Question

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\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\)

Answer 1

ĐKXĐ: x\(\ne\) -2; x\(\ne\) -4; x\(\ne\) -6; x\(\ne\) -8;

\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\) \(\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)

\(\Leftrightarrow\left(x+2+\dfrac{2}{x+2}\right)+\left(x+8+\dfrac{8}{x+8}\right)=\)

\(\left(x+4+\dfrac{4}{x+4}\right)+\left(x+6+\dfrac{6}{x+6}\right)\)

\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)

=> 2.(x+4)(x+8)(x+6) + 8(x+2)(x+4)(x+6)=4(x+2)(x+6)(x+8)

+ 6(x+2)(x+4)(x+8)

<=>(2x+8)(x² + 14x+64) + (8x+48)(x²+6x+8) - (4x+8)(x² + 14x+64)

-(6x+48)(x²+6x+8)

<=> (x² + 14x+64)(2x+8 -4x -8) + (x²+6x+8)(8x+48+6x-48)=0

<=> -2x(x² + 14x+64)+ 2x(x²+6x+8) = 0

<=> -2x³ -28x² -128x+ 2x³ +12x² +16x = 0

<=> -16x² - 112x = 0

<=> -x(16x+112) = 0

<=>\(\left[{}\begin{matrix}x=0\\16x+112=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=7\left(tmđk\right)\end{matrix}\right.\)

vậy S={0;7}

Answer 2

sửa bài:

<=>﴾2x+8﴿﴾x² + 14x+48﴿ + ﴾8x+48﴿﴾x² +6x+8﴿ ‐ ﴾4x+8﴿﴾x² + 14x+48﴿

‐﴾6x+48﴿﴾x² +6x+8﴿
<=> ﴾x² + 14x+48﴿﴾2x+8 ‐4x ‐8﴿ + ﴾x² +6x+8﴿﴾8x+48+6x‐48﴿=0
<=> ‐2x﴾x² + 14x+48﴿+ 2x﴾x² +6x+8﴿ = 0
<=> ‐2x³ ‐28x² ‐96x+ 2x³ +12x² +16x = 0
<=> ‐16x² ‐ 80x = 0
<=> ‐x﴾16x+80﴿ = 0

<=>\(\left[{}\begin{matrix}x=0\\16x+80=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

vậy : S={0;-5}