Đk: x>=0​

\(\Leftrightarrow\frac{x^2-1}{\sqrt[3]{\left(x^2+26\right)^2}+3\sqrt[3]{x^2+26}+9}+3\frac{x-1}{\sqrt{x}+1}+\frac{x-1}{\sqrt{x+3}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{\sqrt[3]{\left(x^2+26\right)^2}+3\sqrt[3]{x^2+26}+9}+\frac{3}{\sqrt{x}+1}+\frac{1}{\sqrt{x+3}+2}\right)=0\)

Với đk x>=0 ta có\(\frac{x+1}{\sqrt[3]{\left(x^2+26\right)^2}+3\sqrt[3]{x^2+26}+9}+\frac{3}{\sqrt{x}+1}+\frac{1}{\sqrt{x+3}+2}>0\)

pt <=> x-1=0<=>x=1 (tm)

Do có quá ít câu hỏi nên bạn nào trả lời được, mình sẽ xóa khỏi mục "Câu hỏi hay" nhé!

a, ĐK: \(x\le-1,x\ge3\)

\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)

\(\Leftrightarrow\left(2\sqrt{x^2-2x-3}+3\right).\left(\sqrt{x^2-2x-3}-1\right)=0\)

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

\(\Leftrightarrow x^2-2x-3=1\)

\(\Leftrightarrow x^2-2x-4=0\)

\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)

b, ĐK: \(-2\le x\le2\)

Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)

Khi đó phương trình tương đương:

\(3t-t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)

Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm

Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)

c, ĐK: \(0\le x\le9\)

Đặt \(\sqrt{9x-x^2}=t\left(0\le t\le\dfrac{9}{2}\right)\)

\(pt\Leftrightarrow9+2\sqrt{9x-x^2}=-x^2+9x+m\)

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

\(\Leftrightarrow-t^2+2t+9=m\)

Khi \(m=9,pt\Leftrightarrow-t^2+2t=0\Leftrightarrow\left[{}\begin{matrix}t=0\\t=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}9x-x^2=0\\9x-x^2=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=9\left(tm\right)\\x=\dfrac{9\pm\sqrt{65}}{2}\left(tm\right)\end{matrix}\right.\)

Phương trình đã cho có nghiệm khi phương trình \(m=f\left(t\right)=-t^2+2t+9\) có nghiệm

\(\Leftrightarrow minf\left(t\right)\le m\le maxf\left(t\right)\)

\(\Leftrightarrow-\dfrac{9}{4}\le m\le10\)

Tham khảo:

1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24

\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+3+4x+4\sqrt{x\left(x+3\right)}=4+x\left(x+3\right)+4\sqrt{x\left(x+3\right)}\)

\(\Leftrightarrow5x+3=4+x^2+3x\)

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

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

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)

\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\) \(\left(Đk:-3\le x\le3\right)\)

\(\sqrt{4\left(x+3\right)+8\sqrt{9-x^2}+4\left(3-x\right)}=x+2\sqrt{3-x}+4\)

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

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

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

\(4\left(x+3\right)=x^2+8x+14\)

\(x^2+4x+2=0\)

\(\Delta=16-8=8\)

\(\Delta>0\)=> phương trình có 2 nghiệm phân biệt

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

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