Question

Bài 3 : Tìm x:

a) 4x\(^2\)-1=0

b) (x-1)\(^2\)=\(\frac{9}{16}\)

c) \(\sqrt{x}=4\)

d) \(\sqrt{x+1}=2\)

Answer 1

dê mà

Answer 2

a) \(4x^2-1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-1=0\\2x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-\frac{1}{2}\end{array}\right.\)

b) \(\left(x-1\right)^2=\frac{9}{16}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=\frac{3}{4}\\x-1=-\frac{3}{4}\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{7}{4}\\x=\frac{1}{4}\end{array}\right.\)

c) \(\sqrt{x}=4\left(ĐK:x\ge0\right)\)

\(\Leftrightarrow x=16\)

d) \(\sqrt{x+1}=2\left(ĐKx\ge-1\right)\)

\(\Leftrightarrow x+1=4\)

\(\Leftrightarrow x=3\)

Answer 3

a) \(\Leftrightarrow4x^2=1\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)

b) \(\Leftrightarrow x-1=\frac{3}{4}\Leftrightarrow x=\frac{3}{4}+1=\frac{7}{4}\)

c) điều kiện : \(x\ge0\)

\(\sqrt{x}=4\Leftrightarrow x=16\)

 

d) điều kiện : \(x+1\ge0\Leftrightarrow x\ge-1\)

\(\sqrt{x+1}=2\Leftrightarrow x+1=4\Leftrightarrow x=3\)