Question

[Thiên Tử]

giúp nhanh vs đang cần gấp lắm luôn

Answer 1

[Trần Thiên Kim]

6.

a. \(\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}=2\)

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

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

Xét \(x< 1\):

(*) \(\Leftrightarrow1-x+3-x=2\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\left(ktm\right)\)

Xét \(1\le x< 3\) :

(*) \(\Leftrightarrow x-1+3-x=2\)

\(\Leftrightarrow2=2\left(vô.số.nghiệm\right)\)

Xét \(x\ge3\) :

(*) \(\Leftrightarrow x-1+x-3=2\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\left(tm\right)\)

Vậy pt đã cho có nghiệm thỏa \(1\le x\le3\).

b. \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\) (ĐK: \(1\ge x\ge\dfrac{1}{2}\))

\(\Leftrightarrow x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{x^2-\sqrt{\left(2x-1\right)^2}}=2\)

\(\Leftrightarrow2x+2\sqrt{x^2-2x+1}=2\)

\(\Leftrightarrow2\sqrt{\left(x-1\right)^2}=2-2x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=1-x\\x-1=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\0=0\left(vô.số.nghiệm\right)\end{matrix}\right.\)

Vậy pt đã cho có nghiệm thỏa \(1\ge x\ge\dfrac{1}{2}\)