ĐKXĐ: \(x>4\)

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

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

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

\(\Leftrightarrow\dfrac{x+5}{x-4}-\dfrac{x-2}{x+3}=0\)

\(\Leftrightarrow\dfrac{(x+5)\left(x+3\right)-\left(x-2\right)\left(x-4\right)}{(x-4)\left(x+3\right)}=0\)

\(\Leftrightarrow(x+5)\left(x+3\right)-\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow x^2+8x+15-x^2+6x-8=0\)

\(\Leftrightarrow14x-7=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}+2}=\dfrac{3-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\dfrac{3}{\sqrt{x}+2}-1\)

Có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)\(\Leftrightarrow\dfrac{3}{\sqrt{x}+2}-1\le\dfrac{1}{2}\)\(\Leftrightarrow A\le\dfrac{1}{2}\)

Dấu "=" xảy ra khi x=0 (tm)

Vậy \(A_{max}=\dfrac{1}{2}\)

Bài 2:

Đk: \(x\ge3;y\ge5;z\ge4\)

Pt\(\Leftrightarrow\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}+\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}+\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}=20\)

Áp dụng AM-GM có:

\(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\ge2\sqrt{\sqrt{x-3}.\dfrac{4}{\sqrt{x-3}}}=4\)

\(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\ge6\)

\(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\ge10\)

Cộng vế với vế \(\Rightarrow VT\ge20\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-3}=\dfrac{4}{\sqrt{x-3}}\\\sqrt{y-5}=\dfrac{9}{\sqrt{y-5}}\\\sqrt{z-4}=\dfrac{25}{\sqrt{z-4}}\end{matrix}\right.\)\(\Leftrightarrow x=7;y=14;z=29\) (tm)

Vậy...

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

\(\Leftrightarrow\left|x+5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)

2) \(ĐK:x\ge2\)

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

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

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

\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

4) \(ĐK:x\ge0\)

\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)

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

\(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\left(x\ge1\right)\)

\(< =>5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)

\(< =>30\sqrt{x-1}-15\sqrt{x-1}=36+6\sqrt{x-1}\)

\(< =>9\sqrt{x-1}=36\\ < =>\sqrt{x-1}=4\\ < =>x-1=16\\ < =>x=17\left(tm\right)\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{1}{3}\sqrt{x-1}-\sqrt{x-1}=6\)

=>\(1.5\cdot\sqrt{x-1}=6\)

=>\(\sqrt{x-1}=4\)

=>x-1=16

=>x=17

\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\) (ĐK: \(x\ge0,x\ne1\))

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

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

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

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

\(\Leftrightarrow0=-2\) (vô lý)

⇒ Phương trình vô nghiệm

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

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

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

\(\Leftrightarrow-2=0\) (sai)

\(đk:x\ge0;x\ne1\)

\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ \Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)\\ \Rightarrow x-2\sqrt{x}+\sqrt{x}-2=x-\sqrt{x}\\ \Rightarrow-\sqrt{x}-2+\sqrt{x}=0\\ \Rightarrow-2=0\left(voli\right)\)

Vậy phương trình vô nghiệm