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Jika [tex]x = 5[/tex] dan [tex]y = 4[/tex], maka nilai dari [tex]x² + x²y² + xy² + y²[/tex] adalah ....
[tex]ingat!~~ab² ≠ (ab)²[/tex].
[tex]\begin{aligned} nilai &= x² + x²y² + xy² + y² \\&= 5² + 5²4² + 5(4²) + 4² \\&= 25 + 25(16) + 5(16) + 16 \\&= 25 + 400 + 80 + 16 \\&= 25 + 400 + 96 \\&= 25 + 496 \\&= \boxed{\bold{\underline{521}}} \end{aligned}[/tex]
Cara Lain
[tex]\begin{aligned} nilai &= x² + x²y² + xy² + y² \\&= x²×(1 + y²) + y²×(x + 1) \\&= 5² × (1 + 4²) + 4² × (5 + 1) \\&= 25(17) + 16 (6) \\&= 425 + 96 \\&= \boxed{\bold{\underline{521}}} \end{aligned}[/tex]
Opsi Jawaban : A
[tex]\begin{array}{lr}\texttt{Selamat Menunaikan Ibadah Puasa}\end{array}[/tex]
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 12 - 04 - 2023}}[/tex]
Jawaban:
A. 521
Penjelasan dengan langkah-langkah:
Dik:
x = 5
x² = 5² = 5
y = 4
y² = 4² = 16
Maka:
x² + x²y² + xy² + y²
= 25 + (25.16) + (5.16) + 16
= 25 + 400 + 80 + 16
= 425 + 80 + 16
= 505 + 16
= 521
Jawabannya A.
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Algebra
[Subtitusi Aljabar]
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Jika [tex]x = 5[/tex] dan [tex]y = 4[/tex], maka nilai dari [tex]x² + x²y² + xy² + y²[/tex] adalah ....
Penyelesaian Soal
[tex]ingat!~~ab² ≠ (ab)²[/tex].
[tex]\begin{aligned} nilai &= x² + x²y² + xy² + y² \\&= 5² + 5²4² + 5(4²) + 4² \\&= 25 + 25(16) + 5(16) + 16 \\&= 25 + 400 + 80 + 16 \\&= 25 + 400 + 96 \\&= 25 + 496 \\&= \boxed{\bold{\underline{521}}} \end{aligned}[/tex]
Cara Lain
[tex]\begin{aligned} nilai &= x² + x²y² + xy² + y² \\&= x²×(1 + y²) + y²×(x + 1) \\&= 5² × (1 + 4²) + 4² × (5 + 1) \\&= 25(17) + 16 (6) \\&= 425 + 96 \\&= \boxed{\bold{\underline{521}}} \end{aligned}[/tex]
Opsi Jawaban : A
[tex]\begin{array}{lr}\texttt{Selamat Menunaikan Ibadah Puasa}\end{array}[/tex]
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 12 - 04 - 2023}}[/tex]
Jawaban:
A. 521
Penjelasan dengan langkah-langkah:
Dik:
x = 5
x² = 5² = 5
y = 4
y² = 4² = 16
Maka:
x² + x²y² + xy² + y²
= 25 + (25.16) + (5.16) + 16
= 25 + 400 + 80 + 16
= 425 + 80 + 16
= 505 + 16
= 521
Jawabannya A.