\(\left|x+1\right|=x\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+1=x\)

\(\Leftrightarrow1=0\left(VLý\right)\)

Vậy \(S=\varnothing\)

\(|x+1|=x\) (đk: \(x\ge-1\))

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

<=> \(\left[{}\begin{matrix}0=-1\left(VLí\right)\\x=\dfrac{-1}{2}\left(TM\right)\end{matrix}\right.\)

Vậy nghiệm PT là \(S=\left\{\dfrac{-1}{2}\right\}\)

minh dang can gap nhe

a) \(|x|+x=\frac{1}{3}\)

\(|x|=\frac{1}{3}-x\)

Ta có: \(|x|\ge0\)

\(\Rightarrow\frac{1}{3}-x\ge0\)

\(\Rightarrow x\ge\frac{1}{3}>0\)

\(\Rightarrow|x|=x\)

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

\(\Rightarrow2x=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{6}\)

Vậy \(x=\frac{1}{6}\)

con phan b

\(\left(x-3\right)\left(x-5\right)+1=0\)

\(\Leftrightarrow x^2-5x-3x+15+1=0\)

\(\Leftrightarrow x^2-8x+16=0\)

\(\Leftrightarrow\left(x-4\right)^2=0\)

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)

Vậy \(x=4\)

\(\left(x-3\right)\left(x-5\right)+1=0\)

\(\Rightarrow x^2-3x-5x+15+1=0\)

\(\Rightarrow x^2-8x+16=0\)

\(\Rightarrow x^2-2x.4+4^2=0\)

\(\Rightarrow\left(x-4\right)^2=0\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

\(\left(x-3\right)\left(x-5\right)+1=0\)

\(\Leftrightarrow x^2-5x-3x+15+1=0\)

\(\Leftrightarrow x^2-8x+16=0\)

\(\Leftrightarrow\left(x-4\right)^2=0\)

\(\Leftrightarrow x=4\)

\(x+\frac{x}{3}=24\)

\(\frac{3x}{3}+\frac{x}{3}=24\)

\(\frac{4x}{3}=24\)

4x=24x3

4x=72

x=72:4

x=18

x+\(\frac{x}{3}\)=24

\(\frac{3x}{3}\)+\(\frac{x}{3}\)=24

\(\frac{3x+x}{3}\)=24

\(\frac{4x}{3}\)=24

4x=24x3

x=18