a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)

\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)

\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)

d. \(\sqrt{9x^2+12x+4}=4\)

<=> \(\sqrt{\left(3x+2\right)^2}=4\)

<=> \(|3x+2|=4\)

<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)

\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)

\(\Leftrightarrow x=1\)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)

\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)

a) Áp dụng bđt AM-GM có:

\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)

\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)

Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)

Vậy...

b)Đk:\(x\ge2\)

Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)

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

Do \(x\ge2\Rightarrow x-1>0\)

Chia cả hai vế của pt cho x-1 ta được:

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

\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2+x-2-x^2+3x-2\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy S={2}

c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)

Thay x=3 vào pt thấy thỏa mãn

Vậy S={3}

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

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

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

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

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

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

câu a

Học tại nhà - Toán - Bài 110035

b,  ĐK \(x\ge-4\)

PT 

<=> \(\left(x-\sqrt{x+4}\right)+\left(\sqrt{2x^2-10x+17}-2x+3\right)=0\)

<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}+\frac{-2x^2+2x+8}{\sqrt{2x^2-10x+17}+2x-3}=0\)với \(x+\sqrt{x+4}\ne0\)

<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}-\frac{2\left(x^2-x-4\right)}{\sqrt{2x^2-10x+17}+2x-3}=0\)

<=> \(\orbr{\begin{cases}x^2-x-4=0\\\frac{1}{x+\sqrt{x+4}}-\frac{2}{\sqrt{2x^2-10x+17}+2x-3}=0\left(2\right)\end{cases}}\)

Giải (2)

=> \(2x+2\sqrt{x+4}=2x-3+\sqrt{2x^2-10x+17}\)

<=> \(\sqrt{2x^2-10x+17}=2\sqrt{x+4}+3\)

<=> \(2x^2-10x+17=4\left(x+4\right)+9+12\sqrt{x+4}\)

<=> \(x^2-7x-4=6\sqrt{x+4}\)

<=> \(\left(x-6\right)^2+5x-40=6\sqrt{6\left(x-6\right)-5x+40}\)

Đặt x-6=a;\(\sqrt{6\left(x-6\right)-5x+40}=b\)

=> \(\hept{\begin{cases}a^2+5x-40=6b\\b^2+5x-40=6a\end{cases}}\)

=> \(a^2-b^2+6\left(a-b\right)=0\)

<=> \(\orbr{\begin{cases}a=b\\a+b+6=0\end{cases}}\)

+ a=b

=> \(x-6=\sqrt{x+4}\)

=> \(\hept{\begin{cases}x\ge6\\x^2-13x+32=0\end{cases}}\)=> \(x=\frac{13+\sqrt{41}}{2}\)

+ a+b+6=0

=> \(x+\sqrt{x+4}=0\)(loại)

Vậy \(S=\left\{\frac{13+\sqrt{41}}{2};\frac{1+\sqrt{17}}{2}\right\}\)