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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}
Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\left(\dfrac{x^2-10x-27}{1973}-1\right)+\left(\dfrac{x^2-10x-29}{1971}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1973}+\dfrac{x^2-10x-2000}{1971}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1973}+\dfrac{1}{1971}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)=0\) vì \(\left(\dfrac{1}{1973}+\dfrac{1}{1971}-\dfrac{1}{29}-\dfrac{1}{27}\right)\ne0\)
\(\Leftrightarrow x^2-50x+40x-2000=0\)
\(\Leftrightarrow x\left(x-50\right)+40\left(x-50\right)=0\)
\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-50=0\\x+40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)
Vậy: Giá trị x thỏa mãn là: \(x=-40;50\)
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
vì \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
Nên \(x^2-10x-2000=0\)
<=> \(x^2-50x+40x-2000=0\)
<=> \(x\left(x-50\right)+40\left(x-50\right)=0\)
<=> \(\left(x-50\right)\left(x+40\right)=0\)
<=> \(x=50\) hoặc \(x=-40\)
Vậy tập nghiệm của phương trình là \(S=\left\{50;-40\right\}\)
\(\text{a) }\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\\ \Leftrightarrow\dfrac{2-x-2002}{2002}=\left(\dfrac{1-x}{2003}-1\right)+\left(1-\dfrac{x}{2004}\right)\\ \Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2003-x}{2003}-\dfrac{2004-x}{2004}=0\\ \Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\\ \Leftrightarrow2004-x=0\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\right)\\ \Leftrightarrow x=2004\)
Vậy phương trình có nghiệm \(x=2004\)
\(\text{b) }\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\left(\text{ Chữa đề }\right)\\ \Leftrightarrow\left(\dfrac{x^2-10x-29}{1971}-1\right)+\left(\dfrac{x^2-10x-27}{1973}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\\ \Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-2000}{29}-\dfrac{x^2-10x-2000}{27}=0\\ \Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\\ \Leftrightarrow x^2-10x-2000=0\left(\text{Vì }\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\right)\\ \Leftrightarrow x^2-20x+10x-2000=0\\ \Leftrightarrow x\left(x-20\right)+10\left(x-20\right)=0\\ \Leftrightarrow\left(x+10\right)\left(x-20\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+10=0\\x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=20\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{-10;20\right\}\)
Thêm (-1) vào từng số hạng=> tử số các số hạng là: \(\left(x^2-10x-2000\right)\)
\(\Leftrightarrow x^2-10x-2000=0\Leftrightarrow\left(x-5\right)^2=2025=45^2\)
\(\orbr{\begin{cases}x=50\\x=-40\end{cases}}\)
\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\left(\dfrac{x^2-10x-29}{1971}-1\right)+\left(\dfrac{x^2-10x-27}{1973}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\)
\(\Leftrightarrow x^2-10x-2000=0\)
\(\Leftrightarrow x^2-10x+25-2025=0\)
\(\Leftrightarrow\left(x-5\right)^2=2025\)
=>x-5=45 hoặc x-5=-45
=>x=50 hoặc x=-40
\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)=> \(\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)=>\(\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\) => \(\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)\)
=> \(x^2-10x-2000=0\)
Tự giải ra nhé hi hi
`(x^2-10x-29)/1971+(x^2-10x-27)/1973=(x^2-10x-1971)/1929+(x^2-10x-1973)/1927`
`<=>(x^2-10x-29)/1971-1+(x^2-10x-27)/1973-1=(x^2-10x-1971)/1929-1+(x^2-10x-1973)/1927-1`
`<=>(x^2-10x-200)/1971+(x^2-10x-200)/1973=(x^2-10x-200)/1971+(x^2-10x-200)/1927`
`<=>(x^2-10x-200)(1/1971+1/1973-1/1929-1/1927)=0`
`<=>x^2-10x-200=0` do `1/1971+1/1973-1/1929-1/1927<0`
`<=>x^2-20x+10x-200=0`
`<=>x(x-20)+10(x-20)=0`
`<=>(x-20)(x+10)=0`
`<=>` \(\left[ \begin{array}{l}x=20\\x=-10\end{array} \right.\)
Vậy `S={20,-10}`