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a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)
Ta co \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\cdot\left(x+1\right)}\)
\(\Rightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\cdot\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\)\(2\cdot\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow\)\(2\cdot\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow\)\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(x=17\)
a) 3 x (x + 1) + 4 x (x + 2) = 18
<=> 3x + 3 + 4x + 8 = 18
<=> 7x + 11 = 18
<=> 7x = 18 - 11
<=> 7x = 7
<=> x = 7 : 7 = 1
=> x = 1
b) 4 x (x + 1) + 7 x (2x + 3) = 61
<=> 4x + 4 + 14x + 21 = 61
<=> 18x + 25 = 61
<=> 18x = 61 - 25
<=> 18x = 36
<=> x = 36 : 18
<=> x = 2
=> x = 2
đề chính xác là
a/ \(13-2\times\left(36-5\times x\right)=1\)
b/ \(53-10\times\left(40-7\times x\right)=3\)
c/ \(\frac{3}{4}-\frac{1}{6}\div x=0,375\)
\(\left(74,25+\frac{3}{4}\right)\times x-131,5=18,5\)
\(75\times x=18,5+131,5\)
\(75\times x=150\)
\(x=150:75\)
\(x=2\)
\(\frac{3}{4}=0,75\)
(74,25+0,75)x\(x\)-131,5=18,5
75 x \(x\)-131,5=18,5
75x\(x\) =18,5+131,5
75x\(x\) =150
\(x\) =150:75
\(x\) =2
( 5 . x + 4 ) x ( 12 : x - 1 ) = 0
Vậy nếu kết quả bằng không thì trong hai tích sau đây sẽ có 1 k = 0
mà k đầu tiên ko thể = 0 vì 5 . x ( x = số nào cúng đc bé nhất là 0 đi ) thì = 0 lại cộng thêm 4 ko thể = 0
Vậy k 2 = 0
12 : x = 0 + 1 = 1
vậy x = 12 : 1
x = 12
\(P=\frac{1}{5x8}+\frac{1}{8x11}+.....+\frac{1}{602x605}\)
\(\Rightarrow3P=\frac{3}{5x8}+\frac{3}{8x11}+......+\frac{3}{602x605}\)
\(\Rightarrow3P=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-.....+\frac{1}{602}-\frac{1}{605}\)
\(\Rightarrow3P=\frac{1}{5}-\frac{1}{605}\)
\(\Rightarrow3P=\frac{24}{121}\)
\(\Rightarrow P=\frac{24}{121}:3\)
\(\Rightarrow P=\frac{8}{121}\)
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\right)\cdot x=2009\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\right)\cdot x=2009\)
\(\left(1-\frac{1}{2010}\right)\cdot x=2009\)
\(\frac{2009}{2010}\cdot x=2009\)
\(x=2009:\frac{2009}{2010}\)
\(x=2010\)