Question

Tìm x biết:

a) \(\sqrt{16x^2-8x+1}-x=3\)

b) \(\sqrt{x^2+x+\frac{1}{4}}-3x=2\)

Answer 1

Chia 2 trường hợp:

+) T/h 1: ta có:

+) T/h 2: ta có:

Vậy ....................

Answer 2

(sorry nha, bị lỗi :<)

\(a.\sqrt{16x^2-8x+1}-x=3\\ \Leftrightarrow\sqrt{\left(4x\right)^2-2\cdot4x\cdot1+1}-x=3\\ \Leftrightarrow\sqrt{\left(4x-1\right)^2}-x=3\\ \Leftrightarrow\left|4x-1\right|-x=3\left(1\right)\)

Chia 2 trường hợp:

+) T/h 1: \(4x-1\ge0\Leftrightarrow x\ge\frac{1}{4}\)ta có:

\(\left(1\right)\Leftrightarrow4x-1-x=3\Leftrightarrow3x=4\Leftrightarrow x=\frac{4}{3}\left(tm\right)\)

+) T/h 2: \(4x-1< 0\Leftrightarrow x< \frac{1}{4}\)ta có:

\(\left(1\right)\Leftrightarrow1-4x-x=3\Leftrightarrow-5x=2\Leftrightarrow x=-\frac{2}{5}\left(tm\right)\)

Vậy ....................

Answer 3

\(b.\sqrt{x^2+x+\frac{1}{4}}-3x=2\\ \Leftrightarrow\sqrt{x^2+2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2}-3x=2\\ \Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}-3x=2\\ \Leftrightarrow\left|x+\frac{1}{2}\right|-3x=2\left(1\right)\)

Chia 2 trường hợp:

+) T/h 1: \(x\ge-\frac{1}{2}\)ta có:

\(\left(1\right)\Leftrightarrow x+\frac{1}{2}-3x=2\Leftrightarrow-2x=\frac{3}{2}\Leftrightarrow x=-\frac{3}{4}\left(ktm\right)\)

+) T/h 2: \(x< -\frac{1}{2}\)ta có:

\(\left(1\right)\Leftrightarrow-x-\frac{1}{2}-3x=2\Leftrightarrow-4x=\frac{5}{2}\Leftrightarrow x=-\frac{5}{8}\left(tm\right)\)

Vậy ....................