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Ta có: \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
\(\left(x+\frac{1}{2}\right)\times\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow x+\frac{1}{2}=0\)
\(x=0-\frac{1}{2}\)
\(x=-\frac{1}{2}\)
\(\Rightarrow\left(\frac{2}{3}-2x\right)=0\)
\(2x=\frac{2}{3}-0\)
\(2x=\frac{2}{3}\)
\(x=\frac{2}{3}\div2\)
\(x=\frac{1}{3}\)
Vạy tồn tại hai giá trị \(-\frac{1}{2}\) và \(\frac{1}{3}\)
\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}.\frac{3}{14}=\frac{5}{14}\)
( -1/7 + 2/7 - 5/11 ) - ( 1/4 - 5/7 )
= ( 1/7 - 5/11 ) - ( 1/4 - 5/7 )
= 1/7 - 5/11 - 1/4 + 5/7
= 6/7 - 5/11 - 1/4
= 47/308
A=3/1.3+3/3.5+3/5.7+............+3/49.51
A=3/1-3/3=3/3-3/5+3/5-3/7+...............+3/49-3/51
A=1-1/3+1/3-1/5+1/5-1/7+.....................+1/39-1/51
A=1-1/51
A=50/51
A\(=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...\frac{1}{49.51}\right) \)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...\frac{2}{49.51}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
=\(\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}\)
\(=\frac{25}{17}\)
[(x+1)/2]2=26/25-17/25=9/25
=> (x+1)/2=3/5 => x=1/5 hoặc x=-11/5
hoặc (x+1)/2=-3/5
Đề vầy đúng không :
( x+1/2)^2 + 17/25 = 26 / 25
(x+1/2)^2=26/25-17/25
(x+1/2)^2=9/25
=> x+1/2=3/5
=> x = 3/5-1/2=1/10