Question

Giúp mình với ạ

Answer 1

a: Ta có: \(\sqrt{4x-3}=5\)

\(\Leftrightarrow4x-3=25\)

\(\Leftrightarrow4x=28\)

hay x=7

b: Ta có: \(\sqrt{2x+1}=7\)

\(\Leftrightarrow2x+1=49\)

\(\Leftrightarrow2x=48\)

hay x=24

c: Ta có: \(\sqrt{3x+2}-5=0\)

\(\Leftrightarrow3x+2=25\)

\(\Leftrightarrow3x=23\)

hay \(x=\dfrac{23}{3}\)

d: Ta có: \(\sqrt{1-4x}+3=0\)

\(\Leftrightarrow\sqrt{1-4x}=-3\)(Vô lý

Answer 2

e: Ta có: \(\sqrt{\left(x-1\right)^2}=3\)

\(\Leftrightarrow\left|x-1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

f: Ta có: \(\sqrt{x^2-6x+9}=2\)

\(\Leftrightarrow\left|x-3\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

g: Ta có: \(\sqrt{4x^2-4x+1}-3=0\)

\(\Leftrightarrow\left|2x-1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

h: Ta có: \(\sqrt{x^2-10x+25}=1\)

\(\Leftrightarrow\left|x-5\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=1\\x-5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\)

Answer 3

a) \(\sqrt{4x-3}=5\)

\(\Rightarrow4x-3=25\\ \Rightarrow4x=28\\ \Rightarrow x=7\)

b) \(\sqrt{2x+1}=7\)

\(\Rightarrow2x+1=49\\ \Rightarrow2x=48\\ \Rightarrow x=24\)

c) \(\sqrt{3x+2}-5=0\)

\(\Rightarrow\sqrt{3x+2}=5\\ \Rightarrow3x+2=25\\ \Rightarrow3x=23\\ \Rightarrow x=\dfrac{23}{3}\)

d) \(\sqrt{1-4x}+3=0\)

\(\Rightarrow\sqrt{1-4x}=-3\left(vôlí\right)\)

e)\(\sqrt{\left(x-1\right)^2}=3\)

\(\Rightarrow x-1=3\\ \Rightarrow x=4\)

f) \(\sqrt{x^2-6x+9}=0\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}=0\\ \Rightarrow x-3=0\\ \Rightarrow x=3\)

g) \(\sqrt{4x^2-4x+1}-3=0\)

\(\Rightarrow\sqrt{\left(2x-1\right)^2}=3\\ \Rightarrow2x-1=3\\ \Rightarrow2x=4\\ \Rightarrow x=2\)

h) \(\sqrt{x^2-10x+25}=1\)

\(\Rightarrow\sqrt{\left(x-5\right)^2}=1\\ \Rightarrow x-5=1\\ \Rightarrow x=6\)