Ta có : \(\left(x-\frac{1}{2}\right)\left(1+5x\right)=0\)

=> \(\left[{}\begin{matrix}x-\frac{1}{2}=0\\1+5x=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{5}\end{matrix}\right.\)

Vậy phương trình trên có tập nghiệm là \(S=\left\{\frac{1}{2};-\frac{1}{5}\right\}\)

/5x-4/=/x+2/

5x-4=+-(x+2)

5x-4=x+2

5x-x=2+4

4x=6

x=3/2

+5x-4=-x-2

5x+x=-2+4

6x=2

x=1/3

Mình không chắc đâu

/5x-4/=/x+2/

TH1:5x-4=x+2

4+2=5x-x

6=4x

=>x=1,5

TH2:5x-4=-x+2

4+2=5x+x

6=6x

=>x=1

TH3:-5x-4=x+2

4+2=-5x-x

6=-6x

=>x=-1

Vậy x\(\in\)\(\left\{-1;1;1,5\right\}\)

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

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

\(\frac{-1}{20}.x=\frac{-2}{3}\)

\(x=\frac{-2}{3}:\frac{-1}{20}\)

x =\(\frac{-40}{3}\)

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

=> \(x.\left(\frac{3}{4}-\frac{4}{5}\right)\) = \(\frac{-2}{3}\)

=> \(x.\frac{-1}{20}\) = \(\frac{-2}{3}\)

=> \(x=\frac{-2}{3}:\frac{-1}{20}\)

=> \(x=\frac{40}{3}\)

\(=b\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(b-5\right)\)

S={x|x>3}

\(\Rightarrow2^{x+3}.2^4=2^6\\ \Rightarrow2^{x+3}=2^2\\ \Rightarrow x+3=2\\ \Rightarrow x=-1\)

2^x+3. 4² = 64

2^x+3 . 2⁴ = 2⁶

=> 2^x+3 = 2²

x + 3 = 2

x = 2 - 3 = -1