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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 \(4\frac{12}{13}\)
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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/ \(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(=\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\frac{1}{13\cdot19}+\frac{1}{19\cdot25}+\frac{1}{25\cdot31}+\frac{1}{31\cdot37}\)
\(=\frac{7-1}{1\cdot7}+\frac{13-7}{7\cdot13}+....+\frac{37-31}{31\cdot37}\)
\(=\frac{1}{6}\cdot\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+....+\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac{1}{6}\cdot\left(1-\frac{1}{37}\right)\)
\(=\frac{1}{6}\cdot\frac{36}{37}\)
\(=\frac{6}{37}\)
Ta có : \(2\frac{x}{7}=\frac{2.7+x}{7}=\frac{14+x}{7}\)
Nên : \(\frac{14+x}{7}=\frac{2x+9}{7}\)
<=> 14 + x = 2x + 9
=> 14 - 9 = 2x - x
=> x = 5
Vậy x = 5
14+x7 =2x+97
<=> 14 + x = 2x + 9
=> 14 - 9 = 2x - x
=> x = 5
Vậy x = 5
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
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
a) Ta có:+) \(\frac{12}{16}=\frac{-x}{4}\) <=> 12.4 = 16.(-x)
<=> 48 = -16x
<=> x = 48 : (-16) = -3
+) \(\frac{12}{16}=\frac{21}{y}\) <=> 12y = 21.16
<=> 12y = 336
<=> y = 336 : 12 = 28
+) \(\frac{12}{16}=\frac{z}{-80}\) <=> 12. (-80) = 16z
<=> -960 = 16z
<=> z = -960 : 16 = -60
b) Ta có: \(\frac{x+3}{7+y}=\frac{3}{7}\) <=> (x + 3).7 = 3(7 + y)
<=> 7x + 21 = 21 + 3y
<=> 7x = 3y
<=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x+y}{3+7}=\frac{20}{10}=2\)
=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{7}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.7=14\end{cases}}\)
Vậy ...
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 ~
Trả lời :
a)x=4
b)x=2
\(\downarrow\)