\(2x+3=2\sqrt{x+1}+\sqrt{2x+1}\left(đk:x\ge-\frac{1}{2}\right)\) (*)
Đặt \(2\sqrt{x+1}=a\left(a\ge0\right)\) , \(\sqrt{2x+1}=b\left(b\ge0\right)\)
Có \(a^2-b^2=4\left(x+1\right)-2x-1=4x+4-2x-1=2x+3\)
Có \(2x+3=a+b\)
=> \(a^2-b^2=a+b\)( do \(a^2-b^2=2x+3\))
<=> \(\left(a+b\right)\left(a-b\right)-\left(a+b\right)=0\)
<=> \(\left(a+b\right)\left(a-b-1\right)=0\)
=> \(\left[{}\begin{matrix}a=-b\\a=b+1\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}2\sqrt{x+1}=-\sqrt{2x+1}\\2\sqrt{x+1}=\sqrt{2x+1}+1\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}4\left(x+1\right)=2x+1\\4\left(x+1\right)=2x+1+2\sqrt{2x+1}+1\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}4x+4-2x-1=0\\4x+4-2x-1-1=2\sqrt{2x+1}\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}2x+3=0\\2x+2=2\sqrt{2x+1}\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=-\frac{3}{2}\left(ktm\right)\\x+1=\sqrt{2x+1}\end{matrix}\right.\)
=> \(x+1=\sqrt{2x+1}\)
<=> x2+2x+1=2x+1
<=> x2=0
<=>x=0(t/m pt (*))
Vậy pt (*) có tập nghiệm \(S=\left\{0\right\}\)
b, \(2+\sqrt{3-8x}=6x+\sqrt{4x-1}\) (*) (đk: \(\frac{1}{4}\le x\le\frac{3}{8}\))
<=>\(2-6x=\sqrt{4x-1}-\sqrt{3-8x}\)
Đặt \(\sqrt{3-8x}=a\left(a\ge0\right)\) , \(\sqrt{4x-1}=b\left(b\ge0\right)\)
Có \(\left\{{}\begin{matrix}a^2-b^2=3-8x-4x+1\\2-6x=b-a\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=4-12x\\2-6x=b-a\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=2\left(2-6x\right)\\2-6x=b-a\end{matrix}\right.\)
=> \(\left(a+b\right)\left(a-b\right)=2\left(b-a\right)\)
<=> \(\left(a+b\right)\left(a-b\right)-2\left(b-a\right)=0\)
<=> \(\left(a-b\right)\left(a+b+2\right)=0\)
=> a-b=0(do a+b+2 >0 với \(a;b\ge0\))
<=> a=b <=> \(\sqrt{3-8x}=\sqrt{4x-1}\)<=> \(3-8x=4x-1\)
<=> \(3+1=4x+8x\)<=> \(4=12x\)
<=> \(x=\frac{1}{3}\)
Vậy pt (*) có tập nghiệm \(S=\left\{\frac{1}{3}\right\}\)
