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= 7.( \(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-....-\frac{1}{70}\))
= 7.( \(\frac{1}{10}-\frac{1}{70}\))
= 7.(\(\frac{7}{70}-\frac{1}{70}\))
= 7.\(\frac{6}{70}\)
= \(\frac{3}{5}\)
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.
Ta có: \(\frac{1}{1^2}=\frac{1}{1\cdot1};\frac{1}{2^2}<\frac{1}{1\cdot2};...;\frac{1}{50^2}<\frac{1}{49\cdot50}\)
=>\(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}<1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}=1+1-\frac{1}{50}=2-\frac{1}{50}=1,98\)
hay A<1,98 mà 1,98<2 nên A<2
Vậy A<2
theo bài ra ta có:
15+a/29+a=3/5
=>(15+a).5=(29+a).3
=>75+5a=87+3a
=>5a-3a=87-75
=>2a=12
=>a=6
vậy a=6
tick nhé
Câu 1 :
Đk: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{2x-1}=5\\ \Leftrightarrow x-1+2\sqrt{\left(x-1\right)\left(2x-1\right)}+2x-1=25\\ \Leftrightarrow2\sqrt{2x^2-3x+1}=27-3x\\ \)
\(\Leftrightarrow\begin{cases}27-3x\ge0\\4\left(2x^2-3x+1\right)=9x^2-162x+729\end{cases}\) \(\Leftrightarrow\begin{cases}x\le9\\x^2-150x+725=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x\le9\\x=145hoặcx=5\end{cases}\)
với x= 5 thoản mãn điều kiện, x=145 loại
Vậy \(S=\left\{5\right\}\)
Tổng quát: \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\) (với mọi số tự nhiên n khác 0)
Ta có: \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}<\frac{1}{2}\) (vì \(\frac{1}{100}>0\) )
=>đpcm
\(\frac{9.25-63}{9.10+153}\)=\(\frac{9.25-9.7}{9.10+9.17}\)=\(\frac{9.\left(25-7\right)}{9.\left(10+17\right)}\)=\(\frac{9.18}{9.27}\)=\(\frac{1.2}{1.3}\)=\(\frac{2}{3}\)
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)\(\frac{1}{a}-\frac{1}{b}=\frac{b}{ab}-\frac{a}{ab}=\frac{b-a}{ab}\)
\(\frac{10.11+50.55+70.77}{11.12+55.60+77.84}=\frac{10.11.\left(1+5.5+7.7\right)}{11.12.\left(1+5.5+7.7\right)}=\frac{10}{12}=\frac{5}{6}\)