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\(DK:x\in\left[\frac{7}{2};5\right]\)
PT\(\Leftrightarrow\left(\sqrt{x-3}-1\right)+\left(\sqrt{5-x}-1\right)+\left(\sqrt{2x-7}-1\right)-\left(x-4\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\frac{x-4}{\sqrt{x-3}+1}-\frac{x-4}{\sqrt{5-x}+1}+\frac{2\left(x-4\right)}{\sqrt{2x-7}+1}-\left(x-4\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(\frac{1}{\sqrt{x-3}+1}-\frac{1}{\sqrt{5-x}+1}+\frac{1}{\sqrt{2x-7}+1}-2x+1\right)=0\)
Vi \(\frac{1}{\sqrt{x-3}+1}-\frac{1}{\sqrt{5-x}+1}+\frac{1}{\sqrt{2x-7}+1}-2x+1\ne0\)(voi moi \(x\in\left[\frac{7}{2};5\right]\)
\(\Rightarrow x=4\)
Vay nghiem cua PT la \(x=4\)
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Câu 1 :
Đk: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{2x-1}=5\\ \Leftrightarrow x-1+2\sqrt{\left(x-1\right)\left(2x-1\right)}+2x-1=25\\ \Leftrightarrow2\sqrt{2x^2-3x+1}=27-3x\\ \)
\(\Leftrightarrow\begin{cases}27-3x\ge0\\4\left(2x^2-3x+1\right)=9x^2-162x+729\end{cases}\) \(\Leftrightarrow\begin{cases}x\le9\\x^2-150x+725=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x\le9\\x=145hoặcx=5\end{cases}\)
với x= 5 thoản mãn điều kiện, x=145 loại
Vậy \(S=\left\{5\right\}\)
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Kết quả bằng 2 nha bạn k đúng cho mk ná!^^
Chúc bạn hok tốt =))
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Ta thấy: \(\left(\sqrt{a}+\sqrt{b}\right)^2=a+b+2\sqrt{ab}\)
\(\left(\sqrt{a+b}\right)^2=a+b\)
Nếu: \(2\sqrt{ab}>0\left(a,b>0\right)\text{ thì: }\left(\sqrt{a}+\sqrt{b}\right)^2>\left(\sqrt{a+b}\right)^2\)
<=>\(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\)
\(B=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}+....+\frac{1}{\sqrt{2013}+\sqrt{2015}}\)
\(=\frac{1}{2}.\left(\frac{2}{\sqrt{1}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{5}}+...+\frac{2}{\sqrt{2013}-\sqrt{2014}}\right)\)
\(=\frac{1}{2}.\left(-1+\sqrt{3}-\sqrt{3}+\sqrt{5}-...-\sqrt{2013}+\sqrt{2015}\right)\)
=\(\frac{\sqrt{2015}-1}{2}\)
Xét hiệu: B-A=\(\frac{\sqrt{2015}-1}{2}-\sqrt{481}=\frac{\sqrt{2015}-1}{2}-\frac{\sqrt{1924}}{2}=\frac{\sqrt{2015}-\left(\sqrt{1}+\sqrt{1924}\right)}{2}>\frac{\sqrt{2015}-\sqrt{1+1924}}{2}\)
\(=\frac{\sqrt{2015}-\sqrt{1925}}{2}>0\Rightarrow A>B\)
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\(\int\limits^1_{\sqrt{ }3}\)\(\sqrt{\left(1+x^2\right)}\)\(dx\)
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\(\begin{cases}\sqrt{xy}+\frac{1}{\sqrt{xy}}=\frac{5}{2}\\\sqrt{x}+\sqrt{y}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{9}{2}\end{cases}\)
<=>\(\begin{cases}xy+1=\frac{5\sqrt{xy}}{2}\\\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{x}+\sqrt{y}=\frac{9\sqrt{xy}}{2}\end{cases}\)
Đặt P=\(\sqrt{xy}\);S=\(\sqrt{x}+\sqrt{y}\)(S2\(\ge\)4P)
Ta có HPT: \(\begin{cases}P^2+1=\frac{5P}{2}\\S.P+P=\frac{9P}{2}\end{cases}\)
Tới đây dễ tự làm
ai trả lời đầu tiên cho
= 2,076249377
~ Học tốt ~