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1/21+1/28+1/36+...+2/[x.(x+1)]=2/9
=>2/42+2/56+2/72+...+2/[x.(x+1)]=2/9
=>2.(1/42+1/56+1/72+...+1/[x.(x+1)])=2/9
=>2.(1/6-1/7+1/7-1/8+1/8-1/9+...+1/x-1/(x+1))=2/9
=>1/6-1/(x+1)=1/9
=>1/(x+1)=1/18
=>x+1=18
=>x=17
Đặt \(A=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{21.2}+\frac{1}{28.2}+\frac{1}{36.2}...+\frac{2}{x.\left(x+1\right).2}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6.4}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}-\frac{1}{x+1}\)
\(\Rightarrow A=\left(\frac{1}{6}-\frac{1}{x+1}\right):\frac{1}{2}\)
Theo bài ra ta có :
\(\left(\frac{1}{6}-\frac{1}{x+1}\right):\frac{1}{2}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}.\frac{1}{2}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{18}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{2}{18}\)
\(\frac{1}{x+1}=\frac{3}{18}-\frac{2}{18}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=18-1\)
\(\Rightarrow x=17\)
Vậy x = 17
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
Câu 1,
x+y=-1/3 ; y+z=5/4 ; x+z= 4/3
=> 2(x+y+z)=9/4
=> x+y+z=9/8
Ta lại có: x+y=-1/3
=> z=9/8 -(-1/3)=35/24
Ta lại có: z+y=5/4
=> y=-5/24
=> x=.....
Câu 2:
\(-4\le x\le-\frac{11}{18}\)
a) \(x=\frac{9}{10}\)
b) \(x=\frac{-4}{3}\)
c) \(x=\frac{1}{42}\)
d) \(x=\frac{-47}{10}\)
ko có thời gian nên mình chỉ cho đáp án thôi nhé
thông cảm cho mình ngen
đúng thì k đấy
chúc bạn học giỏi
Giải toán trên mạng - Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
Em tham khảo nhé!
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=>\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2=\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}=\frac{2009}{4018}-\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)
=> \(1\cdot2009=1\left(x+1\right)\)
=> \(x+1=2009\Rightarrow x=2009-1=2008\)
Vậy x = 2008
Chúc bn hk tốt !
Tìm x
\(a,2x-25\%=\frac{1}{2}\)
\(b,\left(\frac{3x}{7}+1\right).\left(-0,25\right)=\frac{-1}{28}\)
\(\)
\(\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}\le x\le\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)\)
\(taco:\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}=\frac{-7}{6}:\frac{-1}{4}=\frac{14}{3}\)
\(\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)=\left(\frac{-5}{6}+\frac{-16}{3}\right)\cdot\left(-14\right)=\frac{-37}{6}\cdot\left(-14\right)=\frac{259}{3}\)
TU DO \(=>X=\frac{14}{3};\frac{15}{3};,,,;\frac{259}{3}\)
CHUC BAN HOC TOT :))
=2/42+2/56+....+2/x.(x+1)
=2/6.7+2/7.8+.....+2/x.(x+1)
=2.(1/6.7-1/7.8+....+1/x.(x+1)
=2.(1/6-1/7+1/7-1/8+...+1/x-1/x+1)
=2.(1/6-1/x+1)
=2.(x-5/6x+6)
=2x-10/6x+6
bn vt thieu de nha = bao nhieu bn tu tinh