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2. \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{4}\)
\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-3}{2}\right)=\dfrac{-21}{4}.\dfrac{2}{7}\)
\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-15}{10}\right)=\dfrac{-3}{2}\)
\(\Leftrightarrow x.\dfrac{6}{5}=\dfrac{-3}{2}\)
\(\Leftrightarrow x=\dfrac{-3}{2}:\dfrac{6}{5}\)
\(\Leftrightarrow x=\dfrac{-3}{2}.\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{-5}{4}\)
3.\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=1\\2x-\dfrac{3}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1+\dfrac{3}{4}\\2x=\left(-1\right)+\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=\dfrac{-7}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}.\dfrac{1}{2}\\x=\dfrac{-7}{3}.\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{-7}{6}\end{matrix}\right.\)
vậy \(x\in\left\{\dfrac{7}{6};\dfrac{-7}{6}\right\}\)
1: =>7/3x=3+1/3-8-2/3=-5-1/3=-16/3
=>x=-16/3:7/3=-7/16
2: =>1/3|x-2|=4/5+3/7=28/35+15/35=43/35
=>|x-2|=129/35
=>x-2=129/35 hoặc x-2=-129/35
=>x=199/35 hoặc x=-59/35
2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)
= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)
=1-\(\dfrac{1}{101}\)=...........
mk làm vậy thôi nha
thông cảm
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)
=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)
tương tự
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
\(A=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+.......+\left(\dfrac{1}{2}\right)^{10}\)
\(\Leftrightarrow A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+.......+\dfrac{1}{2^{10}}\)
\(\Leftrightarrow2A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^9}\)
\(\Leftrightarrow2A-A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{10}}\right)\)
\(\Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{2^{10}}\)
1: \(\Leftrightarrow\left(\dfrac{2}{5}x+2\right)=-\dfrac{3}{2}\cdot\left(-4\right)=6\)
=>2/5x=4
hay x=10
2: \(x\left(x+1\right)=0\)
=>x=0 hoặc x+1=0
=>x=0 hoặc x=-1
3: \(2x-16=x+40\)
=>2x-x=40+16
=>x=56
4: \(\dfrac{x}{4}=\dfrac{9}{x}\)
nên \(x^2=36\)
=>x=6 hoặc x=-6
\(=1-\dfrac{1}{1+\dfrac{2}{1+1}}=1-\dfrac{1}{1+1}=1-\dfrac{1}{2}=\dfrac{1}{2}\)
\(1-\dfrac{1}{1+\dfrac{2}{1-\dfrac{3}{1-4}}}\)
\(=1-\dfrac{1}{1+\dfrac{2}{1+\dfrac{3}{3}}}\)
\(=1-\dfrac{1}{1+\dfrac{2}{1+1}}\)
\(=1-\dfrac{1}{1+1}\)
\(=1-\dfrac{1}{2}\)
\(=\dfrac{1}{2}\)
#AvoidMe