Giải PT sau :\(\sqrt{x+1}+2\left(x+1\right)=x-1+\sqrt{1-x}+3\sqrt{1-x^2}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
27 tháng 11 2021
\(a,ĐK:1\le x\le3\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\\\sqrt{3-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)
\(PT\Leftrightarrow a+b-ab=1\Leftrightarrow a+b-ab-1=0\\ \Leftrightarrow\left(a-1\right)\left(1-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\3-x=1\end{matrix}\right.\Leftrightarrow x=2\left(tm\right)\)
\(b,ĐK:0\le x\le9\\ PT\Leftrightarrow9+2\sqrt{x\left(9-x\right)}=-x^2+9x+9\\ \Leftrightarrow2\sqrt{-x^2+9x}-\left(-x^2+9x\right)=0\\ \Leftrightarrow\sqrt{-x^2+9x}\left(2-\sqrt{-x^2+9x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x^2+9x=0\\\sqrt{-x^2+9x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\\x^2-9x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=9\left(n\right)\\x=\dfrac{9+\sqrt{65}}{2}\left(n\right)\\x=\dfrac{9-\sqrt{65}}{2}\left(n\right)\end{matrix}\right.\)
18 tháng 6 2021
Đk:\(x\ge-1\)
Đặt \(\left(a,b,c\right)=\left(x;\sqrt{x+1};\sqrt{2}\right)\)
Pt tt: \(a^3+b^3+c^3=\left(a+b+c\right)^3\)
\(\Leftrightarrow a^3+b^3+c^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(\Leftrightarrow0=3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)
\(\Leftrightarrow3\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\)
\(\Leftrightarrow3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\b+c=0\\a+c=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x+1}=0\\\sqrt{x+1}+\sqrt{2}=0\left(vn\right)\\x+\sqrt{2}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}=-x\\x=-\sqrt{2}\left(ktm\right)\end{matrix}\right.\)\(\Rightarrow\)\(\sqrt{x+1}=-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le0\\x+1=x^2\end{matrix}\right.\)\(\Rightarrow x=\dfrac{1-\sqrt{5}}{2}\) (tm)
Vậy...
ĐK : \(-1\le x\le1\)
+ Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+1}\ge0\\b=\sqrt{1-x}\ge0\end{matrix}\right.\) thì pt đã cho trở thành :
\(a+2a^2=-b^2+b+3ab\)
\(\Rightarrow a+2a^2+b^2-b-3ab=0\)
\(\Rightarrow a\left(2a-b+1\right)-b\left(2a-b+1\right)=0\)
\(\Rightarrow\left(a-b\right)\left(2a-b+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=b\\2a=b-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{1-x}\\2\sqrt{x+1}+1=\sqrt{1-x}\end{matrix}\right.\)
+ TH1 : \(\sqrt{x+1}=\sqrt{1-x}\Leftrightarrow x+1=1-x\Leftrightarrow x=0\) ( TM )
+ TH2 : \(2\sqrt{x+1}+1=\sqrt{1-x}\)
\(\Leftrightarrow4x+5+4\sqrt{x+1}=1-x\)
\(\Leftrightarrow4\sqrt{x+1}=-5x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}-5x-4\ge0\\16\left(x+1\right)=\left(-5x-4\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\frac{4}{5}\\16x+16=25x^2+40x+16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le-\frac{4}{5}\\25x^2+24x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le-\frac{4}{5}\\x\left(25x+24\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le-\frac{4}{5}\\\left[{}\begin{matrix}x=0\left(KTM\right)\\x=-\frac{24}{25}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)