Question

Tìm x

1,(-14)+(2x–4²)=/-10/×/5/

2,(-100)–/x+20/=-150

3)3x+16=4x–61

4,2(x–5)–3(x+6)=0

Answer 1

\(\text{1,(-14)+(2x–4^2)=/-10/×/5/}\)

\(-14-16+2x=10.5\)

\(-30+2x=50\)

\(2x=50--30\)

\(2x=80\)

\(x=80:2\)

\(x=40\)

\(\text{2,(-100)–/x+20/=-150}\)

\(|x+20|=-100--150\)

\(|x+20|=50\)

\(\Rightarrow\orbr{\begin{cases}x+20=50\\x+20=-50\end{cases}\Leftrightarrow\orbr{\begin{cases}x=30\\x=-70\end{cases}}}\)

\(\Rightarrow x\in\left\{-70;30\right\}\)

\(\text{3)3x+16=4x–61}\)

\(3x-4x=-61-16\)

\(-x=-77\)

\(\Rightarrow x=77\)

\(\text{4,2(x–5)–3(x+6)=0}\)

\(2x-10-3x-18=0\)

\(2x-3x=0+18+10\)

\(-x=28\)

\(x=-28\)

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