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a: \(B=\left(-\dfrac{1}{5}-\dfrac{5}{7}+\dfrac{-3}{35}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{2}\right)+\dfrac{1}{41}\)
\(=\dfrac{-7-25-3}{35}+\dfrac{3+2+1}{6}+\dfrac{1}{41}=\dfrac{42}{41}-1=\dfrac{1}{41}\)
Câu 1 :\(P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{100}=\frac{1}{100}\)
Thay a,b,c lần lượt vào biểu thức...
Tính được kết quả:
a) A= \(-\frac{7}{10}\)
b) B= \(-\frac{2}{7}\)
c) C= 0
\(\dfrac{1}{k^2}<\dfrac{1}{k(k-1)}=\dfrac{1}{k-1}-\dfrac{1}{k}\)
Ap dung:
\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\ldots+\dfrac{1}{n^2}<1+\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\ldots+\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)=2-\dfrac{1}{n}<2\)
Không chép lại đề nhé
Ta có:
P=\(\frac{50-49}{49}+\frac{50-48}{48}+...+\frac{50-2}{2}+\frac{50-1}{1}\)
P=\(\frac{50}{49}-\frac{49}{49}+\frac{50}{48}-\frac{48}{48}+...+\frac{50}{2}-\frac{2}{2}+\frac{50}{1}-\frac{1}{1}\)
P=\(\left(\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\right)+\frac{50}{1}-\left(\frac{49}{49}+\frac{48}{48}+...+\frac{2}{2}+\frac{1}{1}\right)\)
P=\(50\cdot\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)+50-49\) (chỗ này gộp nha)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{48}+\frac{1}{49}\right)+1\)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)+\frac{50}{50}\)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)\)
=>P=50S
=>\(\frac{S}{P}=\frac{S}{50S}=\frac{1}{50}\)
Vừa nãy mình nói nhầm, Sorry.
d) 7/18 . x - 2/3 = 5/18
7/18 . x = 5/18+2/3
= 17/18 : 7/18
= 17/7
e) 4/9 - 7/8 . x = -2/3
7/8 . x = 4/9 - -2/3
= 10/9 : 7/8
= 80/63
f) 1/6 + -5/7 : x = -7/18
-5/7 : x = -7/18 - 1/6
-5/7 : x = -5/9
= -5/7 : -5/9
= 9/7
Sau đó bạn thử lại kết quả nha!
a)\(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1}{a\left(a+1\right)}-\frac{a}{\left(a+1\right)a}=\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}\)
b)\(\frac{1}{a}-\frac{1}{b}=\frac{b}{ab}-\frac{a}{ab}=\frac{b-a}{ab}\)
Ta có: Vế phải bằng: \(\frac{1}{n}\) - \(\frac{1}{n+1}\) = \(\frac{n+1}{n\left(n+1\right)}\) - \(\frac{n}{n\left(n+1\right)}\) = \(\frac{1}{n\left(n+1\right)}\)= \(\frac{1}{n}\) - \(\frac{1}{n+1}\) =>đpcm.
a) \(\frac{1}{n}\) - \(\frac{1}{n+1}\) = \(\frac{n+1}{n\left(n+1\right)}\) - \(\frac{n}{n\left(n+1\right)}\) = \(\frac{1}{n\left(n+1\right)}\) = \(\frac{1}{n}\) . \(\frac{1}{n+1}\) =>đpcm
b) A= \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{8}\) - \(\frac{1}{9}\) +\(\frac{1}{9}\)
= \(\frac{1}{2}\) + \(\frac{1}{9}\)= \(\frac{11}{18}\)