tìm x, biết
1/3.4+1/4.5+1/5.6+1/6.7+...+1/x.(x+1)=3/10
tìm x biết:
1/3.4+1/4.5+1/5.6+1/6.7+....+1/x(x+1)=3/10
1/3.4+1/4.5+1/5.6+1/6.7+....+1/x(x+1)=3/10
<=> \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(x+1\right)x}=\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
<=> \(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)=> x+1=30=>x=29
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{30}\)
\(\Rightarrow x+1=30\)
\(x=30-1=29\)
Tìm x:
1/3.4+1/4.5+1/5.6+1/6.7+....+1/x(x+1)=3/10
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tìm x biết:
1/3.4+1/4.5+1/5.6+1/6.7+....+1/x(x+1)
\(\frac{1}{3.4}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{3}-\frac{1}{x+1}\)
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{x\left(x+1\right)}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+1}\)
\(=\frac{1}{3}-\frac{1}{x+1}\)
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+..........+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
Theo đề suy ra
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
=> \(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)
=>x+1=30
=>x=29
Tìm \(x\in Z\), biết:
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
1/3.4+1/4.5+1/5.6+.....+1/x(x+1)=3/10
1/3-1/4+1/4-1/5+1/5-........-1/x+1/x-1/x+1=3/10
=>1/3-1/x+1=3/10
1/x+1=3/10-1/3=1/30
=>x+1=30
x=30-1
x=29
Ta có :
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
=>\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
=>\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
=>\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)
=>\(\frac{1}{x+1}=\frac{1}{30}\)
=>\(x+1=30\)
=>\(x=30-1\)
=>\(x=29\)
Vậy \(x=29\)
Tìm \(x\in Z\), biết:
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
Quá dễ:
=> 1/3 - 1/4 + 1/4 - 1/5 + ....+ 1/x - 1/x+1 = 3/10
=> 1/3 - 1/x+1 = 3/10
=> 1/x+1 = 1/3 - 3/10
Còn lại tự làm nhá!
<=> 1/3 - 1/(x+1) = 3/10
<=> 1/(x+1) = 1/30
=> x+1 = 30
<=> x= 29
= > 1/3 - 1/4 +1/4 -1/5 + ..... + 1/x - 1/x+1 = 3/10
= > 1/3 - 1/x+1 = 3/10
= > 1/x+1 = 1/3 - 3/10
= > 1/x+1 =1/30
= > x+1 = 30
= > x = 29
Vậy x = 29
1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\\ =\dfrac{1}{2}-\dfrac{1}{7}\\ =\dfrac{5}{14}\)
Tính hợp lý
A = 1/1.2 x 4/2.3 x 9/3.4 x 16/4.5 x 25/5.6 x 36/6.7
Cảm ơn mọi người nhé
ta phân tích thành
\(\frac{1}{1\cdot2}\)x\(\frac{2\cdot2}{2\cdot3}\)x\(\frac{3\cdot3}{3\cdot4}\)x......x\(\frac{5\cdot5}{5\cdot6}\)x\(\frac{6\cdot6}{6\cdot7}\)
Tử nhân tử mẫu nhân mẫu ta có
1x2x2x3x3x.........x5x5x6x6
1x2x2x3x3x4x....x5x6x6x7
Rút gọn ta có
\(\frac{1}{7}\)
Vậy A=\(\frac{1}{7}\)