Jika x^(1/2) + x^(−1/2) = 2, maka nilai dari x^(2)+x^(−2) adalah

Jawaban yang benar adalah 2. Ingat! Sifat bilangan berpangkat. (aᵇ)ᶜ=aᵇᶜ. Diketahui: x^(1/2) + x^(−1/2) = 2 Perhatikan perhitungan berikut ini. [x^(1/2) + x^(−1/2)]² = 2² [x^(1/2)]²+2.x^(1/2). x^(−1/2)+[x^(−1/2)]² = 4 x^(1/2 . 2)+2.x^(1/2−1/2)+x^(−1/2 . 2) = 4 x^(1)+2.x^(0)+x^(−1) = 4 x+2.1+x^(−1) = 4 x+2+x^(−1) = 4 x+x^(−1) = 4-2 x+x^(−1) = 2 Kuadratkan kedua ruas. [x+x^(−1)]² = 2² x²+2.x.x^(−1)+[x^(−1)]² = 4 x²+2x^(1−1)+x^(−2) = 4 x²+2x^(0)+x^(−2) = 4 x²+2.1+x^(−2) = 4 x²+2+x^(−2) = 4 x²+x^(−2) = 4-2 x²+x^(−2) = 2 Jadi, nilai dari x^(2)+x^(−2)=2.