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b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)
Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)
(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)
a./
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)
(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
a, \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Leftrightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Leftrightarrow x=11\)
b,\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Leftrightarrow\frac{1}{7}x-\frac{2}{7}=0\)hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\)hoặc \(\frac{1}{3}x+\frac{4}{3}=0\)
+) \(\frac{1}{7}x-\frac{2}{7}=0\Leftrightarrow\frac{1}{7}x=\frac{2}{7}\Leftrightarrow x=2\)
+)\(-\frac{1}{5}x+\frac{3}{5}=0\Leftrightarrow-\frac{1}{5}x=-\frac{3}{5}\Leftrightarrow x=3\)
+)\(\frac{1}{3}x+\frac{4}{3}=0\Leftrightarrow\frac{1}{3}x=-\frac{4}{3}\Leftrightarrow x=-4\)
c, \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x=\frac{4}{9}\)
\(\Leftrightarrow x=\frac{8}{9}\)
a/ \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Rightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Rightarrow x=11\)
b/ \(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{1}{7}x=\frac{2}{7}\Rightarrow x=2\)
hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow-\frac{1}{5}x=-\frac{3}{5}\Rightarrow x=3\)
hoặc \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{1}{3}x=-\frac{4}{3}\Rightarrow x=-4\)
Vậy x = 2, x = 3, x = -4
c/ \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x=\frac{4}{9}\Rightarrow x=\frac{8}{9}\)
Vậy x = 8/9
1) \(\left(x-1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}\right)=0\)
mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}\ne0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
2) \(\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-5}{95}-1=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{95}=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
x - 100 = 1
x = 101
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
Cộng vế theo vế
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có \(\frac{7}{8}< 1\)
Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
a, (x2 - 5)(x2 - 24) < 0
=> x2 - 5 và x2 - 24 trái dấu
Mà x2 - 5 > x2 - 24 => \(\hept{\begin{cases}x^2-5>0\\x^2-24>0\end{cases}\Rightarrow5< x^2< 24}\)
Vì x \(\in\)Z nên x2 = 9;16
+) x2 = 9 => x = 3 hoặc x = -3
+) x2 = 16 => x = 4 hoặc x = -4
Vậy...
b,
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
=> x + 1 = 0 => x = 0 - 1 => x = -1
\(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)
\(\Rightarrow\left(\frac{x+1}{14}+1\right)+\left(\frac{x+2}{13}+1\right)=\left(\frac{x+3}{12}+1\right)+\left(\frac{x+4}{11}+1\right)\)
\(\Rightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)
\(\Rightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)
\(\Rightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)
Mà \(\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)\ne0\)
=> x + 15 = 0 => x = 0 - 15 => x = -15
\(-\frac{9}{11}\cdot\frac{3}{8}-\frac{9}{11}\cdot\frac{5}{8}+\frac{17}{11}=-\frac{9}{11}\left(\frac{3}{8}+\frac{5}{8}\right)+\frac{17}{11}=-\frac{9}{11}\cdot1+\frac{17}{11}=1\)
\(\frac{2}{1.3}+....+\frac{2}{53.55}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{53}-\frac{1}{55}=1-\frac{1}{55}=\frac{54}{55}\)
\(x+5-\frac{1}{2}=3\frac{1}{2}\)
\(x+5=3.5+0.5=4\)
\(x=4-5=-1\)
\(3^{x+1}=27=3^3\)
\(x+1=3\)
vậy x=2