Fuentes Savage
08/05/2023 · Junior High School
If \( x=4 \) and \( y=-2 \), determine the value of \( -2(x+y)^{3} \) Factorise completely: i. \( 16 x^{4}-1 \) ii. \( 3 a^{2}-24 a+45 \) Write down the restriction(s) and solve for \( x: \frac{5}{6}-\frac{3}{2 x}=\frac{3}{x}-\frac{2}{3} \)
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The value of \( -2(x+y)^{3} \) is -16. The factorization of \( 16x^4 - 1 \) is \( (4x^2 - 1)(4x^2 + 1) \), and the factorization of \( 3a^2 - 24a + 45 \) is \( (3a - 15)(a - 3) \). The solution to the equation \( \frac{5}{6} - \frac{3}{2x} = \frac{3}{x} - \frac{2}{3} \) is \( x = 3 \).
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