Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
a/ ĐKXĐ:...
\(\Leftrightarrow4x^2-4x\sqrt{2x-1}-3x^2+6x-3=0\)
\(\Leftrightarrow4x\left(x-\sqrt{2x-1}\right)-3\left(x-1\right)^2=0\)
\(\Leftrightarrow\frac{4x\left(x-1\right)^2}{x+\sqrt{2x-1}}-3\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\frac{4x}{x+\sqrt{2x-1}}=3\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4x=3x+3\sqrt{2x-1}\)
\(\Leftrightarrow x=3\sqrt{2x-1}\)
\(\Leftrightarrow x^2-18x+9=0\) \(\Rightarrow9\pm6\sqrt{2}\)
Vậy pt có 3 nghiệm....
b/ ĐKXĐ:...
\(\Leftrightarrow4x^2-4x\sqrt{4x-3}-x^2+4x-3=0\)
\(\Leftrightarrow4x\left(x-\sqrt{4x-3}\right)-\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\frac{4x\left(x^2-4x+3\right)}{x+\sqrt{4x-3}}-\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\Rightarrow x=...\\\frac{4x}{x+\sqrt{4x-3}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4x=x+\sqrt{4x-3}\)
\(\Leftrightarrow3x=\sqrt{4x-3}\)
\(\Leftrightarrow9x^2-4x+3=0\) (vô nghiệm)
Vậy...
c/
\(\Leftrightarrow x^2+2x+1+2\left(x^2+3\right)=3\left(x+1\right)\sqrt{x^2+3}\)
\(\Leftrightarrow\left(x+1\right)^2+2\left(x^2+3\right)-3\left(x+1\right)\sqrt{x^2+3}=0\)
Đặt \(\left\{{}\begin{matrix}x+1=a\\\sqrt{x^2+3}=b\end{matrix}\right.\)
\(\Rightarrow a^2+2b^2-3ab=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\2\sqrt{x^2+3}=x+1\left(x\ge-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=x^2+2x+1\\4x^2+12=x^2+2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=2\\3x^2-2x+11=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=1\)
d/
\(\Leftrightarrow2\left(2x^2+3\right)+2\left(x^2+2x+1\right)-5\left(x+1\right)\sqrt{2x^2+3}=0\)
\(\Leftrightarrow2\left(2x^2+3\right)+2\left(x+1\right)^2-5\left(x+1\right)\sqrt{2x^2+3}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+3}=a\\x+1=b\end{matrix}\right.\)
\(\Rightarrow2a^2+2b^2-5ab=0\)
\(\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{2x^2+3}=x+1\\\sqrt{2x^2+3}=2\left(x+1\right)\end{matrix}\right.\) (\(x\ge-1\))
\(\Leftrightarrow\left[{}\begin{matrix}8x^2+12=x^2+2x+1\\2x^2+3=4x^2+8x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}7x^2-2x+11=0\left(vn\right)\\2x^2+8x+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-4+\sqrt{14}}{2}\\x=\frac{-4-\sqrt{14}}{2}\left(l\right)\end{matrix}\right.\)
e/ ĐKXĐ: \(x^2+x-6\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\end{matrix}\right.\)
\(\Leftrightarrow5\left(x^2+x-6\right)-\left(x-3\right)\sqrt{x^2+x-6}+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow5\left(x^2+x-6\right)-\left(x-3\right)\sqrt{x^2+x-6}+\left(x-3\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x-6}=a\\x-3=b\end{matrix}\right.\)
\(\Rightarrow5a^2-ab+b^2=0\)
\(\Leftrightarrow\frac{19}{4}a^2+\left(\frac{a}{2}-b\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+x-6=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow ptvn\)