ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=14\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)

\(\Leftrightarrow\left|\sqrt{2x-5}+1\right|+\left|\sqrt{2x-3}+3\right|=14\)

\(\Leftrightarrow2\sqrt{2x-5}=10\)

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

\(\Leftrightarrow2x-5=25\)

\(\Leftrightarrow x=15\)

Hong Ra On chuyên gì thế hả sao gọi mình là sao

\(\sqrt{x+2-3\sqrt{2x-5}}+\sqrt{x-2+\sqrt{2x-5}}=2\sqrt{2}\)

\(\left\{{}\begin{matrix}x\ge\dfrac{5}{2};y=\sqrt{2x-5};y\ge0\\\sqrt{\dfrac{\left(y-3\right)^2}{2}}+\sqrt{\dfrac{\left(y+1\right)^2}{2}}=2\sqrt{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x\ge\dfrac{5}{2};y=\sqrt{2x-5};y\ge0\\\left|\dfrac{\left(y-3\right)}{\sqrt{2}}\right|+\left|\dfrac{\left(y+1\right)}{\sqrt{2}}\right|=\left|\dfrac{4}{\sqrt{2}}\right|=2\sqrt{2}=VP\end{matrix}\right.\)đẳng thức khi

\(7\ge x\ge\dfrac{5}{2}\)

kết luận

nghiệm của pt là : \(7\ge x\ge\dfrac{5}{2}\)

ko

1: ĐKXĐ: 5/2<=x<=4

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

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

=>\(\frac{x-2-1}{\sqrt{x-2}+1}+\frac{4-x-1}{\sqrt{4-x}+1}+\frac{2x-5-1}{\sqrt{2x-5}+1}=2x^2-6x+x-3\)

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

=>\(\left(x-3\right)\left(\frac{1}{\sqrt{x-2}+1}-\frac{1}{\sqrt{4-x}+1}+\frac{2}{\sqrt{2x-5}+1}\right)-\left(x-3\right)\left(2x+1\right)=0\)

=>\(\left(x-3\right)\left(\frac{1}{\sqrt{x-2}+1}-\frac{1}{\sqrt{4-x}+1}+\frac{2}{\sqrt{2x-5}+1}-2x-1\right)=0\)

=>x-3=0

=>x=3(nhận)

Đặt \(2x-5=t^2\)ta có \(x=\frac{t^2+5}{2}\)thay giá trị của x vào phương trình đã cho được:

\(\sqrt{\frac{t^2+5}{2}-2+t}+\sqrt{\frac{t^2+5}{2}+2+3t}=7\sqrt{2}\)

hay \(\sqrt{t^2+5-2+2t}+\sqrt{t^2+5+4+6t}=14\)

\(\sqrt{t^2+2t+1}+\sqrt{t^2+6t+9}=14\)

\(\sqrt{\left(t+1\right)^2}+\sqrt{\left(t+3\right)^2}=14\)

\(t+1+t+3=14\)

\(2t+4=14\)

2t=10

t=5

Từ đó \(x=\frac{25+5}{2}=15\)

có một chút thiếu sót và sai nha ! cảm ơn bnaj đã tả lời câu hỏi này !

\(\sqrt{13}-\sqrt{12}=\frac{1}{\sqrt{13}+\sqrt{12}}\) ; \(\sqrt{7}-\sqrt{6}=\frac{1}{\sqrt{7}+\sqrt{6}}\)

Mà \(\sqrt{13}+\sqrt{12}>\sqrt{7}+\sqrt{6}\Rightarrow\frac{1}{\sqrt{13}+\sqrt{12}}< \frac{1}{\sqrt{7}+\sqrt{6}}\)

\(\Rightarrow\sqrt{13}-\sqrt{12}< \sqrt{7}-\sqrt{6}\)

ĐKXĐ: \(x\ge\frac{5}{2}\)

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

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

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

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

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

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

Dấu "=" xảy ra khi và chỉ khi \(1-\sqrt{2x-5}\ge0\Rightarrow2x-5\le1\Rightarrow x\le3\)

Vậy nghiệm của pt là \(\frac{5}{2}\le x\le3\)

a: ĐKXĐ: x>=-2

\(\sqrt{5x+10}=8-x\)

=>\(\begin{cases}8-x\ge0\\ \left(8-x\right)^2=5x+10\end{cases}\Rightarrow\begin{cases}x\le8\\ x^2-16x+64=5x+10\end{cases}\)

=>\(\begin{cases}-2\le x\le8\\ x^2-21x+54=0\end{cases}\Rightarrow\begin{cases}-2\le x\le8\\ \left(x-3\right)\left(x-18\right)=0\end{cases}\)

=>x=3

b: ĐKXĐ: \(4x^2+x-12\ge0\)

=>\(x^2+\frac14x-3\ge0\)

=>\(x^2+2\cdot x\cdot\frac18+\frac{1}{64}-\frac{193}{64}\ge0\)

=>\(\left(x+\frac18\right)^2\ge\frac{193}{64}\)

=>\(\left[\begin{array}{l}x+\frac18\ge\frac{\sqrt{193}}{8}\\ x+\frac18\le-\frac{\sqrt{193}}{8}\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\frac{\sqrt{193}-1}{8}\\ x\le\frac{-\sqrt{193}-1}{8}\end{array}\right.\)

\(\sqrt{4x^2+x-12}=3x-5\)

=>\(\begin{cases}3x-5\ge0\\ \left(3x-5\right)^2=4x^2+x-12\end{cases}\Rightarrow\begin{cases}3x\ge5\\ 9x^2-30x+25-4x^2-x+12=0\end{cases}\)

=>\(\begin{cases}x\ge\frac53\\ 5x^2-31x+37=0\end{cases}\)

\(\Delta=\left(-31\right)^2-4\cdot5\cdot37=221\) >0

=>Phương trình có hai nghiệm phân biệt là

\(\left[\begin{array}{l}x=\frac{31-\sqrt{221}}{2\cdot5}=\frac{31-\sqrt{221}}{10}\left(loại\right)\\ x=\frac{31+\sqrt{221}}{10}\left(nhận\right)\end{array}\right.\)