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a) 10-x-5=-5-7-11
=> 5 - x = -23
=> x = 28
b) |x| -3=0
=> |x| = 3
=> x = 3 hoặc x -3
c) ( 7-|x| ) .(2x-4)=0
=> 7 - |x| = 0 hoặc 2x - 4 = 0
=> |x| = 7 hoặc 2x = 4
=> x = 7 hoặc x = - 7 hoặc x = 2
c)2+3x=-15-19
=> 2 + 3x = -34
=> 3x = 36
=> x = 12
a,
A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99
=1−(3+5+7+...+99)=1−(3+5+7+...+99)
=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498
b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98
c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99
các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha
a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)
-\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)
-6 < \(x\) < -4
vì \(x\) \(\in\) Z nên \(x\) = -5
\(\frac{6}{11}\cdot\frac{3}{14}+\frac{-5}{16}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-3}{16}\)
\(=\frac{6}{11}\cdot\frac{3}{14}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-5}{16}+\frac{-3}{16}\)
\(=\frac{3}{14}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+\left(\frac{-5}{16}+\frac{-3}{16}\right)\)
\(=\frac{3}{14}\cdot\frac{6+5}{11}+\frac{-5+\left(-3\right)}{16}\)
\(=\frac{3}{14}\cdot\frac{11}{11}+\frac{-8}{16}\)
\(=\frac{3}{14}\cdot1+\frac{-1}{2}\)
\(=\frac{3}{14}+\frac{-1}{2}\)
\(=\frac{3}{14}+\frac{-7}{14}\)
\(=\frac{3+\left(-7\right)}{14}\)\(=\frac{-4}{14}=\frac{-2}{7}\)
\(\frac{-5}{6}+\left(7x-\frac{1}{2}\right)\cdot\frac{2}{9}=-1\frac{1}{3}\)
\(\frac{-5}{6}+\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}\)
\(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}+\frac{5}{6}\)
\(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{1}{2}\)
\(\frac{14}{9}\cdot x=-\frac{1}{2}+\frac{1}{9}\)
\(\frac{14}{9}\cdot x=-\frac{7}{18}\)
\(x=-\frac{7}{18}:\frac{14}{9}\)
\(x=-\frac{1}{4}\)
1)C= 1/5+1/10+1/20+1/40+...+1/1280
\(=\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
Đặt cái trong ngoặc là A ta có:\(2A=2+1+...+\frac{1}{2^7}\)
\(2A-A=\left(2+1+...+\frac{1}{2^7}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
\(A=2-\frac{1}{2^8}\).Thay A vào ta được:\(C=\frac{1}{5}\left(2-\frac{1}{2^8}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)
2)D= 2/1*3+2/3*5+2/5*10+2/7*9+2/9*11+2/11*18+2/13*15
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
3)E= 4/3*7+4/7*11+4/11*15+4/15*19+4/19*23+4/23*27
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}\)
\(=\frac{8}{27}\)
4)G= 1/2+1/6+1/12+1/20+1/30+1/42+...+1/110
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
5)H= 3/1*2+3/2*3+3/3*4+3/4*5+...+3/9*10
\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=3\left(1-\frac{1}{10}\right)\)
\(=3\times\frac{9}{10}\)
\(=\frac{27}{10}\).Lần sau bạn đăng ít một thôi nhé
1: \(\left(15-6\dfrac{13}{18}\right):11\dfrac{1}{27}-2\dfrac{1}{8}:1\dfrac{11}{40}\)
\(=\left(15-6-\dfrac{13}{18}\right):\left(11+\dfrac{1}{27}\right)-\dfrac{17}{8}:\dfrac{51}{40}\)
\(=\left(9-\dfrac{13}{18}\right):\dfrac{298}{27}-\dfrac{17}{8}\cdot\dfrac{40}{51}\)
\(=\dfrac{149}{18}\cdot\dfrac{27}{298}-\dfrac{5}{3}\)
\(=\dfrac{3}{4}-\dfrac{5}{3}=\dfrac{9-20}{12}=-\dfrac{11}{12}\)
2:
a: \(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\)
=>\(\dfrac{7}{4}x-5=\dfrac{10}{3}\)
=>\(\dfrac{7}{4}x=5+\dfrac{10}{3}=\dfrac{25}{3}\)
=>\(x=\dfrac{25}{3}:\dfrac{7}{4}=\dfrac{25}{3}\cdot\dfrac{4}{7}=\dfrac{100}{21}\)
b: \(3\dfrac{1}{3}x+16=13,25\)
=>\(\dfrac{10}{3}x=13,25-16=-2,75=-\dfrac{11}{4}\)
=>\(x=-\dfrac{11}{4}:\dfrac{10}{3}=-\dfrac{11}{4}\cdot\dfrac{3}{10}=-\dfrac{33}{40}\)
c: \(2x-1=\left(-4\right)^2\)
=>2x-1=16
=>2x=17
=>\(x=\dfrac{17}{2}\)