BT3: Tìm x, biết
5) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)-2x=\dfrac{1}{2}\)
Tìm x biết:
a, \(\left|3x-4\right|\le3\)
b, \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)-2x=\dfrac{1}{2}\)
b) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)-2x=\dfrac{1}{2}\)
\(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}+.......-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\)
\(\Leftrightarrow\left(1-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{99}{100}-2x=\dfrac{1}{2}\)
\(\Leftrightarrow-2x=\dfrac{1}{2}-\dfrac{99}{100}\)
\(\Leftrightarrow-2x=\dfrac{-49}{100}\)
\(\Leftrightarrow x=\dfrac{-49}{100}:\left(-2\right)\)
\(\Leftrightarrow x=\dfrac{49}{200}\)
Vậy ...............................
1) Tìm x biết:
\(\left(1-\dfrac{3}{10}-x\right):\left(\dfrac{19}{10}-1-\dfrac{2}{5}\right)+\dfrac{4}{5}=1\)
2) Tính nhanh
a)\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{10.11.12}\)
b)\(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\)
câu b bài 2:
\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
câu a bài 2:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)
\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)
Tính hợp lí:
a) (100-9)(99-9)(98-9)...(1-9).
b) \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)\)
c) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)
10.4. Tính tổng
a) \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)
b) \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\)
c) \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +...........\(\dfrac{1}{99.100}\)
d) \(\dfrac{3}{1.2}\) + \(\dfrac{3}{2.3}\) +.........\(\dfrac{1}{99.100}\)
giúp em
a)
`1/1-1/2`
`=2/2-1/2`
`=1/2`
b)
`1/(1*2)+1/(2*3)`
`=1/1-1/2+1/2-1/3`
`=1/1-1/3`
`=3/3-1/3`
`=2/3`
c)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
d)
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?
\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)
Bài 2: Tìm \(x\) biết:
\(x\)\(\times\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=1\)
\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)
Bài 2: Tìm \(x\) biết:
\(x\)\(\times\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=1\)
\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)
x-\(\dfrac{20}{11.13}-\dfrac{20}{13.15}-\dfrac{20}{15.17}-...-\dfrac{20}{53.55}=\dfrac{3}{11}\)
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(\dfrac{-2}{3}\left|x\right|+1\dfrac{1}{2}=\dfrac{2}{5}\)
các bạn phại đổi kiểu chữ để làm bài này (VNI) , (TELEX)
x - \(\dfrac{20}{11.13}-\dfrac{20}{13.15}-\dfrac{20}{15.17}-........-\dfrac{20}{53.55}=\dfrac{3}{11}\)
x - \(\left(\dfrac{20}{11.13}+\dfrac{20}{13.15}+\dfrac{20}{15.17}+...+\dfrac{20}{53.55}\right)=\dfrac{3}{11}\)
Đặt A = \(\dfrac{20}{11.13}+\dfrac{20}{13.15}+\dfrac{20}{15.17}+...+\dfrac{20}{53.55}\)
A = \(\dfrac{20}{2}\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+\dfrac{2}{15.17}+....+\dfrac{2}{53.55}\right)\)
A = \(\dfrac{20}{2}\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)
A = \(10.\left(\dfrac{1}{11}-\dfrac{1}{55}\right)\)
A = 10 . 4/55
A = 8/11
TA Có : x - 8/11 = 3/11
x = 3/11 + 8/11
x = 11/11
x = 1
Vậy x = 1
bài1: Tìm x
\(\left|x+\dfrac{1}{1.2}\right|+\left|x+\dfrac{1}{2.3}\right|+......+\left|x+\dfrac{1}{99.100}\right|=100x\)
Help me, please! Thanks.
Do mỗi số hạng ở vế trái nằm trong dấu giá trị tuyệt đối mà vế phải 100 là số dương nên x cũng là số dương
Do x dương nên ta có:
\(x+\dfrac{1}{1.2}+x+\dfrac{1}{2.3}+...+x+\dfrac{1}{99.100}=100x\)
Dãy trên có 99 số hạng nên
\(99x+\left(x-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(1-\dfrac{1}{100}=x\Rightarrow x=\dfrac{99}{100}\)
Vậy \(x=\dfrac{99}{100}\)
Bài 1 : Tìm 2 số biết hiệu của chúng bằng 5 và 50% số lớn = 1 nửa số bé.
Bài 2 : tính giá trị biểu thức: A = \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{98.99}+\dfrac{9}{99.100}\)
Bài 3 : Tìm x
a , \(\dfrac{x}{3}-\dfrac{1}{8}=\dfrac{5}{8}\)
bài 2:
\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)
bài 3:
\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)
\(=>x=3\)