Question

[8.8_11_Phạm Nguyễn Khang]

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Answer 1

[Shinichi Kudo]

7/ \(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)

\(\dfrac{148-x}{25}-1+\dfrac{169-x}{23}-2+\dfrac{186-x}{21}-3+\dfrac{199-x}{19}-4=0\)

\(\dfrac{123-x}{25}+\dfrac{123-x}{23}+\dfrac{123-x}{21}+\dfrac{123-x}{19}=0\)

\(\left(123-x\right)\left(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\right)=0\)

Vì \(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\ne0\)

=> 123-x=0

           x=123

8/\(\dfrac{1}{x-1}-\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-4}=0\)             ĐKXĐ: \(\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\\x\ne4\end{matrix}\right.\)

\(\dfrac{1}{x-1}+\dfrac{1}{x-4}-\dfrac{1}{x-2}-\dfrac{1}{x-3}=0\)

\(\dfrac{2x-5}{\left(x-1\right)\left(x-4\right)}-\dfrac{2x-5}{\left(x-2\right)\left(x-3\right)}=0\)

\(\left(2x-5\right)\left(\dfrac{1}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{\left(x-2\right)\left(x-3\right)}\right)=0\)

=> 2x-5=0                hoặc     \(\dfrac{1}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{\left(x-2\right)\left(x-3\right)}=0\)

      2x=5                                 \(\dfrac{x^2-5x+6-x^2+5x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}=0\)

        \(x=\dfrac{5}{2}\)                             \(\dfrac{2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}=0\)

                                                 => 2=0 (vô lí)

Answer 2

[8.8_11_Phạm Nguyễn Khang]

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