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\(\Leftrightarrow\frac{x-241}{17}-1+\frac{x-220}{19}-2+\frac{x-195}{21}-3+\frac{x-166}{23}-4=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right).\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Leftrightarrow x-258=0\)
\(\Leftrightarrow x=258\)
\(\Leftrightarrow\frac{x-241}{17}-1+\frac{x-220}{19}-2+\frac{x-195}{21}-3+\frac{x-166}{23}-4=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\text{Mà }\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\ne0\text{ nên }x-258=0\Leftrightarrow x=258\)
d, 2x2-5x-3 = 0
\(\Leftrightarrow\)2x2-6x+x-3= 0
\(\Leftrightarrow\)(2x2-6x) +(x-3) = 0
\(\Leftrightarrow\)2x(x-3) + (x-3) = 0
\(\Leftrightarrow\)(x-3) (2x+1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S =\(\left\{3;\frac{-1}{2}\right\}\)
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=-10\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=-10\)
\(.....................\)
đến đây thì dễ rồi :)
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=10\)
\(\Leftrightarrow\) \(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}-10=0\)
\(\Leftrightarrow\) \(\left(\frac{x-241}{17}-1\right)+\left(\frac{x-220}{19}-2\right)+\left(\frac{x-195}{21}-3\right)+\left(\frac{x-166}{23}-4\right)=0\)
\(\Leftrightarrow\) \(\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-166}{23}=0\)
\(\Leftrightarrow\) \(\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{21}\right)=0\)
\(\Leftrightarrow\) \(x-258=0\) \(\Leftrightarrow\) \(x=258\)
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=10\)
\(\Leftrightarrow\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=10-1-2-3-4\)
\(\Leftrightarrow\left(\frac{x-241}{17}-1\right)+\left(\frac{x-220}{19}-2\right)+\left(\frac{x-195}{21}-3\right)+\left(\frac{x-166}{23}-4\right)=10-1-2-3-4\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{20}+\frac{x-258}{21}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\right)=0\)
\(\Leftrightarrow x-258=0\).Do \(\frac{1}{17}+\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\ne0\)
\(\Leftrightarrow x=258\)
\(\Leftrightarrow\left(\frac{x+14}{86}+1\right)+\left(\frac{x+15}{85}+1\right)+\left(\frac{x+16}{84}+1\right)+\left(\frac{x+17}{83}+1\right)+\left(\frac{166}{4}-4\right)=0\)
\(\Leftrightarrow\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(x+100\right)=0\Rightarrow x=-100\left(\text{vì }\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)\ne0\)
a)
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\\ \Leftrightarrow\frac{201-x}{99}+\frac{99}{99}+\frac{203-x}{97}+\frac{97}{97}+\frac{205-x}{95}+\frac{95}{95}+4=4\\ \Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\) (*)
Do \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)\ne0\)
nên (*) \(\Leftrightarrow300-x=0\\ \Leftrightarrow x=300\)
b)
\(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\\ \Leftrightarrow\frac{2-x}{2002}+\frac{2002}{2002}-1+1=\frac{1-x}{2003}+\frac{2003}{2003}-\frac{x}{2004}+\frac{2004}{2004}\\ \Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\\ \Leftrightarrow\frac{2004-x}{2002}-\frac{2004-x}{2003}+\frac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\) (*)
Do \(\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)\ne0\)
nên (*) \(\Leftrightarrow2004-x=0\)
\(\Leftrightarrow x=2004\)
c) \(\left|2x-3\right|=2x-3\) (1)
ĐKXĐ: \(\\ 2x-3\ge0\)
\(\Leftrightarrow x\ge\frac{3}{2}\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}2x-3=2x-3\\2x-3=-2x+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}0x=0\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\forall x\in R\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\frac{3}{2}\right\}\)
1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
\(\frac{x-5}{1990}-1+\frac{x-15}{1980}-1=\frac{x-1980}{15}-1+\frac{x-1990}{5}-1\)
\(\frac{x-1995}{1990}+\frac{x-1995}{1980}-\frac{x-1995}{15}-\frac{x-1995}{5}=0\)
\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\right)=0\)
Mà \(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\ne0\)
Nên \(x-1995=0\Leftrightarrow x=1995\)
\(\Leftrightarrow\frac{58,93x+78,73\left(166-x\right)}{524,477}=\frac{63,19.166}{524,477}\)
\(\Leftrightarrow58,93x+13069,18-78,73x=10489,54\)
\(\Leftrightarrow-19,8x=-2579,64\)
\(\Leftrightarrow x=\frac{2579,64}{19,8}=\frac{21497}{165}\)