IH
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
16 tháng 6 2019
\(\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\) (ĐK: \(x\ge\frac{-1}{2}\) )
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[2x\left(x^2+1\right)+\left(x^2+1\right)\right]\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow2x+1=\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+1\right)-\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2+1-1\right)=0\)
\(\Leftrightarrow x^2\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=0\end{cases}}\) (nhận)
Vậy .....
16 tháng 6 2019
\(\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[x^2\left(2x+1\right)+2x+1\right]\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\left|x+\frac{1}{2}\right|}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)(1)
Vì VT > 0 nên VP >0
\(\Leftrightarrow\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\ge0\)
\(\Leftrightarrow x\ge-\frac{1}{2}\)
Khi đó \(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{x^2-x-\frac{3}{4}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow x^2-x-\frac{3}{4}=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\2x-3=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\end{cases}}\)
Cần cù bù thông minh , phá tung pt dưới ra được cái phương trình bậc 5, sau đó dùng Wolfram|Alpha: Computational Intelligence để tính nghiệm rồi phân tích nhân tử =))
N
29 tháng 8 2019
\(DK:x\ge0\)
\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+1}}{x-x-1}+\frac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}+\frac{\sqrt{x+2}-\sqrt{x+3}}{x+2-x-3}=1\)
\(\Leftrightarrow-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}=1\)
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x+3}=1+\sqrt{x}\)
\(\Leftrightarrow x+3=x+2\sqrt{x}+1\)
\(\Leftrightarrow x=1\)
Vay nghiem cua PT la \(x=1\)
24 tháng 9 2016
1) ĐẶT \(\sqrt{x^2+1993}=y\)
==> \(1993=y^2-x^2\)
khi đó pt trở thành \(x^4+y=y^2-x^2\)
<=> \(\left(x^4-y^2\right)+\left(x^2+y\right)=0\)
<=> \(\left(x^2+y\right)\left(x^2-y\right)+\left(x^2+y\right)=0\)
<=> \(\left(x^2+y\right)\left(x^2-y+1\right)=0\)
đến đây bạn giải nốt nhé
30 tháng 7 2019
\(a,\frac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\)
\(\Rightarrow3x+2=2\sqrt{x+2}.\sqrt{x+2}\)
\(\Rightarrow3x+2=2\left(x+2\right)\)
\(\Rightarrow3x+2=2x+4\)
\(\Rightarrow3x-2x=4-2\)
\(\Rightarrow x=2\)
\(b,\sqrt{4x^2-1}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-2\right)=0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{2x+1}=0\\\sqrt{2x-1}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x+1=0\\\sqrt{2x-1}=2\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-1\\2x-1=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{2}\end{cases}}}\)
\(c,\sqrt{x-2}+\sqrt{4x-8}-\frac{2}{5}\sqrt{\frac{25x-50}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}-\frac{2}{5}\sqrt{\frac{25\left(x-2\right)}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\frac{2}{5}.\frac{5\sqrt{x-2}}{2}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\sqrt{x-2}=4\)
\(\Rightarrow2\sqrt{x-2}=4\)
\(\Rightarrow\sqrt{x-2}=2\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
\(d,\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
\(\Rightarrow\sqrt{x+4}=\sqrt{1-2x}+\sqrt{1-x}\)
\(\Rightarrow x+4=1-2x+2\sqrt{\left(1-2x\right)\left(1-x\right)}+1-x\)
\(\Rightarrow x+4=2-3x+2\sqrt{1-3x+2x^2}\)
\(\Rightarrow x+4-2+3x=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x+2=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow2x+1=\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x^2+4x+1=1-3x+2x^2\)
\(\Rightarrow4x^2-2x^2+4x+3x+1-1=0\)
\(\Rightarrow2x^2+7x=0\)
\(\Rightarrow x\left(2x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-7}{2}\end{cases}}}\)
\(e,\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\)
\(\frac{2x\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{2x\left(\sqrt{3}-1\right)}{3-1}=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+\sqrt{3}\right)-x\left(\sqrt{3}-1\right)=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+\sqrt{3}x-\sqrt{3x}+x=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+x=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+1\right)=\sqrt{5}+1\)
\(\Rightarrow x=1\)
\(ĐKXĐ:-\frac{1}{4}\le x\le2\)
\(PT\Leftrightarrow x+\sqrt{x+\frac{1}{4}+\sqrt{x+\frac{1}{4}}+\frac{1}{4}}=2\)
\(\Leftrightarrow x+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)
\(\Leftrightarrow x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)
\(\Leftrightarrow\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2=2\)
\(\Leftrightarrow\sqrt{x+\frac{1}{4}}+\frac{1}{2}=\sqrt{2}\Leftrightarrow x+\frac{1}{4}=\left(\sqrt{2}-\frac{1}{2}\right)^2=\frac{9-4\sqrt{2}}{4}\)
\(\Rightarrow x=\frac{9-4\sqrt{2}-1}{4}=\frac{8-4\sqrt{2}}{4}=2-\sqrt{2}\) (TMĐKXĐ)
Vậy PT trên có nghiệm là \(x=2-\sqrt{2}\)