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\(\sqrt{\dfrac{1}{4}+\dfrac{1}{\left(2n-1\right)^2}+\dfrac{1}{\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(2n-1\right)^2\left(2n+1\right)^2+4\left(2n-1\right)^2+4\left(2n+1\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)
\(=\sqrt{\dfrac{\left(4n^2-1\right)^2+4\left(4n^2-4n+1\right)+4\left(4n^2+4n+1\right)}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)
\(=\sqrt{\dfrac{16n^4+24n^2+9}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(4n^2+3\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\dfrac{4n^2+3}{2\left(2n-1\right)\left(2n+1\right)}\)
\(=\dfrac{\left(4n^2-1\right)+4}{2\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(=\dfrac{1}{2}+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)
Do đó:
\(P=\left(\dfrac{1}{2}+\dfrac{1}{1}-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{399}-\dfrac{1}{401}\right)\)
\(=\dfrac{1}{2}.200+1-\dfrac{1}{401}=\dfrac{40500}{401}\)
\(\Rightarrow Q=400\)
Lời giải:
\(=\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\frac{-\sqrt{5}(\sqrt{7}-\sqrt{3})}{\sqrt{7}-\sqrt{3}}=\frac{4(\sqrt{5}+1)}{5-1}-\sqrt{5}=(\sqrt{5}+1)-\sqrt{5}=1\)
\(\dfrac{4}{\sqrt{5}-1}+\dfrac{\sqrt{15}-\sqrt{35}}{\sqrt{7}-\sqrt{3}}\)
\(=\sqrt{5}+1-\sqrt{5}\)
=1
Ta có: \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\cdot\sqrt{4.5}+\dfrac{2}{5}\sqrt{50}\right):\dfrac{4}{15}\sqrt{\dfrac{1}{8}}\)
\(=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{9}{4}\sqrt{2}+2\sqrt{2}\right):\dfrac{\sqrt{2}}{15}\)
\(=0\)
a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
\(đk:u\ge0\)
\(=4.5\sqrt{u}-\dfrac{15}{2}.\dfrac{4}{3}.\sqrt{u}-\dfrac{2}{u}.\dfrac{13}{2}.u\sqrt{u}\)
\(=20\sqrt{u}-10\sqrt{u}-13\sqrt{u}=-3\sqrt{u}\)
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
ĐKXĐ: \(x\ne4;y\ne-1\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x-4}=u\\\dfrac{1}{y+1}=u\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}3u+2v=\dfrac{15}{12}\\2u-v=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3u+2v=\dfrac{15}{12}\\4u-2v=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7u=-\dfrac{11}{4}\\v=2u+2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=-\dfrac{11}{28}\\v=\dfrac{17}{14}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-4}=-\dfrac{11}{28}\\\dfrac{1}{y+1}=\dfrac{17}{14}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-4=-\dfrac{28}{11}\\y+1=\dfrac{14}{17}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{16}{11}\\y=-\dfrac{3}{17}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3}{x-4}+\dfrac{2}{y+1}=\dfrac{15}{12}\\\dfrac{2}{x-4}-\dfrac{1}{y+1}=-2\end{matrix}\right.\)
Đặt: \(\left\{{}\begin{matrix}a=\dfrac{1}{x-4}\\b=\dfrac{1}{y+1}\end{matrix}\right.\)
Hệ phương trình: \(\Leftrightarrow\left\{{}\begin{matrix}3a+2b=\dfrac{15}{12}\\2a-b=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a+2b=\dfrac{15}{12}\\4a-2b=-4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}7a=-\dfrac{11}{4}\\2a-b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{11}{28}\\2\cdot\left(-\dfrac{11}{28}\right)-b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{11}{28}\\-\dfrac{11}{14}-b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{11}{28}\\b=\dfrac{17}{14}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-4}=-\dfrac{11}{28}\\\dfrac{1}{y+1}=\dfrac{17}{14}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-4=\dfrac{1}{-\dfrac{11}{28}}\\y+1=\dfrac{1}{\dfrac{17}{14}}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{16}{11}\\y=-\dfrac{3}{17}\end{matrix}\right..}\)
\(\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{399}=4.\left(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{399}\right)=4.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{19.21}\right)=4.\left[\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\right]=4.\left[\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\right]=2.\left(\dfrac{7-1}{21}\right)=\dfrac{12}{21}=\dfrac{4}{7}\)