3\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d,h,i,k cững tương tự....

a, \(\left(2x-1\right)=-8\)

\(2x=-8+1\)

\(2x=-7\)

\(x=-7:2\)

\(x=-3,5\)

a) (2x - 1) = -8

⇒ 2x = -8 + 1

⇒ 2x = -7

b) (3x - 2)\(^2\) = \(\frac{1}{49}\)

Ta có: \(\frac{1}{49}\) = \(\frac{1}{7}\). \(\frac{1}{7}\) hoặc \(\frac{1}{49}\) = \(\frac{-1}{7}\). \(\frac{-1}{7}\)

TH1: 3x - 2 = \(\frac{1}{7}\) TH2: 3x - 2 = \(\frac{-1}{7}\)

⇒ 3x = \(\frac{1}{7}\)+2 ⇒ 3x = \(\frac{-1}{7}\)+2

⇒ 3x = \(\frac{15}{7}\) ⇒ 3x = \(\frac{13}{7}\)

⇒ x = \(\frac{5}{7}\) ⇒ x = \(\frac{13}{21}\)

Vậy: x = \(\frac{5}{7}\) hoặc x = \(\frac{13}{21}\)

a) \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

b)\(\orbr{\begin{cases}3x=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)

c)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)

d)\(\orbr{\begin{cases}x^2\\x+4=0\end{cases}=0\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)

e)\(\orbr{\begin{cases}\left(x+1\right)^2\\3x-5=0\end{cases}=0}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)

g)\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varphi\)

h)Tương tự các câu trên

i) x = 0

k)\(\left(\frac{3}{4}\right)^x=1=\left(\frac{3}{4}\right)^0\Rightarrow x=0\)

l)\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}=\left(\frac{2}{5}\right)^3\)

=> x + 1 = 3 => x = 2

x.(x+1)=0

suy ra x=0 hoac x+1=0

                               x=0-1

                              x=-1

vay x=0 hoac  x=-1

mấy câu sau cũng làm tương tự

a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)

        \(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)

        \(\frac{2}{5}-x=-3\)

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

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

                   \(x=\frac{3}{5}-\frac{15}{5}\)

                   \(x=-\frac{12}{5}\)

Vay \(x=-\frac{12}{5}\) 

b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)

        \(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)

        \(-3+\frac{3}{x}=\frac{-25}{12}\)

                     \(\frac{3}{x}=\frac{-25}{12}+3\)

                      \(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)

                      \(\frac{3}{x}=\frac{5}{6}\)

                      \(\frac{18}{6x}=\frac{5x}{6x}\)

Đèn dây , bạn tự làm tiếp nhé , de rồi chứ

1) \(\left(\frac{2x}{3}-3\right):\left(-10\right)=\frac{2}{5}\)

\(\Leftrightarrow-\frac{\frac{2x}{3}-3}{10}=\frac{2}{5}\)

\(\Leftrightarrow-\left(\frac{\frac{2x}{3}}{10}-\frac{3}{10}\right)=\frac{2}{5}\)

\(\Leftrightarrow-\left(\frac{2x}{3\times10}-\frac{3}{10}\right)=\frac{2}{5}\)

\(\Leftrightarrow-\left(\frac{2x}{30}-\frac{3}{10}\right)=\frac{2}{5}\)

\(\Leftrightarrow-\frac{x}{15}+\frac{3}{10}=\frac{2}{5}\)

\(\Leftrightarrow\frac{3}{10}-\frac{x}{15}=\frac{2}{5}\)

\(\Leftrightarrow-\frac{x}{15}=\frac{2}{5}-\frac{3}{10}\)

\(\Leftrightarrow-\frac{x}{15}=\frac{1}{10}\)

\(\Leftrightarrow-x=\frac{15}{10}\)

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

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

Vậy \(x=-\frac{3}{2}\)

2) \(\left|2x-1\right|+1=4\)

\(\Leftrightarrow\left|2x-1\right|=3\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

Vậy \(x\in\left\{2;-1\right\}\)

Câu a:

(\(x\) - \(\frac{-7}{8}\)) + 2,3 = 3,24

\(x+\frac78\) = 3,24 - 2,3

\(x\) + \(\frac78\) = 0,94

\(x\) = 0,94 - 7/8

\(x\) = 0,065

Vậy x = 0,065

Câu b:

x.(4,6 + 3/5) = 7,2 - 8,15

x.(4,6 + 3/5) = - 0,95

x.5,2 = - 0,95

x = -0,95 : 5,2

x = - 19/104

Vậy x = -19/104

a)

\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)

b)

\(\frac{1}{4}-(2x-1)^2=0\)

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

\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)

c)

\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)

\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)

\(\Leftrightarrow 5-x=\frac{-3}{4}\)

\(\Leftrightarrow x=\frac{23}{4}\)

d)

\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)

\(\Rightarrow x=3,8:2=1,9\)

e)

\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)

\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)

\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)

\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)

f)

\(5^{(x+5)(x^2-4)}=1\)

\(\Leftrightarrow (x+5)(x^2-4)=0\)

\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)

g)

\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)

\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)

h)

\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)

\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)

a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)

     \(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) =>   \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\)  =>   \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) =>   \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)

Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)

b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)

=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=>  \(\frac{27x}{4}=\frac{27}{40}\)

\(27x.40=27.4\)

\(1080.x=108\)

             \(x=\frac{1}{10}\)

Vậy \(x=\frac{1}{10}\)

c) \(\left|x-1\right|+4=6\)

\(\left|x-1\right|=6-4\)

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

\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=>  \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Vậy \(x=\left[3,-1\right]\)

d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)

e) \(\left(x^2-3\right)^2=16\)

\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)

\(x^2=7=>x=\sqrt{7}\)

Vậy \(x=\sqrt{7}\)

f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)

               \(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\) 

               \(\frac{2}{5}x=-\frac{4}{15}\)

          \(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)

Vậy \(x=-\frac{2}{3}\)

g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)

\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)

\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)

Vậy \(x=-3\)

k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)

\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)

\(\frac{2}{5}x=\frac{4}{15}\)

      \(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)

Vậy \(x=\frac{2}{15}\)

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

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

\(\frac{3}{5}x=\frac{5}{14}\)

\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)

Vậy \(x=\frac{25}{42}\)