Question

\(\sqrt{4x^2-20x+25}+2x=5\)

Answer 1

`\(\sqrt{4x^2-20x+25}+2x=5\)

\(\Leftrightarrow\sqrt{\left(2x-5\right)^2}=5-2x\)

\(\Leftrightarrow\left|2x-5\right|=5-2x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+2x=5+5=10\\2x-5=-5+2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{4}=\dfrac{5}{2}\\2x-5=2x-5\end{matrix}\right.\)(luôn đúng)

 

Answer 2

\(\Leftrightarrow\left|2x-5\right|=-2x+5\)

\(\Leftrightarrow2x-5< =0\)

hay x<=5/2

Answer 3

`=> sqrt (2x-5)^2 + 2x=  5`

`=> |2x - 5| = 5 - 2x`

`=> 2x - 5 = 5 - 2x` hoặc `5 -2x = 5 - 2x`

`TH_1 => x = 2,5`

`TH_2` luôn đúng.

Answer 4

ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2-20x+25}+2x=5\)

\(\Leftrightarrow\sqrt{\left(2x\right)^2-2\cdot2x\cdot5+5^2}+2x=5\)

\(\Leftrightarrow\sqrt{\left(2x-5\right)^2}=5-2x\)

\(\Leftrightarrow\left|2x-5\right|=5-2x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=5-2x\\2x-5=2x-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\0x=0\end{matrix}\right.\)\(\Leftrightarrow x\in R\)

Vậy phương trình có nghiệm là: \(x\in R\)

Answer 5

= 4x^2 - 20x + 25 = 25 - 20x + 4x^2

= \(x\le\dfrac{5}{2}\)