ĐKXĐ: \(x\ge\dfrac{5}{2}\)

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

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

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

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

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

\(\Leftrightarrow2x-5=25\)

\(\Leftrightarrow x=15\)

\(\sqrt{4x^2-20x+25}+2x=5\\ < =>\sqrt{\left(2x-5\right)^2}+2x=5\\ < =>\left|2x-5\right|+2x=5 \\ < =>\left[{}\begin{matrix}2x-5+2x=5\left(x\ge\dfrac{5}{2}\right)\\2x-5+2x=-5\left(x< \dfrac{5}{3}\right)\end{matrix}\right.< =>\left[{}\begin{matrix}4x=10< =>x=\dfrac{5}{2}\left(tmdk\right)\\4x=0< =>x=0\left(ktmdk\right)\end{matrix}\right.\\ =>x=\dfrac{5}{2}\)

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

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

\(\Leftrightarrow5-2x\ge0\) (tính chất: \(\left|A\right|=A\Leftrightarrow A\ge0\))

\(\Leftrightarrow x\le\dfrac{5}{2}\)

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

\(\sqrt{4x^2-20x+25}+2x=5\left(đk:x\le\dfrac{5}{2}\right)\)

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

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

a) ĐK: \(x^2\leq 5\)

Ta có: \(\sqrt{5-x^2}=x-1\)

\(\Rightarrow \left\{\begin{matrix} x-1\geq 0\\ (\sqrt{5-x^2})^2=(x-1)^2\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\geq 1\\ 5-x^2=x^2-2x+1\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\geq 1\\ 2x^2-2x-4=0\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\geq 1\\ x^2-x-2=0\end{matrix}\right.\)\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ (x-2)(x+1)=0\end{matrix}\right.\)

\(\Rightarrow x=2\)

b)

ĐK: \(x\geq \frac{5}{2}\)

Nhân cả 2 vế của pt với $\sqrt{2}$ thu được:

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

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

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

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

\(\Rightarrow |\sqrt{2x-5}-3|=3-\sqrt{2x-5}(*)\)

Nếu \(x\geq 7\Rightarrow |\sqrt{2x-5}-3|=\sqrt{2x-5}-3\)

$(*)$ trở thành: \(\sqrt{2x-5}-3=3-\sqrt{2x-5}\)

\(\Rightarrow \sqrt{2x-5}=3\Rightarrow x=7\) (thỏa mãn)

Nếu \(\frac{5}{2}\leq x< 7\Rightarrow |\sqrt{2x-5}-3|=3-\sqrt{2x-5}\)

$(*)$ trở thành:

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

Vậy pt có nghiệm $x=7$ hoặc $\frac{5}{2}\leq x< 7$

Hay PT có nghiệm thuộc \([\frac{5}{2}; 7]\)

c)
ĐK: \(x,y\geq 1\)

Áp dụng BĐT Cô-si cho các số không âm

\(\sqrt{y-1}=\sqrt{1(y-1)}\leq \frac{1+(y-1)}{2}=\frac{y}{2}\)

\(\Rightarrow x\sqrt{y-1}\leq \frac{xy}{2}\)

\(\sqrt{x-1}=\sqrt{1(x-1)}\leq \frac{1+(x-1)}{2}=\frac{x}{2}\)

\(\Rightarrow 2y\sqrt{x-1}\leq xy\)

Do đó:
\(x\sqrt{y-1}+2y\sqrt{x-1}\leq \frac{xy}{2}+xy=\frac{3}{2}xy\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix} y-1=1\\ x-1=1\end{matrix}\right.\Leftrightarrow x=y=2\)

Vậy pt có nghiệm $x=y=2$

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

Cái căn thứ 2 hình như thiếu số 3 đúng không?

HD: nhân 2 vế vs \(\sqrt{2}\). Sau đó phân tích trong căn thành hằng đẳng thức

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}\right)^2+2\sqrt{2x-3}\cdot1+1^2}+\sqrt{\left(\sqrt{2x-3}\right)+2\sqrt{2x-3}\cdot4+4^2}=5\)

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

\(\Leftrightarrow\sqrt{2x-3}=0\Leftrightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\)

ĐKXĐ: \(x\ge\dfrac{3}{2}\)

\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+2.4.\sqrt{2x-3}+16}=5\)

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

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

\(\Leftrightarrow2\sqrt{2x-3}=0\)

\(\Leftrightarrow2x-3=0\Rightarrow x=\dfrac{3}{2}\)