Question

[Thầy Cao Đô]

Câu 2. (3 điểm). Tìm x biết:

a) $\dfrac{3}{8}-\dfrac{1}{6}x=\dfrac{1}{4}$;                 

b) ${{\left( x-1 \right)}^{2}}=\dfrac{1}{4}$;                  

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

Answer 1

[Phan Tuấn Anh]

A. 3/5 

B. 2/8 

C. 2 

HT

Answer 2

[༺ღ¹⁷⁰⁶²⁰¹⁰H𝚘̷àทջ✎﹏ᑭh𝚘̷ทջღ²ᵏ¹⁰༻ღtea...]

a) \(\frac{3}{8}-\frac{1}{6}x=\frac{1}{4}\)

\(\frac{1}{6}x=\frac{3}{8}-\frac{1}{4}\)

\(\frac{1}{6}x=\frac{1}{8}\)

\(x=\frac{1}{8}\div\frac{1}{6}\)

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

Vậy \(x=\frac{3}{4}\)

b) \(\left(x-1\right)^2=\frac{1}{4}\)

\(\left(x-1\right)^2=\left(\pm\frac{2}{4}\right)^2\)

TH1: 

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

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

\(x-1=\frac{1}{2}\)

\(x=\frac{1}{2}+1\)

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

TH2:

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

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

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

\(x=\frac{1}{2}\)

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

c) \(\left(x-\frac{-1}{2}\right).\left(x+\frac{1}{3}\right)=0\)

TH1:

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

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

\(x=\frac{-1}{2}\)

TH2:

\(x+\frac{1}{3}=0\)

\(x=0-\frac{1}{3}\)

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

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