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Hoàng Thị Mai Trang

Giải phương trình sau:

a)\(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3=0\)

b)\(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)

c)\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)

Natsu Dragneel
17 tháng 2 2020 lúc 10:53

a)\(\frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{95}+1=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)

\(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\Rightarrow300-x=0\Rightarrow x=300\)

b)\(\frac{2-x}{2002}+1=\frac{1-x}{2003}+2-\frac{x}{2004}\)

\(\Leftrightarrow\frac{2004-x}{2002}=\frac{1-x}{2003}+1+1-\frac{x}{2004}\)

\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}+\frac{2004-x}{2004}\)

\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)

\(\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\ne0\Rightarrow2004-x=0\Rightarrow x=2004\)

c)\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}-2=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}-2\)

\(\Leftrightarrow\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}=\frac{x^2-10x-2000}{29}+\frac{x^2-10x-2000}{27}\)

\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\right)=0\)

\(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\ne0\)

\(\Rightarrow x^2-10x-2000=0\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)

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