Question

[v]

giúp mình với

Answer 1

[l҉o҉n҉g҉ d҉z҉]

\(\frac{1}{x^2+2x}+\frac{1}{x^2+6x+8}+\frac{1}{x^2+10x+24}+\frac{1}{x^2+14x+48}=\frac{4}{105}\)

ĐKXĐ : x ≠ 0 ; x ≠ -2 ; x ≠ -4 ; x ≠ -6 ; x ≠ -8

<=> \(\frac{1}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+8\right)}=\frac{4}{105}\)

<=> \(\frac{1}{2}\left[\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+6\right)\left(x+8\right)}\right]=\frac{4}{105}\)

<=> \(\frac{1}{2}\left(\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+8}\right)=\frac{4}{105}\)

<=> \(\frac{1}{2}\left(\frac{1}{x}-\frac{1}{x+8}\right)=\frac{4}{105}\)

<=> \(\frac{x+8}{x\left(x+8\right)}-\frac{x}{x\left(x+8\right)}=\frac{8}{105}\)

<=> \(\frac{8}{x\left(x+8\right)}=\frac{8}{105}\)

=> x( x + 8 ) = 105

<=> x² + 8x - 105 = 0

<=> x² - 7x + 15x - 105 = 0

<=> x( x - 7 ) + 15( x - 7 ) = 0

<=> ( x - 7 )( x + 15 ) = 0

<=> x = 7 hoặc x = -15 (tm)

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