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\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{2010}{2012}:2\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{1005}{2012}\)
\(\Rightarrow\dfrac{1}{\left(x+1\right)}=\dfrac{1}{4}-\dfrac{1005}{2012}\)
\(\dfrac{1}{\left(x+1\right)}=\dfrac{-251}{1006}\)
\(\Rightarrow1:\left(x+1\right)=\dfrac{-251}{1006}\)
\(\left(x+1\right)=1:\dfrac{-251}{1006}\)
\(x+1=\dfrac{-1006}{251}\)
\(x=\dfrac{-1006}{251}-1\)
\(x=\dfrac{-1257}{251}\)
Nếu bạn tìm \(x\in Z\) hay \(x\in N\) thì \(x=\varnothing\) (không có x thoả mãn)
\(\Leftrightarrow2\left(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1005}{1006}\)
\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1005}{2012}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{-251}{1006}\)
=>x+1=-1006/251
hay x=-1257/251
b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=93-3=90\)
\(\Rightarrow x=90:2=45\)
a) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow\)\(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+2^2+2^3\right)=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+4+8\right)=480\)
\(\Leftrightarrow\)\(2^x.15=480\)
\(\Rightarrow\)\(2^x=480:15\)
\(\Leftrightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5.
Bài 1:
\(a,\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2-\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{24+2-3}{12}\right)=\dfrac{7}{46}\)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\dfrac{23}{12}=\dfrac{7}{46}\)
\(x+\dfrac{1}{4}-\dfrac{1}{3}=\dfrac{7}{46}.\dfrac{23}{12}\)
\(x+\dfrac{1}{4}-\dfrac{1}{3}=\dfrac{7}{24}\)
\(x+\dfrac{1}{4}=\dfrac{7}{24}+\dfrac{1}{3}\)
\(x+\dfrac{1}{4}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}-\dfrac{1}{4}=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\)
\(b,\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{7}{10}\)
\(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\)
\(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{1}{6}\)
\(\dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\)
\(\dfrac{13}{21}+x=\dfrac{2}{7}\)
\(x=\dfrac{2}{7}-\dfrac{13}{21}=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
Bài 2:
\(a,\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)
\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(\dfrac{121}{12}-\dfrac{19}{2}\right)\)
\(=\dfrac{77}{18}:\dfrac{7}{12}\)
\(=\dfrac{22}{3}\)
\(b,1\dfrac{5}{18}-\dfrac{5}{18}.\left(\dfrac{1}{15}+1\dfrac{1}{12}\right)\)
\(=\dfrac{23}{18}-\dfrac{5}{18}.\dfrac{69}{60}\)
\(=\dfrac{23}{18}-\dfrac{23}{72}\)
\(=\dfrac{23}{24}\)
\(c,-\dfrac{1}{7}.\left(9\dfrac{1}{2}-8,75\right):\dfrac{2}{7}+0,625:1\dfrac{2}{3}\)
\(=\dfrac{-1}{7}.\dfrac{3}{4}:\dfrac{2}{7}+\dfrac{5}{8}:\dfrac{5}{3}\)
\(=-\dfrac{3}{8}+\dfrac{5}{8}:\dfrac{5}{3}\)
\(=-\dfrac{3}{8}+\dfrac{3}{8}\)
\(=\dfrac{0}{8}=0\)
Chúc bạn học tốt
ukm
bn có thể giải cho mik mấy bài mà mik vừa đăng đc ko mik đang cần gấp
Ta có:
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{10}\right)=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{9}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1.2.3.....9}{2.3.4.....10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow x=\dfrac{2010}{10}\)
\(\Leftrightarrow x=201\)
Vậy x = 201
1.Tính hợp lý:
a. 1152 - (374 + 1152) + (374 - 65) = 1152 - 374 - 1152 + 374 - 65 = ( 1152 - 1152 ) + ( -65) + ( 374 - 374 ) = 0 + ( - 65) + 0 = -65
Bài 1 : Tính hợp lý : c. \(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\) = \(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\) = \(\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\) = \(\dfrac{3^{29}.2^3}{2^2.3^{28}}\) = 6
\(\Leftrightarrow\dfrac{3}{7}\left(\dfrac{46}{3}+\dfrac{12}{5}\right)< =x< =\left(\dfrac{10}{3}:7-\dfrac{13}{2}\right)\cdot\dfrac{-7}{3}\)
\(\Leftrightarrow\dfrac{38}{5}< =x< =\dfrac{253}{18}\)
mà x là số nguyên
nên \(x\in\left\{8;9;10;11;12;13;14\right\}\)
https://hoc24.vn/question/246430.html
\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{2010}{2012}:2\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{1005}{2012}\)
\(\dfrac{1}{\left(x+1\right)}=\dfrac{1}{4}-\dfrac{1005}{2012}\)
\(\dfrac{1}{\left(x+1\right)}=\dfrac{-251}{1006}\)
\(\Rightarrow1:\left(x+1\right)=\dfrac{-251}{1006}\)
\(\left(x+1\right)=1:\dfrac{-251}{1006}\)
\(x+1=\dfrac{-1006}{251}\)
\(x=\dfrac{-1006}{251}-1\)
\(x=\dfrac{-1257}{251}\)
Vì \(x\in N\) nên \(x=\varnothing\) (không có giá trị nào của x thoả mãn)