Tìm a biết rằng 12×3+13×4+14×5+⋯+1a×(a+1)=299600.
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 2
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
5 tháng 10 2017
a) \(\left(19x+2\cdot5^2\right)\div14=\left(13-8\right)^2-4^2\)
\(\left(19x+2\cdot25\right)\div14=5^2-16\)
\(\left(19x+50\right)\div14=25-16=9\)
\(19x+50=9\cdot14=126\)
\(19x=126-50=76\)
\(x=76\div19=4\)
b) \(2\cdot3^x=10\cdot3^{12}+8\cdot\left(3^3\right)^4\)
\(2\cdot3^x=\left(10+8\right)\cdot3^{12}\)
\(2\cdot3^x=18\cdot3^{12}\)
\(\Rightarrow2\cdot3^x=2\cdot3^2\cdot3^{12}\Rightarrow2\cdot3^x=2\cdot3^{15}\Rightarrow x=15\)
EP
5 tháng 10 2017
a, \left(19.x+2.5^2\right)\div14=\left(13-8\right)^2-4^2(19.x+2.52)÷14=(13−8)2−42
\left(19.x+2.25\right)\div14=5^2-4^2(19.x+2.25)÷14=52−42
\left(19.x+2.25\right)\div14=25-16(19.x+2.25)÷14=25−16
\left(19.x+50\right)\div14=9(19.x+50)÷14=9
\left(19.x+50\right)=9.14(19.x+50)=9.14
19.x+50=12619.x+50=126
19.x=126-5019.x=126−50
19.x=7619.x=76
\Rightarrow x=76\div19⇒x=76÷19
\Rightarrow x=4⇒x=4
Vậy x = 4
b, 2.3^x=10.3^{12}+8.27^42.3x=10.312+8.274
2.3^x=10.3^{12}+8.\left(3^3\right)^42.3x=10.312+8.(33)4
2.3^x=10.3^{12}+8.3^{12}2.3x=10.312+8.312
2.3^x=\left(10+8\right).3^{12}2.3x=(10+8).312
2.3^x=18.3^{12}2.3x=18.312
2.3^x=2.3^3.3^{12}2.3x=2.33.312
2.3^x=2.3^{15}2.3x=2.315
\Rightarrow x=15⇒x=15
Vậy x = 15
7 tháng 3 2021
Bài 2:
a) Ta có: \(f\left(2\right)-f\left(-1\right)=7\)
\(\Leftrightarrow2\left(m-1\right)-\left(-1\right)\cdot\left(m-1\right)=7\)
\(\Leftrightarrow2m-2+m-1=7\)
\(\Leftrightarrow3m-3=7\)
\(\Leftrightarrow3m=10\)
hay \(m=\dfrac{10}{3}\)
8 tháng 6 2023
\(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+...+\dfrac{1}{a\times\left(a+4\right)}=\dfrac{50}{609}\)
\(\dfrac{1}{4}\times\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+...+\dfrac{4}{a\times\left(a+4\right)}\right)=\dfrac{50}{609}\)
\(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{a}-\dfrac{1}{a\times4}=\dfrac{50}{609}\div\dfrac{1}{4}\)
\(\dfrac{1}{3}-\dfrac{1}{a\times4}=\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{3}-\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{203}\)
\(a\times4=203\)
\(a=\dfrac{203}{4}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 6 2023
\(\dfrac{1}{3\times7}\)+\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\)
4\(\times\)( \(\dfrac{1}{3\times7}\) +\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\)) = \(\dfrac{50}{609}\) \(\times\)4
\(\dfrac{4}{3\times7}\)+ \(\dfrac{4}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{4}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\) \(\times\) 4
\(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\)-\(\dfrac{1}{15}\)+...+\(\dfrac{1}{a}\)-\(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{3}\) - \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{203}\)
a + 4 = 203
\(a\) = 203 - 4
\(a\) = 199
Đáp số: \(a\) = 199
20 tháng 1
a: \(\dfrac{4}{5}-\dfrac{5}{6}< =\dfrac{x}{30}< =\dfrac{1}{3}-\dfrac{3}{10}\)
=>\(\dfrac{24-25}{30}< =\dfrac{x}{30}< =\dfrac{10-9}{30}\)
=>\(\dfrac{-1}{30}< =\dfrac{x}{30}< =\dfrac{1}{30}\)
=>-1<=x<=1
mà x nguyên
nên \(x\in\left\{-1;0;1\right\}\)
b: \(\dfrac{a}{7}+\dfrac{1}{14}=\dfrac{-1}{b}\)
=>\(\dfrac{2a+1}{14}=\dfrac{-1}{b}\)
=>\(\left(2a+1\right)\cdot b=-14\)
mà 2a+1 lẻ (do a là số nguyên)
nên \(\left(2a+1\right)\cdot b=1\cdot\left(-14\right)=\left(-1\right)\cdot14=7\cdot\left(-2\right)=\left(-7\right)\cdot2\)
=>\(\left(2a+1;b\right)\in\left\{\left(1;-14\right);\left(-1;14\right);\left(7;-2\right);\left(-7;2\right)\right\}\)
=>\(\left(a;b\right)\in\left\{\left(0;-14\right);\left(-1;14\right);\left(3;-2\right);\left(-4;2\right)\right\}\)