a) pt đã cho \(\Leftrightarrow x^2-8=0\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{8}\left(N\right)\\x=\sqrt{8}\left(N\right)\end{matrix}\right.\)
Kl: x= +- căn 8
b) đk :....tự làm....
pt đã cho \(\Leftrightarrow x-1+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{\left(x-1\right)\left(x-7\right)}\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\\b=\sqrt{7-x}\end{matrix}\right.\)
...... từ từ, VT có căn(7-x) và VP có căn(x-7) thật à??...
a)
\(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)
\(\Leftrightarrow x^2+3x+1-3\left(x+3\right)=\left(x+3\right)\sqrt{x^2+1}-3\left(x+3\right)\)
\(\Leftrightarrow x^2-8=\left(x+3\right)\left(\sqrt{x^2+1}-3\right)\)
\(\Leftrightarrow x^2-8=\left(x+3\right).\dfrac{\left(\sqrt{x^2+1}-3\right).\left(\sqrt{x^2+1}+3\right)}{\sqrt{x^2+1}+3}\)
\(\Leftrightarrow x^2-8=\dfrac{\left(x+3\right)\left(x^2-8\right)}{\sqrt{x^2+1}+3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-8=0\\\sqrt{x^2+1}+3=x+3\end{matrix}\right.\)\(\begin{matrix}\left(1\right)\\\left(2\right)\end{matrix}\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x_1=-2\sqrt{2}\\x_2=2\sqrt{2}\end{matrix}\right.\) \(\left(2\right)\Leftrightarrow\sqrt{x^2+1}=x\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+1=x^2\end{matrix}\right.\) vô nghiệm
a) Đặt \(\sqrt{x^2+1}=t\) #, ta có pt:
\(t^2+3x=\left(x+3\right)t\)
\(\Leftrightarrow t^2-t\left(x+3\right)+3x=0\)
Giải \(\Delta=\left(x+3\right)^2-12x=x^2+6x+9-12x=\left(x-3\right)^{^{ }2}\) theo x
\(\Rightarrow\left\{{}\begin{matrix}t_1=\dfrac{-x-3+x-3}{2}=-3\\t_2=\dfrac{-x-3-x+3}{2}=-x\end{matrix}\right.\)
Thế t1=-3 vào # ta được:
\(\sqrt{x^2+1}=-3\)
<=> \(x^2+1=9\) <=> \(x^2=8\) <=> \(x=2\sqrt{2};x=-2\sqrt{2}\)
Thế t2=-x vào #, ta được:
\(\sqrt{x^2+1}=-x\)
<=> \(x^2+1=x^2\)
<=> \(0x=-1\) => pt vn
Vậy...
câu b ai giải giúp với:
\(x+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{-x^2+8x-7}+1\)
