\(\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)

⇔ \(\sqrt{2x-5+2.3\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\)

⇔ \(\text{ |}\sqrt{2x-5}+3\text{ |}+\text{ |}\sqrt{2x-5}-1\text{ |}=4\)

⇔ \(\sqrt{2x-5}+3+\text{ |}\sqrt{2x-5}-1\text{ |}=4\) ( x ≥ \(\dfrac{5}{2}\) ) ( 1)

+) Với : \(\sqrt{2x-5}\text{≥}1\) ⇔ x ≥ 3 , ta có :

\(\left(1\right)\text{⇔}\sqrt{2x-5}+3+\sqrt{2x-5}-1=4\)

\(\text{⇔}2\sqrt{2x-5}=2\)

\(\text{⇔}x=3\left(TM\right)\)

+) Với : \(\sqrt{2x-5}< 1\) ⇔ x < 3 , ta có :

\(\left(1\right)\text{⇔}\sqrt{2x-5}+3+1-\sqrt{2x-5}=4\)

\(\text{⇔}4=4\) ( luôn đúng với : \(3>x\text{≥}\dfrac{5}{2}\) )

KL...............

\(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{2x-5-6\sqrt{2x-5}+9}+\sqrt{2x-5+2\sqrt{2x-5}+1}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}+1\right)^2}=4\)

\(\Leftrightarrow\left|\sqrt{2x-5}-3\right|+\left|\sqrt{2x-5}+1\right|=4\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-5}-3+\sqrt{2x-5}+1=4\\\sqrt{2x-5}-3+\sqrt{2x-5}+1=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{2x-5}-2=4\\2\sqrt{2x-5}-2=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{2x-5}=6\\2\sqrt{2x-5}=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-5}=3\\\sqrt{2x-5}=-1\left(L\right)\end{cases}}\)

\(\Leftrightarrow2x-5=9\)

\(\Leftrightarrow x=7\)

điều kiện 2x-5+3 >=0 và 2x-5-1>=0

<=>x>=1 và x>=3

=> x>=1

từ pt đã cho ta có

căn 2x-5+6(2x-5)+9 + căn 2x-5-2(2x-5)+1 = 4

<=>(2x-5+3)+(2x-5-1)=4

<=>4x-8=4

<=> 4x=12

<=>x=3(TMDKXD)

vậy x=3

\(ĐKXĐ:x\ge\frac{5}{2}\)

Ta có: \(\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)

\(\Leftrightarrow\left|\sqrt{2x-5}+3\right|+\left|\sqrt{2x-5}-1\right|=4\)

\(\Leftrightarrow\left|\sqrt{2x-5}+3\right|+\left|1-\sqrt{2x-5}\right|=4\)(1)

Có : \(VT\ge\left|\sqrt{2x-5}+3+1-\sqrt{2x-5}\right|=4\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{2x-5}+3\ge0\\1-\sqrt{2x-5}\ge0\end{cases}\Leftrightarrow-3\le\sqrt{2x-5}\le1}\)

                                                                             \(\Leftrightarrow0\le2x-5\le1\)

                                                                              \(\Leftrightarrow5\le2x\le6\)

                                                                             \(\Leftrightarrow\frac{5}{2}\le x\le3\)

KẾt hợp với ĐKXĐ được \(\frac{5}{2}\le x\le3\)

Vậy pt có nghiệm nằm trong khoảng \(\frac{5}{2}\le x\le3\)

ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{\left(2x-5\right)+2.\sqrt{2x-5}\cdot3+9}+\sqrt{\left(2x-5\right)-2\sqrt{2x-5}.1+1}=4\)\(\Leftrightarrow\sqrt{\left(2x-5+3\right)^2}+\sqrt{\left(2x-5-1\right)^2}=4\)

\(\Leftrightarrow\left|2x-2\right|+\left|2x-6\right|=4\)

\(\Leftrightarrow\left|x-1\right|+\left|x-3\right|=4\)

Xét x<1:

=>1-x+3-x=4

=>-2x=0

=>x=0

Xét \(1\le x< 3\)

=>x-1+3-x=4

=>0x=2(vô lý)

Xét \(x\ge3\)

=>x-1+x-3=4

=>2x=-2

=>x=-1

a/ \(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4\)

\(\Leftrightarrow x+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=4\)

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

Làm nốt

b/ \(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

\(\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)

Làm nốt

a/ \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=4\)

\(\Leftrightarrow x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=4\)

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

Làm nốt

b/ \(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)

Hắc Hường Mashiro Shiina Nguyễn Thanh Hằng Hiếu Cao Huy Phùng Khánh Linh Mến Vũ Hung nguyen Aki Tsuki Thiên Chỉ Hạc Trần Quốc Lộc trả lời giùm mình với

1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

2) ĐKXĐ: \(x\ge3\)

\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)

4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)