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x is a negative number such that k + k-1 = -2, then what is the value of $\frac{k^2+4k-2}{k^2+k-5}$? |
7 1 -7 -1 |
1 |
k + k-1 = -2 Put k = -1 $\frac{k^2+4k-2}{k^2+k-5}$ = $\frac{-1^2+4×1-2}{-1^2+-1-5}$ = \(\frac{-5}{-5}\) = 1 |
