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Ta có:
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
....
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}\)
\(-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
=> đpcm
\(a,\left(4\frac{1}{2}-\frac{2}{5}x\right):1\frac{3}{4}=\frac{11}{14}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right):\frac{7}{4}=\frac{11}{4}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{11}{4}\cdot\frac{7}{4}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{77}{16}\)
\(\Rightarrow\frac{9}{2}-\frac{2}{5}x=\frac{77}{16}\)
\(\Rightarrow-\frac{2}{5}x=\frac{77}{16}-\frac{9}{2}\)
\(\Rightarrow-\frac{2}{5}x=\frac{5}{16}\)
\(\Rightarrow x=\frac{5}{16}:\left(-\frac{2}{5}\right)\)
\(\Rightarrow x=-\frac{25}{32}\)
\(b,\frac{2}{3}\cdot x-\frac{2}{5}x=\frac{9}{3}\)
\(\Rightarrow x\left(\frac{2}{3}-\frac{2}{5}\right)=\frac{8}{3}\)
\(\Rightarrow x\cdot\frac{4}{15}=\frac{8}{3}\)
\(\Rightarrow x=\frac{8}{3}:\frac{4}{15}\)
\(\Rightarrow x=10\)
\(c,\frac{-2}{3}|x|+1\frac{1}{2}=\frac{2}{5}\)
\(\Rightarrow\frac{-2}{3}|x|+\frac{3}{2}=\frac{2}{5}\)
\(\Rightarrow\frac{-2}{3}|x|=\frac{2}{5}-\frac{3}{2}\)
\(\Rightarrow\frac{-2}{3}|x|=-\frac{11}{10}\)
\(\Rightarrow|x|=\frac{-11}{10}:\frac{-2}{3}\)
\(\Rightarrow|x|=\frac{33}{20}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{33}{20}\\x=-\frac{33}{20}\end{cases}}\)
\(d,|2x-\frac{1}{3}|+\frac{1}{6}=\frac{3}{4}\)
\(\Rightarrow|2x-\frac{1}{3}|=\frac{3}{4}-\frac{1}{6}\)
\(\Rightarrow|2x-\frac{1}{3}|=\frac{7}{12}\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=\frac{7}{12}\\2x-\frac{1}{3}=-\frac{7}{12}\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{11}{12}\\2x=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{24}\\x=-\frac{1}{8}\end{cases}}}\)
Bài làm:
Xét: \(\frac{1}{5^2}>\frac{1}{5.6}\) ; \(\frac{1}{6^2}>\frac{1}{6.7}\) ; ... ; \(\frac{1}{100^2}>\frac{1}{100.101}\)
=> \(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\) (1)
Lại có: \(\frac{1}{5^2}< \frac{1}{4.5}\) ; \(\frac{1}{6^2}< \frac{1}{5.6}\) ; ... ; \(\frac{1}{100^2}< \frac{1}{99.100}\)
=> \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\) (2)
Từ (1) và (2) => \(\frac{1}{6}< A< \frac{1}{4}\)
(1/12+3 1/6-30,75).x -8 = (3/5+0,415+1/200):0,01
(1/12+19/6-123/4).x-8=(3/5+83/200+1/200):1/100
-55/2.x-8=51/50:1/100
-55/2.x-8=102
-55/2.x=102+8=110
x=110:-55/2=-4
a) Vì n\(\inℕ\)nên n + 1 \(\inℕ\)và 2n + 3\(\inℕ\).
Gọi d \(\in\)ƯCLN ( n + 1 , 2n + 3 )
\(\Rightarrow n+1⋮d\)và \(2n+3⋮d\)
\(\Rightarrow\left(2n+3\right)-2\left(n+1\right)⋮d\)
\(\Rightarrow2n+3-2n-2⋮d\)
\(\Rightarrow1⋮d\Rightarrow d\in\left\{1;-1\right\}\)
\(\Rightarrow\frac{n+1}{2n+3}\)là phân số tối giản .
Vậy \(\frac{n+1}{2n+3}\)tối giản \(\forall n\inℕ\).
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow2B=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{101}\)
\(\Rightarrow2B-B=\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{101}\right]-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\right]\)
\(\Rightarrow B=1-\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow B=1-\frac{1}{2^{100}}\)
\(\Rightarrow B< 1\)(đpcm)
_Chúc bạn học tốt_