d: \(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=8\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4=8\)
=>8x=8
hay x=1(nhận)
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`(x+2)/(x-2)-(x-2)/(x+2)=8/(x^2-4)(x\ne+-2)`
`<=>(x+2)^2/((x-2)(x+2))-(x-2)^2/((x-2)(x+2))-8/((x-2)(x+2))=0`
`<=>(x^2+4x+4-x^2+4x-4-8)/((x-2)(x+2))=0`
`<=>(8x-8)/((x-2)(x+2))=0`
`=>8x-8=0<=>x=1`
Vậy `S={1}`
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