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\(\frac{1}{2}\times\left(x-\frac{4}{5}\right)+\frac{3}{4}x=\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{2}{5}+\frac{3}{4}x=\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{3}{4}x=\frac{5}{12}+\frac{2}{5}\)
\(\Leftrightarrow\frac{5}{4}x=\frac{49}{60}\)
\(\Leftrightarrow x=\frac{49}{75}\)
Vậy \(x=\frac{49}{75}\)
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
Ta có: S = \(\dfrac{1}{3}+\dfrac{3}{3.7}+\dfrac{5}{3.7.11}+...+\dfrac{2n+1}{3.7.11...\left(4n+3\right)}\)
⇒ 2S = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+...+\dfrac{4n+2}{3.7.11...\left(4n+3\right)}\)
⇒ 2S + \(\dfrac{1}{3.7.11...\left(4n+3\right)}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+...+\dfrac{4n+3}{3.7.11...\left(4n+3\right)}\)
Đến đây nó sẽ rút gọn liên tục và sau nhiều lần rút gọn ta có:
2S + \(\dfrac{1}{3.7.11...\left(4n+3\right)}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+\dfrac{1}{3.7.11}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{11}{3.7.11}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{1}{3.7}\) = \(\dfrac{2}{3}+\dfrac{7}{3.7}=\dfrac{2}{3}+\dfrac{1}{3}=1\)
Suy ra 2S < 1 ⇒ S < \(\dfrac{1}{2}\)(đpcm)
a) 1 + 3 + 5 + ... + 13
= (13 + 1).[(13 - 1) : 2 + 1] : 2
= 14 . 7 : 2
= 49
= 7²
b) 3² + 4² + 12²
= 9 + 16 + 144
= 169
= 13²
TL
S= ( 1+ 3+ 3^2+ 3^3+ 3^4+ 3^5+ 3^6+ 3^7+ 3^8+ 3^9)
3.S=3.( 1+ 3+ 3^2+ 3^3+ 3^4+ 3^5+ 3^6+ 3^7+ 3^8+ 3^9)
3S=3+3^2+3^3+....+3^10
3S-S=3+3^2+3^3+....+3^10-(1+ 3+ 3^2+ 3^3+ 3^4+ 3^5+ 3^6+ 3^7+ 3^8+ 3^9)
2S=3^10-1
S=3^10-1/2
HỌC TỐT NHÉ
\(\frac{2}{5}+\frac{-1}{5}-\frac{3}{4}-\frac{-2}{3}\text{ }\)
\(=\frac{2}{5}+\frac{-1}{5}+\frac{-3}{4}+\frac{2}{3}\)
\(=\left(\frac{2}{5}+\frac{-1}{5}\right)+\left(\frac{-3}{4}+\frac{2}{3}\right)\)
\(=\frac{1}{5}+\left(\frac{-9}{12}+\frac{8}{12}\right)\)
\(=\frac{1}{5}+\frac{-1}{12}\)
\(=\frac{12}{60}+\frac{-5}{60}\)
\(=\frac{7}{60}\)
\(\frac{2}{5}+\left(-\frac{1}{5}\right)-\frac{3}{4}-\left(-\frac{2}{3}\right)\)
\(=\frac{2}{5}-\frac{1}{5}-\frac{3}{4}+\frac{2}{3}\)
\(=\frac{1}{5}-\frac{3}{4}+\frac{2}{3}\)
\(=\frac{12}{60}-\frac{45}{60}+\frac{40}{60}\)
\(=\frac{12}{60}-\left(\frac{45}{60}-\frac{40}{60}\right)\)
\(=\frac{12}{60}-\frac{5}{60}\)
\(=\frac{7}{60}\)