Giải các phương trình sau:
a) \(\sqrt{x}+\sqrt{x+7}+2\sqrt{x^2+7x}=35-2x\) (Đáp số: \(\dfrac{841}{144}\))
b) \(\dfrac{1}{x}+\dfrac{1}{\sqrt{2-x^2}}=2\) (Đáp số: x =1 hoặc x =\(\dfrac{-\sqrt{3}-1}{2}\))
c) \(x^4+4x^3+6x^2+4x+\sqrt{x^2+2x+10}=2\) (Đáp số: x = 1)
d) \(\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\)
Một số câu có đáp án nhưng mình lại giải ko ra đáp án đúng mà ko biết sai ở đâu ý:((( Mai mình cần rồi, cảm ơn mọi người nhé<3
a/ \(x\ge0\)
\(\sqrt{x}+\sqrt{x+7}+2x+7+2\sqrt{x^2+7x}-42=0\)
\(\Leftrightarrow\sqrt{x}+\sqrt{x+7}+\left(\sqrt{x}+\sqrt{x+7}\right)^2-42=0\)
Đặt \(\sqrt{x}+\sqrt{x+7}=t>0\)
\(\Rightarrow t^2+t-42=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-7< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\sqrt{x+7}=6\Leftrightarrow2x+7+2\sqrt{x^2+7x}=36\)
\(\Leftrightarrow2\sqrt{x^2+7x}=29-2x\) \(\Leftrightarrow\left\{{}\begin{matrix}29-2x\ge0\\4\left(x^2+7x\right)=\left(29-2x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{29}{2}\\144x=841\end{matrix}\right.\) \(\Rightarrow x=\dfrac{841}{144}\)
b/ \(x^2< 2;x\ne0\)
Đặt \(\sqrt{2-x^2}=a>0\) ta được hệ:
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{a}=2\\x^2+a^2=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+a=2ax\\\left(x+a\right)^2-2ax=2\end{matrix}\right.\) \(\Rightarrow4\left(ax\right)^2-2ax-2=0\)
\(\left[{}\begin{matrix}ax=1\\ax=\dfrac{-1}{2}\end{matrix}\right.\Rightarrow\) \(\left[{}\begin{matrix}x+a=2\\x+a=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+\sqrt{2-x^2}=2\left(1\right)\\x+\sqrt{2-x^2}=-1\left(2\right)\end{matrix}\right.\)
- Xét (1): \(1.x+1.\sqrt{2-x^2}\le\sqrt{\left(1^2+1^2\right)\left(x^2+2-x^2\right)}=2\)
Dấu "=" xảy ra khi \(x=\sqrt{2-x^2}\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=1\end{matrix}\right.\) \(\Rightarrow x=1\)
- Xét (2): \(\sqrt{2-x^2}=-1-x\) \(\Leftrightarrow\left\{{}\begin{matrix}-1-x\ge0\\2-x^2=\left(-1-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-1\\2x^2+2x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1-\sqrt{3}}{2}\\x=\dfrac{-1+\sqrt{3}}{2}>-1\left(l\right)\end{matrix}\right.\)
Vậy pt đã cho có 2 nghiệm: \(\left[{}\begin{matrix}x=1\\x=\dfrac{-1-\sqrt{3}}{2}\end{matrix}\right.\)
c/
\(x^4+4x^3+6x^2+4x+1+\sqrt{x^2+2x+1+9}=3\)
\(\Leftrightarrow\left(x+1\right)^4+\sqrt{\left(x+1\right)^2+9}=3\)
Ta có: \(\left\{{}\begin{matrix}\left(x+1\right)^4\ge0\\\sqrt{\left(x+1\right)^2+9}\ge3\end{matrix}\right.\) \(\Rightarrow VT\ge3\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\left(x+1\right)^4=0\\\left(x+1\right)^2=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
Vậy pt có nghiệm duy nhất \(x=-1\)
d/
ĐKXĐ: \(x\ge-1\)
Ta có \(\sqrt{x+3}+\sqrt{x+1}>0\) với mọi \(x\ge-1\) \(\Rightarrow\) nhân 2 vế của pt đã cho với \(\sqrt{x+3}+\sqrt{x+1}\) ta được:
\(x^2+\sqrt{\left(x+3\right)\left(x+1\right)}=x\left(\sqrt{x+3}+\sqrt{x+1}\right)\)
\(\Leftrightarrow x^2-x\sqrt{x+3}+\sqrt{\left(x+3\right)\left(x+1\right)}-x\sqrt{x+1}=0\)
\(\Leftrightarrow x\left(x-\sqrt{x+3}\right)-\sqrt{x+1}\left(x-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{x+1}\right)\left(x-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x+1}=0\\x-\sqrt{x+3}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{x+3}\\x=\sqrt{x+1}\end{matrix}\right.\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\left[{}\begin{matrix}x^2=x+3\\x^2=x+1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-3=0\\x^2-x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{13}}{2}\\x=\dfrac{1-\sqrt{13}}{2}< 0\left(l\right)\\x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}< 0\left(l\right)\end{matrix}\right.\)
Vậy pt đã cho có 2 nghiệm: \(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{13}}{2}\\x=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)
