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22 tháng 5 2019
a, \(3\frac{12}{13}< x\frac{12}{13}< 5\frac{12}{13}\Rightarrow x=4\)
b, \(x\frac{3}{4}=\frac{21989}{7996}=\frac{11}{4}=2\frac{3}{4}\Rightarrow x=2\)
~ Hok tốt ~
25 tháng 6 2017
Sao nhiều quá vại??
mk lm k nổi đâu
Dài quá nhìn lòi bảng họng lun ak
DP
26 tháng 6 2017
Bài : 4
a/ \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+....+\frac{1}{24\cdot25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
b/ \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{99\cdot101}\)
\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{101-99}{99\cdot101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{100}{101}\)
c/ \(\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}+\frac{5^2}{16\cdot21}+\frac{5^2}{21\cdot26}+\frac{5^2}{26\cdot31}\)
\(=\frac{25}{1\cdot6}+\frac{25}{6\cdot11}+\frac{25}{11\cdot16}+\frac{25}{16\cdot21}+\frac{25}{21\cdot26}+\frac{25}{26\cdot31}\)
\(=\frac{6-1}{1\cdot6}+\frac{11-6}{6\cdot11}+....+\frac{31-26}{26\cdot31}\)
\(=\frac{25}{5}\cdot\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{26}-\frac{1}{31}\right)\)
\(=\frac{25}{5}\cdot\left(\frac{1}{1}-\frac{1}{31}\right)\)
\(=\frac{25}{5}\cdot\frac{30}{31}\)
\(=\frac{150}{31}\)
d/ \(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+....+\frac{3}{49\cdot51}\)
\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+....+\frac{51-49}{49\cdot51}\)
\(=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{51}\right)\)
\(=\frac{3}{2}\cdot\frac{50}{51}\)
\(=\frac{25}{17}\)
e/ \(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(=\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\frac{1}{13\cdot19}+\frac{1}{19\cdot25}+\frac{1}{25\cdot31}+\frac{1}{31\cdot37}\)
\(=\frac{7-1}{1\cdot7}+\frac{13-7}{7\cdot13}+....+\frac{37-31}{31\cdot37}\)
\(=\frac{1}{6}\cdot\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+....+\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac{1}{6}\cdot\left(1-\frac{1}{37}\right)\)
\(=\frac{1}{6}\cdot\frac{36}{37}\)
\(=\frac{6}{37}\)
DP
30 tháng 6 2017
\(X-\left(\frac{31}{5}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\right)=\frac{9}{13}\)
\(X-\left(\frac{31}{5}+\frac{31}{3\cdot5}+\frac{31}{5\cdot7}+\frac{31}{7\cdot9}+\frac{31}{9\cdot11}+\frac{31}{11\cdot13}\right)=\frac{9}{13}\)
\(X-\left[\frac{31}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)+\frac{31}{5}\right]=\frac{9}{13}\)
\(X-\left[\frac{31}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)+\frac{31}{5}\right]=\frac{9}{13}\)
\(X-\left[\frac{31}{2}\cdot\frac{10}{39}+\frac{31}{5}\right]=\frac{9}{13}\)
\(X-\frac{1984}{195}=\frac{9}{13}\)
\(\Rightarrow X=\frac{9}{13}+\frac{1984}{195}=\frac{163}{15}\)
29 tháng 6 2018
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne=\)
Nên x + 1 = 0 => x = -1
b) \(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)
\(\Leftrightarrow\frac{x+1}{14}+1+\frac{x+2}{13}+1=\frac{x+3}{12}+1+\frac{x+4}{11}+1\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)
Vì \(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\ne0\)
Nên x +15 = 0 => x = -15
29 tháng 6 2018
a,\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)-\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)=0\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}>\frac{1}{13};\frac{1}{11}>\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}>\frac{1}{13}+\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
b, Bạn cộng thêm 1 vào \(\frac{x+1}{14};\frac{x+1}{13};\frac{x+1}{12};\frac{x+1}{11}\)Mội bên phân số 1 đơn vị rồi áp dụng như bài 1
23 tháng 7 2016
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}\frac{x+2}{13^{13}}\)
=> x + 2 = 0
=> x = 0 - 2
=> x = -2
19 tháng 6 2018
a) Ta có: \(\frac{143}{363}=\frac{143:11}{363:11}=\frac{13}{33}\) (đpcm)
b) Ta có: \(\frac{1212}{1717}=\frac{1212:101}{1717:101}=\frac{12}{17}\) (đpcm)
LN
19 tháng 6 2018
a)13/33 ; 143/363=143:11/363:11=13/33
=>13/33=143/363
b)12/27 ; 1212/2727=1212:101/2727:101=12/27
=>12/27=1212/2727
chúc bạn học tốt nha
26 tháng 3 2020
\(B=81\cdot\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}\right)\cdot\frac{158158158}{711711711}\)
\(B=81\cdot\left(\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)+1}\right)\cdot\frac{2}{9}\)
\(B=81\cdot\left(\frac{12}{4}:\frac{6470}{7653}\right)\cdot\frac{2}{9}\)
Xem lại đề bài bẹn owii -.-
Theo đề bài ta có:
\(\frac{561}{143}< x\frac{12}{13}< \frac{1463}{247}\)
\(\Leftrightarrow\frac{51}{13}< \frac{13x+12}{13}< \frac{77}{13}\)
\(\Leftrightarrow51< 13x+12< 77\)
\(\Leftrightarrow39< 13x< 65\)
\(\Leftrightarrow x=4\)
Vậy x = 4
Ta có :
\(\frac{561}{143}< x\frac{12}{13}< \frac{1463}{247}\)
\(\Leftrightarrow\)\(\frac{51}{13}< \frac{13x+12}{13}< \frac{77}{13}\)
\(\Leftrightarrow\)\(51< 13x+12< 77\)
\(\Leftrightarrow\)\(39< 13x< 65\)
Mà x là phần nguyên nên \(13x\inℤ\)
\(\Rightarrow\)\(13x=52\)
\(\Rightarrow\)\(x=\frac{52}{13}\)
\(\Rightarrow\)\(x=4\)
Vậy \(x=4\) hay hỗn số cần tìm là \(4\frac{12}{13}\)
Chúc bạn học tốt ~