Question

Kuis ⁽¹⁵²⁾
Jika diketahui
[tex]\rm (3^x-27)^2+(625-5^x)^2=(3^x+5^x-625)^2[/tex]

tentukan nilai dari
[tex]\displaystyle\rm \frac{3^{6-x}}{5^{x}-598}[/tex]

Answer 1

Verified answer

aljabar

(3ˣ - 27)² + (625 - 5ˣ)² = (3ˣ + 5ˣ - 625)²

(3ˣ - 27)² + (-(5ˣ - 598) + 27)² = (3ˣ - 27 + 5ˣ - 598)²

a² + (27 - b)² = (a + b)²

27² - 54b = 2ab

3⁶ = 2b(27 + a)

3⁶ = 2b(27 + (3ˣ - 27))

3⁶ = 2b . 3ˣ

3⁶ / (3ˣ b) = 2

3⁶⁻ˣ / (5ˣ - 598) = 2

Answer 2

[tex]\large\text{$\begin{aligned}\frac{3^{6-x}}{5^x-598}&=\boxed{\,\bf2\,}\end{aligned}$}[/tex]
 

Penjelasan dengan langkah-langkah

Diberikan persamaan:

[tex]\begin{aligned}(3^x-27)^2+(625-5^x)^2=(3^x+5^x-625)^2\end{aligned}[/tex]

Kita akan menentukan nilai dari

[tex]\begin{aligned}\frac{3^{6-x}}{5^x-598}\end{aligned}[/tex]

Ambil [tex]A=5^x-598[/tex], maka persamaan yang diketahui di atas ekuivalen dengan:

[tex]\begin{aligned}\left(3^x-27\right)^2+(27-A)^2&=\left(3^x-27+A\right)^2\end{aligned}[/tex]

Sehingga:

[tex]\begin{aligned}\left(3^x-27\right)^2+(27-A)^2&=\left(3^x-27+A\right)^2\\\cancel{\left(3^x-27\right)^2}+(27-A)^2&=\cancel{\left(3^x-27\right)^2}+A^2+2A\left(3^x-27\right)\\27^2+\cancel{A^2}-\cancel{54A}&=\cancel{A^2}+2A\left(3^x\right)-\cancel{54A}\\\end{aligned}[/tex]

Persamaan terakhir menghasilkan:

[tex]\begin{aligned}27^2&=2A\left(3^x\right)\\3^6&=2\left(5^x-598\right)\left(3^x\right)\\3^{6-x}&=2\left(5^x-598\right)\\\therefore\ \frac{3^{6-x}}{5^x-598}&=\boxed{\,\bf2\,}\end{aligned}[/tex]
[tex]\blacksquare[/tex]