Question

Tìm x:                                                                                                                              a)\(\dfrac{1}{-x}=-4\)      b) \(\dfrac{x+7}{15}\dfrac{-24}{36}\)      c)\(\dfrac{x}{-3}=\dfrac{-12}{x}\)       d)\(\dfrac{-2}{x-3}=\dfrac{x-3}{-8}\)    

Answer 1

a: =>1/x=4

hay x=1/4

b: =>x+7=-10

=>x=-17

c: =>x²=36

=>x=6 hoặc x=-6

d: =>(x-3)²=16

=>x-3=4 hoặc x-3=-4

=>x=7 hoặc x=-1

Answer 2

\(a,\dfrac{1}{-x}=-4\\ \Rightarrow\left(-x\right)\left(-4\right)=1\\ \Rightarrow4x=1\\ \Rightarrow x=\dfrac{1}{4}\\ b,\dfrac{x+7}{15}=\dfrac{-24}{36}\\ \Rightarrow\dfrac{x+7}{15}=\dfrac{-2}{3}\\ \Rightarrow3\left(x+7\right)=-2.15\\ \Rightarrow3x+21=-30\\ \Rightarrow3x=-51\\ \Rightarrow x=-17\)

\(c,\dfrac{x}{-3}=\dfrac{-12}{x}\\ \Rightarrow x.x=\left(-3\right)\left(-12\right)\\ \Rightarrow x^2=36\\ \Rightarrow x=\pm6\)

\(d,\dfrac{-2}{x-3}=\dfrac{x-3}{-8}\\ \Rightarrow\left(x-3\right)\left(x-3\right)=\left(-2\right)\left(-8\right)\\ \Rightarrow\left(x-3\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

Answer 3

a.\(\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)

b.\(\Leftrightarrow\left(x+7\right).36=-360\Leftrightarrow x+7=-10\Rightarrow x=-17\)

c.\(x^2=36\Leftrightarrow x=6;-6\)

d.\(-2.-8=\left(x-3\right)\left(x-3\right)\Leftrightarrow16=x^2-6x+9\Leftrightarrow x^2-6x-7=0\Rightarrow x=7;-1\)