($x^{2}$ – 5)(x+3)=0 $\frac{x-12}{77}$ + $\frac{x-11}{78}$ = $\frac{x-74}{15}$ + $\frac{x-73}{16}$

($x^{2}$ – 5)(x+3)=0
$\frac{x-12}{77}$ + $\frac{x-11}{78}$ = $\frac{x-74}{15}$ + $\frac{x-73}{16}$

0 bình luận về “($x^{2}$ – 5)(x+3)=0 $\frac{x-12}{77}$ + $\frac{x-11}{78}$ = $\frac{x-74}{15}$ + $\frac{x-73}{16}$”

  1. Giải thích các bước giải:

     `(x^2-5)(x+3)=0`

    `=>`\(\left[ \begin{array}{l}x^2-5=0\\x+3=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x^2=5\\x=-3\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=\pm\sqrt5\\x=-3\end{array} \right.\) 

     `(x-12)/77+(x-11)/78=(x-74)/15+(x-73)/16`

    `=>((x-12)/77-1)+((x-11)/78-1)=((x-74)/15-1)+((x-73)/16-1)`

    `=>(x-89)/77+(x-89)/78=(x-89)/15+(x-89)/16`

    `=>(x-89)/77+(x-89)/78-(x-89)/15-(x-89)/16=0`

    `=>(x-89)(1/77+1/78-1/15-1/16)=0`

    Mà `1/77+1/78-1/15-1/16 ne 0`

    `=>x-89=0=>x=89`

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  2. `(x^2-5)(x+3)=0`

    `<=>` \(\left[ \begin{array}{l}x^2-5=0\\x+3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\pm\sqrt{5}\\x=-3\end{array} \right.\) 

     Vậy `S={\pm \sqrt{5}, -3}`

    `\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}`

    `=> \frac{x-12}{77}-1+\frac{x-11}{78}-1-(\frac{x-74}{15}-1+\frac{x-73}{16}-1)=0`

    `<=> \frac{x-12-77}{77}+\frac{x-11-78}{78}-\frac{x-74-15}{15}-\frac{x-73-16}{16}=0`

    `<=> (x-89)(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16})=0`

    Do `\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16} \ne 0`

    `=> x-89=0`

    `<=> x=89`

    Vậy `S={89}`

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