tìm x
a/\(\dfrac{x}{2}=\dfrac{32}{x}\)
b/\(\dfrac{5}{25}=5\)
c/\(\dfrac{30^x}{27^x}=27\)
d/5x . 10x=2500
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21 tháng 2 2021
a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
nên \(x^2-10x-2000=0\)
\(\Leftrightarrow x^2+40x-50x-2000=0\)
\(\Leftrightarrow x\left(x+40\right)-50\left(x+40\right)=0\)
\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Vậy: S={-40;50}
26 tháng 4 2018
\(\text{a) }\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\\ \Leftrightarrow\dfrac{2-x-2002}{2002}=\left(\dfrac{1-x}{2003}-1\right)+\left(1-\dfrac{x}{2004}\right)\\ \Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2003-x}{2003}-\dfrac{2004-x}{2004}=0\\ \Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\\ \Leftrightarrow2004-x=0\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\right)\\ \Leftrightarrow x=2004\)
Vậy phương trình có nghiệm \(x=2004\)
26 tháng 4 2018
\(\text{b) }\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\left(\text{ Chữa đề }\right)\\ \Leftrightarrow\left(\dfrac{x^2-10x-29}{1971}-1\right)+\left(\dfrac{x^2-10x-27}{1973}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\\ \Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-2000}{29}-\dfrac{x^2-10x-2000}{27}=0\\ \Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\\ \Leftrightarrow x^2-10x-2000=0\left(\text{Vì }\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\right)\\ \Leftrightarrow x^2-20x+10x-2000=0\\ \Leftrightarrow x\left(x-20\right)+10\left(x-20\right)=0\\ \Leftrightarrow\left(x+10\right)\left(x-20\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+10=0\\x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=20\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{-10;20\right\}\)
NH
15 tháng 12 2022
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)
