tìm x
\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x.\left(x+1\right)}=\frac{13}{90}\)
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Tìm x \(\in\)N
\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x.\left(x+1\right)}=\frac{13}{90}\)
giúp mik nha các bn giải chi tiết ra nhé các bài kia cx vậy nha
\(\text{Ta có: }\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{x.\left(x+1\right)}=\frac{13}{90}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{13}{90}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 = 18
=> x = 17
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1.Tìm x :
a,frac{1}{5.6}+frac{1}{6.7}+...+frac{1}{xleft(x+1right)}frac{13}{90}
b,frac{1}{1.4}+frac{1}{4.7}+frac{1}{7.10}+...+frac{1}{xleft(x+3right)}frac{49}{148}
c,frac{7}{left(x+3right)left(x+10right)}+frac{11}{left(x+10right)left(x+21right)}+frac{1}{left(x+21right)left(x+34right)}frac{x}{left(x+3right)left(x+34right)}
d,frac{3}{left(x-4right)left(x-7right)}+frac{6}{left(x-7right)left(x-13right)}+frac{15}{left(x-13right)left(x-28right)}-frac{1}{x-38}frac{-1}{20}
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1.Tìm x :
a,\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
b,\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
c,\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}\)\(+\frac{1}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
d,\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}\)\(+\frac{15}{\left(x-13\right)\left(x-28\right)}\)\(-\frac{1}{x-38}=\frac{-1}{20}\)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
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\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+..........+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
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
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Tập hợp các giá trị nguyên dương của x thỏa mãn:\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)x<\frac{13}{7}\)có số phần tử là...........
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)x<\frac{13}{7}\)
\(\left(1-\frac{1}{7}\right).x<\frac{13}{7}\)
\(\frac{6}{7}.x<\frac{13}{7}\Leftrightarrow6x<13\Leftrightarrow x<2,1\left(6\right)\)
x nguyên dương => x thuộc {1;2}
Vậy tập hợp có 2 phần tử
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Tìm x, biết:
\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x.\left(x+1\right)}=\frac{1}{2016}\)
Mình cần giải cụ thể ko cần kết quả ~~
\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2016}\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2016}\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{1}{2016}\Leftrightarrow\frac{x+1-5}{5\left(x+1\right)}=\frac{1}{2016}\Leftrightarrow\frac{x-4}{5x+5}=\frac{1}{2016}\Rightarrow2016\left(x-4\right)=5x+5\Leftrightarrow2016x-8064=5x+5\)còn lại tự giải
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=> \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2016}\)
=> \(\frac{1}{5}-\frac{1}{x+1}=\frac{1}{2016}\)
=> \(\frac{1}{x+1}=\frac{1}{5}-\frac{1}{2016}=\frac{2011}{10080}\)
=> \(x+1=1:\frac{2011}{10080}=\frac{10080}{2011}\)
=> x= 8069/2011
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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
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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 \(x=29\)
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Tìm x,y
a, \(\left|3x-1\right|+\left(y-2\right)^2=0\)
b \(\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{x.\left(x+1\right)}=\frac{19}{100}\)
c. \(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{x.\left(x+2\right)}=\frac{49}{99}\)
d. \(\left(2x+1\right).\left(y-2\right)=6\)
e \(^{x^2-3xy+3y-x=1}\)
b \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{19}{100}\)
=>\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)
=>\(\frac{1}{5}-\frac{1}{x+1}\)\(=\frac{19}{100}\)
=>\(\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)
=>\(\frac{1}{x+1}=\frac{1}{100}\)
=> x+1 =100
=>x=99
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b) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{100}\)
\(\Rightarrow x+1=100\)
\(\Rightarrow x=99\)
c) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{49}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{50}{99}\)
\(\Rightarrow50.\left(x+2\right)=99\)
\(\Rightarrow x+2=\frac{99}{50}\)
\(\Rightarrow x=-\frac{1}{99}\)
d) Ta có : 6 = 1.6 = 2.3 = (-2) . (-3)
Lâp bảng xét 6 trường hợp:
| \(2x+1\) | \(1\) | \(6\) | \(2\) | \(3\) | \(-2\) | \(-3\) |
| \(y-2\) | \(6\) | \(1\) | \(3\) | \(2\) | \(-3\) | \(-2\) |
| \(x\) | \(0\) | \(\frac{5}{2}\) | \(\frac{1}{2}\) | \(1\) | \(-\frac{3}{2}\) | \(-2\) |
| \(y\) | \(8\) | \(3\) | \(5\) | \(4\) | \(-1\) | \(0\) |
Vậy các cặp (x,y) \(\inℤ\)thỏa mãn là : (0;4) ; (1; 4) ; (-2 ; 0)
e) \(x^2-3xy+3y-x=1\)
\(\Rightarrow x\left(x-3y\right)+3y-x=1\)
\(\Rightarrow x\left(x-3y\right)-\left(x-3y\right)=1\)
\(\Rightarrow\left(x-3y\right)\left(x-1\right)=1\)
Lại có : 1 = 1.1 = (-1) . (-1)
Lập bảng xét các trường hợp :
| \(x-1\) | \(1\) | \(-1\) |
| \(x-3y\) | \(1\) | \(-1\) |
| \(x\) | \(2\) | \(0\) |
| \(y\) | \(\frac{1}{3}\) | \(\frac{1}{3}\) |
Vậy các cặp(x,y) thỏa mãn là : \(\left(2;\frac{1}{3}\right);\left(0;\frac{1}{3}\right)\)
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Tìm x : \(\left(1.2.3+2.3.4+...+2015.2016.2017\right)-\left(5.6+6.7+...+99.100\right)\) \(=\left(5+10+15+...+2000\right)-\frac{1}{1.3}-\frac{1}{3.5}-...-\frac{1}{97.99}-4x\)
Tập hợp các giá trị nguyên dương của x thỏa mãn:$\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)x<\frac{13}{7}$(11.2 +12.3 +13.4 +14.5 +15.6 +16.7 )x<137 có số phần tử là...........
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