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a)4/5+x=2/3
x=2/3-4/5
x=-2/15
b)-5/6-x=2/3
x=-5/6-2/3
x=-3/2
c)1/2x+3/4=-3/10
1/2x=-3/10-3/4
1/2x=-21/20
x=-21/20:1/2
x=-21/10
d)x/3-1/2=1/5
x/3=1/5+1/2
x/3=7/10
10x/30=21/30
10x=21
x=21:10
x=21/10
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
\(M=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}+\dfrac{1}{120}\)
\(M=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)
\(M=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(M=2.\dfrac{3}{16}\)
\(M=\dfrac{3}{8}\)
Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)
a) \(\dfrac{1}{2}< \dfrac{x}{9}< \dfrac{y}{18}< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{9}{18}< \dfrac{2x}{18}< \dfrac{y}{18}< \dfrac{12}{18}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=10\\y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=11\end{matrix}\right.\)
b) \(\dfrac{-1}{2}< \dfrac{x}{15}< \dfrac{y}{30}< \dfrac{-2}{5}\)
\(\Leftrightarrow\dfrac{-15}{30}< \dfrac{2x}{30}< \dfrac{y}{30}< \dfrac{-12}{30}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=-14\\y=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\y=-13\end{matrix}\right.\)
\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)
\(=-\dfrac{37}{16}\)
\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)
\(=\dfrac{5}{17}+\dfrac{-3}{17}\)
\(=\dfrac{2}{17}\)
\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)
\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)
\(=2-\dfrac{7}{30}\)
\(=\dfrac{53}{30}\)
\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)
\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)
\(=\dfrac{15}{34}\)
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{30}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{29}{30}\)
\(=\dfrac{1.2.3...29}{2.3.4...30}\)
\(=\dfrac{1}{30}\)
\(B=1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.1\dfrac{1}{24}...1\dfrac{1}{168}\)
\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{169}{168}\)
\(=\dfrac{4.9.16.25...169}{3.8.15.24...168}\)
\(=\dfrac{2.3.4...13}{1.2.3...12}.\dfrac{2.3.4...13}{3.4.5...14}\)
\(=13.\dfrac{1}{7}\)
\(=\dfrac{13}{7}\).
A = \(\dfrac{1}{2}x\dfrac{2}{3}x\dfrac{3}{4}x...x\dfrac{29}{30}=1x1x1x...x\dfrac{1}{30}=\dfrac{1}{30}\)
B = \(\dfrac{4}{3}x\dfrac{9}{8}x\dfrac{16}{15}x\dfrac{25}{24}x...x\dfrac{169}{168}=1x1x1x1x...x\dfrac{13}{7}=\dfrac{13}{7}\)
Câu B em chưa rõ cách làm nhanh cho lắm. Nếu ko cần tính nhanh thì chị có thể giải bình thường ra giấy ha.
mọi người thật là nhẫn tâm
chẳng ai giúp mk
TRỜI ƠI!!! AI MS LÀ BN BÈ THỰC SỰ
Ko cs đứa mô trả lời chứ chi
Loại bn bè vs mấy ng chỉ là giả tạo thôi
a) \(\dfrac{\dfrac{7}{10}+\dfrac{3}{5}}{\dfrac{7}{10}+\dfrac{1}{2}}\) = (\(\dfrac{7}{10}+\dfrac{3}{5}\) ) : ( \(\dfrac{7}{10}+\dfrac{1}{2}\) )
= \(\dfrac{7+6}{10}\) : \(\dfrac{7+5}{10}\)
= \(\dfrac{13}{10}:\dfrac{12}{10}\)
= \(\dfrac{13}{10}.\dfrac{10}{12}\) =\(\dfrac{13}{12}\)
b) \(\dfrac{6-\dfrac{1}{\dfrac{1}{2}-\dfrac{1}{3}}}{6+\dfrac{1}{\dfrac{1}{2}-\dfrac{1}{3}}}\)
= ( 6 - \(\dfrac{1}{\dfrac{1}{2}-\dfrac{1}{3}}\) ) : ( 6 + \(\dfrac{1}{\dfrac{1}{2}-\dfrac{1}{3}}\) )
= ( 6 - \(\dfrac{1}{\dfrac{1}{6}}\) ) : ( 6 + \(\dfrac{1}{\dfrac{1}{6}}\) )
= ( 6 - 6 ) : (6 + 6) = \(\dfrac{0}{12}\) =0
1/
a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)
b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993
2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993
2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993
2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993
2.(1 − 1/x+1) = 3984/1993
1 − 1/x + 1= 3984/1993 :2
1 − 1/x+1 = 1992/1993
1/x+1 = 1 − 1992/1993
1/x+1=1/1993
<=>x+1 = 1993
<=>x+1=1993
<=> x+1=1993
<=> x = 1993-1
<=> x = 1992
\(A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{30^2}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{30}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{29}{30}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{31}{30}\)
\(=\dfrac{1}{30}\cdot\dfrac{31}{2}=\dfrac{31}{60}\)
\(B=1+\dfrac{1}{3}+...+\dfrac{1}{120}\)
\(=\dfrac{2}{2}+\dfrac{2}{6}+...+\dfrac{2}{240}\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{240}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(=2\left(1-\dfrac{1}{16}\right)=2\cdot\dfrac{15}{16}=\dfrac{15}{8}\)
Giá trị của A xấp xỉ 0,4431 và giá trị của B chính xác là 2.