\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
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24 tháng 12 2016
Pro kinh không thể lướt qua
Giao luu: x=3
có thể quy đồng làm bt; \(!vt!\ge!vp!\)
DP
4 tháng 6 2019
a, \(2x-\frac{1}{2}=\frac{2x+1}{4}-\frac{1-2x}{8}\)
\(\Leftrightarrow\frac{1}{2}\left(4x-1\right)=\frac{1}{8}\left(6x+1\right)\)
\(\Leftrightarrow4\left(4x-1\right)=6x+1\)
\(\Leftrightarrow10x=5\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
b, \(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\(\Leftrightarrow\frac{x-3}{13}+\frac{x-3}{14}-\frac{x-3}{15}-\frac{x-3}{16}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy x = 3
KN
4 tháng 6 2019
\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\(\Leftrightarrow\frac{x-3}{13}+\frac{x-3}{14}-\frac{x-3}{15}-\frac{x-3}{16}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=0+3\)
\(\Leftrightarrow x=3\)
4 tháng 2 2017
Phương trình 1:
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\(\Rightarrow\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}-10=0\)
\(\Rightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
\(\Rightarrow\frac{x-85-15}{15}+\frac{x-74-26}{13}+\frac{x-67-33}{11}+\frac{x-64-36}{9}=0\)
\(\Rightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Do \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.
4 tháng 2 2017
Phương trình 3:
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
\(\Rightarrow\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)=0\)
\(\Rightarrow\frac{1909-x+91}{91}+\frac{1907-x+93}{93}+\frac{1905-x+95}{95}+\frac{1903-x+97}{97}=0\)
\(\Rightarrow\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
\(\Rightarrow\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Do \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow2000-x=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000.
XO
20 tháng 2 2020
b) \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
=> \(\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
=> \(\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
=> x - 100 = 0
=> x = 100
19 tháng 9 2016
\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\) .Trừ 1 ở mỗi hạng tử của 2 vế ,ta có :
\(\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)\left(x-11\right)=\left(\frac{1}{14}+\frac{1}{15}\right)\left(x-11\right)\)
\(\Rightarrow\left[\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\right]\left(x-11\right)=0\)
\(\frac{1}{12}>\frac{1}{14};\frac{1}{13}>\frac{1}{15}\Rightarrow\frac{1}{12}+\frac{1}{13}>\frac{1}{14}+\frac{1}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
MA
19 tháng 9 2016
\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\)
\(\Leftrightarrow\frac{x+1}{12}-1+\frac{x+2}{13}-1=\frac{x+3}{14}-1+\frac{x+4}{15}-1\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}-\frac{x-11}{14}-\frac{x-11}{15}=0\)
\(\Leftrightarrow\left(x-11\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
Mà: \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\(\frac{x-3}{13}+\frac{x-3}{14}-\frac{x-3}{15}-\frac{x-3}{16}=0\)
\(\left(x-3\right)\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\right)=0\)
\(x+3=0\) vì \(\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\right)\) ≠ 0
\(x=-3\)
Vậy x = -3
\(\frac{x - 3}{13}+\frac{x - 3}{14}=\frac{x - 3}{15}+\frac{x - 3}{16}\)
\(\frac{x - 3}{13} + \frac{x - 3}{14} - \frac{x - 3}{15} - \frac{x - 3}{16} = 0\)
\(\left(\right.x-3\left.\right)\left(\right.\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\left.\right)=0\)
\(x + 3 = 0\) vì \(\left(\right.\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\left.\right)\) ≠ 0
\(x = - 3\)
Vậy x = -3