Question

a,\(\sqrt{x+1}=7\)

b,(x-2).\(\left(x+\dfrac{2}{3}\right)\)>0

c,\(\left(\dfrac{2}{3}x-1\right).\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)

Answer 1

a) \(\sqrt{x+1}=7\Rightarrow x+1=49\Rightarrow x=48\)

b) \(\left(x-2\right).\left(x+\dfrac{2}{3}\right)>0\)

\(\Rightarrow\left(x-2\right).\left(x+\dfrac{2}{3}\right)\) cùng dấu

\(\Rightarrow\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\)

Với \(\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>2\\x>-\dfrac{2}{3}\end{matrix}\right.\Rightarrow x>2\)

Với \(\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 2\\x< -\dfrac{2}{3}\end{matrix}\right.\Rightarrow x< -\dfrac{2}{3}\)

Vậy \(\left[{}\begin{matrix}x>2\\x< -\dfrac{2}{3}\end{matrix}\right.\)

c) \(\left(\dfrac{2}{3}x-1\right).\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\\dfrac{3}{4}x+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Chúc bạn học tốt!!!!

Answer 2

a, \(\sqrt{x+1}=7\\ \Rightarrow x+1=49\\ \Rightarrow x=48\)

b,TH1:

\(\left\{{}\begin{matrix}x-2>0\\x +\dfrac{2}{3}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x>\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow x>2\)

TH2:

\(\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x< \dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow x< \dfrac{-2}{3}\)

=> Vậy 2<x< \(\dfrac{-2}{3}\)

c, TH1:

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

TH2:

\(\dfrac{3}{4}x+\dfrac{1}{2}=0\\ \Rightarrow\dfrac{3}{4}x=\dfrac{-1}{2}\\ \Rightarrow x=\dfrac{-2}{3}\)

Vậy x = \(\dfrac{3}{2};\dfrac{-2}{3}\)