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a.\(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}\)
\(=\dfrac{3.4.5...100}{2.3.4...99}\)
\(=\dfrac{100}{2}=50\)
a,
\(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\\ =\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\\ =\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}\\ =\dfrac{100}{2}=50\)
b,
\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{100}-1\right)\\ =\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-99}{100}\\ =\dfrac{\left(-1\right)\left(-2\right)\left(-3\right)\cdot...\cdot\left(-99\right)}{2\cdot3\cdot4\cdot...\cdot100}\\ =\dfrac{\left(-1\right)\left(-1\right)\left(-1\right)\cdot...\left(-1\right)}{100}\left(\text{có }99\text{ số }-1\right)\\ =\dfrac{\left(-1\right)^{99}}{100}\\ =\dfrac{-1}{100}\)
c,
\(C=\dfrac{4}{30}+\dfrac{4}{70}+\dfrac{4}{126}+...+\dfrac{4}{798}\\ =\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\\ =\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{19\cdot21}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\\ =\dfrac{1}{3}-\dfrac{1}{21}\\ =\dfrac{7}{21}-\dfrac{1}{21}\\ =\dfrac{6}{21}=\dfrac{2}{7}\)
ta có
\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)
_______________________ X ________________________
\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)
= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)
1) \(x+\dfrac{30}{100}x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}x=-\dfrac{131}{100}\)
\(\Leftrightarrow100x+30x=-131\)
\(\Leftrightarrow130x=-131\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
Vậy \(x=-\dfrac{131}{130}\)
b) \(\left(4,5-2x\right)\cdot\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow\left(\dfrac{9}{2}-2x\right)\cdot\left(-\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow-\dfrac{18}{7}+\dfrac{8}{7}x=\dfrac{11}{4}\)
\(\Leftrightarrow-72+32x=77\)
\(\Leftrightarrow32x=77+72\)
\(\Leftrightarrow32x=149\)
\(\Leftrightarrow x=\dfrac{149}{32}\)
Vậy \(x=\dfrac{149}{32}\)
a) Thực hiện phép nhân ở vế phải rồi áp dụng quy tắc chuyển vế.
b) Thực hiện phép nhân ở về phải rồi quy đồng mẫu hai vế.
ĐS. a) ; b) x = -40.
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)
\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)
\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)
\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)
\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)
\(\Rightarrow x=-36\)
Vậy ...
a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)
\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)
Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)
Chúc bạn học tốt!!!
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, (\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)).10 - x = 0
<=> \(\dfrac{5}{6}.10-x=0\)
<=> \(\dfrac{25}{3}-x=0\)
<=> x = \(\dfrac{25}{3}\) (thỏa mãn)
@Hoàng Mạnh Quân
\(=81.\dfrac{12.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}.\dfrac{158}{711}\)
\(=81.\dfrac{12}{4}:\dfrac{5}{6}.\dfrac{2}{9}\)
\(=243:\dfrac{5}{6}.\dfrac{2}{9}\)
\(=\dfrac{1458}{5}.\dfrac{2}{9}\)
\(=\dfrac{324}{5}\)