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\(\Leftrightarrow2\left(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1005}{1006}\)
\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1005}{2012}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{-251}{1006}\)
=>x+1=-1006/251
hay x=-1257/251
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.
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
a, (x + 1) + (x + 4) + ... + (x + 28) = 155
x + 1 + x + 4 + ... + x + 28 = 155
(x + x + x + ... + x) + (1 + 4 + ... + 28) = 155
x . 10 + 145 = 155
x . 10 = 155 - 145
x . 10 = 10
x = 10 : 10
x = 1
a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).
b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=93-3=90\)
\(\Rightarrow x=90:2=45\)
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
\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{2010}{2012}:2\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{1005}{2012}\)
\(\Rightarrow\dfrac{1}{\left(x+1\right)}=\dfrac{1}{4}-\dfrac{1005}{2012}\)
\(\dfrac{1}{\left(x+1\right)}=\dfrac{-251}{1006}\)
\(\Rightarrow1:\left(x+1\right)=\dfrac{-251}{1006}\)
\(\left(x+1\right)=1:\dfrac{-251}{1006}\)
\(x+1=\dfrac{-1006}{251}\)
\(x=\dfrac{-1006}{251}-1\)
\(x=\dfrac{-1257}{251}\)
Nếu bạn tìm \(x\in Z\) hay \(x\in N\) thì \(x=\varnothing\) (không có x thoả mãn)
Cảm ơn nha