Giải PT: \(\dfrac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\dfrac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)
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1 tháng 2 2022
mình nhầm mẫu nhé :v mình làm lại
\(=\left(\dfrac{x-\sqrt{x}-2x+4\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\right):\dfrac{2-\sqrt{x}}{x-1}\)
\(=\dfrac{-x+3\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{2-\sqrt{x}}=\dfrac{\left(2-\sqrt{x}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(2-\sqrt{x}\right)\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
3 tháng 11 2018
Ta có :
\(\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+1}-\sqrt{x+2}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}=-\sqrt{x+1}+\sqrt{x+2}\)
Tương tự :
\(\dfrac{1}{\sqrt{x+2}+\sqrt{x+3}}=-\sqrt{x+2}+\sqrt{x+3}\)
\(\dfrac{1}{\sqrt{x+3}+\sqrt{x+4}}=-\sqrt{x+3}+\sqrt{x+4}\)
....
\(\dfrac{1}{\sqrt{x+2019}+\sqrt{x+2010}}=-\sqrt{x+2019}+\sqrt{x+2010}\)
Từ những ý trên , pt trở thành :
\(-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+4}-.....-\sqrt{x+2019}+\sqrt{x+2020}=11\)
\(\Leftrightarrow\sqrt{x+2020}-\sqrt{x+1}=11\)
\(\Leftrightarrow x+2020-2\sqrt{\left(x+2020\right)\left(x+1\right)}+x+1=121\)
\(\Leftrightarrow2x+1900=2\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x+950=\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x^2+1900x+902500=x^2+2021x+2020\)
\(\Leftrightarrow121x-900480=0\)
\(\Leftrightarrow x=\dfrac{900480}{121}\)
28 tháng 9 2021
1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)
\(\Leftrightarrow\left|x+5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)
2) \(ĐK:x\ge2\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
\(ĐK:0< x\le4\)
Đặt \(\sqrt{2+\sqrt{x}}=a>0;\sqrt{2-\sqrt{x}}=b>0\)
\(\Rightarrow a^2+b^2=2+\sqrt{x}+2-\sqrt{x}=4\)
\(PT\Leftrightarrow\dfrac{a^2}{\sqrt{2}+a}+\dfrac{b^2}{\sqrt{2}-b}=\sqrt{2}\\ \Leftrightarrow\dfrac{a^2\sqrt{2}-a^2b+b^2\sqrt{2}+ab^2}{2+\sqrt{2}\left(a-b\right)-ab}=\sqrt{2}\\ \Leftrightarrow\sqrt{2}\left(a^2+b^2\right)+ab\left(b-a\right)=2\sqrt{2}+2\left(a-b\right)-\sqrt{2}ab\\ \Leftrightarrow4\sqrt{2}-ab\left(a-b\right)=2\sqrt{2}+2\left(a-b\right)-\sqrt{2}ab\\ \Leftrightarrow\left(2+ab\right)\left(a-b\right)=2\sqrt{2}+\sqrt{2}ab\\ \Leftrightarrow\left(2+ab\right)\left(a-b\right)-\sqrt{2}\left(2+ab\right)=0\\ \Leftrightarrow\left(a-b-\sqrt{2}\right)\left(2+ab\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}ab=-2\\a-b=\sqrt{2}\end{matrix}\right.\)
Xét \(\left\{{}\begin{matrix}ab=-2\\a^2+b^2=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=8\\\left(a+b\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=\pm2\sqrt{2}\\a+b=0\end{matrix}\right.\left(loại.vì.a>0;b\ge0\right)\)
Xét \(\left\{{}\begin{matrix}a-b=\sqrt{2}\\a^2+b^2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\b^2+2\sqrt{2}b+2+b^2=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\2b^2+2\sqrt{2}b-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\b^2+b\sqrt{2}-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left[{}\begin{matrix}b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\\b=\dfrac{-\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\left(sd.\Delta\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\left(b\ge0\right)\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2+\sqrt{x}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\\sqrt{2-\sqrt{x}}=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)
Tới đây dễ r nha