Question

bài 1 tìm x
a. 5 - 3(x+4) = -1
b.(x-1) - (x+2) = 0
c.( \(\dfrac{1}{2}\) + x )-( \(\dfrac{1}{3}\) - x) = 0
d. 2x² - 3 = 5
e. x(2x -1) = 0  
g. \(\dfrac{1}{3}\) . x² - \(\dfrac{1}{6}\)=\(\dfrac{7}{6}\)

Answer 1

a. 5 - 3(x + 4) = -1

⇔ 5 - 3x - 12 = -1

⇔ 3x = -1 - 5 + 12

⇔ 3x = 6

⇔ x = 2

Answer 2

\(d,2x^2-3=5\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

\(e,x\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)

Answer 3

a)\(=>3\left(x+4\right)=6=>x+4=2=>x=-2\)

b)\(=>x-1-x-2=0\)

\(=>-3=0\left(vl\right)\) => x ko tồn tại

 

Answer 4

c)\(=>\dfrac{1}{2}+x-\dfrac{1}{3}+x=0=>\dfrac{1}{6}+2x=0\)

\(=>2x=-\dfrac{1}{6}=>x=-\dfrac{1}{12}\)

d)\(=>2x^2=8=>x^2=4=>\left[{}\begin{matrix}x=\sqrt{4}=2\\x=-\sqrt{4}=-2\end{matrix}\right.\)

 

Answer 5

\(g,\Leftrightarrow\dfrac{1}{3}x^2=\dfrac{4}{3}\)

\(\Leftrightarrow x^2=\dfrac{4}{3}:\dfrac{1}{3}\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

Answer 6

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

⇔ \(\dfrac{1}{2}\)+ x - \(\dfrac{1}{3}\)+ x = 0

⇔ \(\dfrac{1}{6}\)+ x = 0

⇔ x = \(\dfrac{-1}{6}\)