Bài làm

4x⁴ + 81 = ( 2x² + ax + b )( 2x² - ax + b )

<=> 4x⁴ + 81 = 4x⁴ - 2ax³ + 2bx² - 2ax³ - a²x² + abx + 2bx² - abx + b²

<=> 4x⁴ + 81 = 4x⁴ + ( 4b - a² )x² + b²

Đồng nhất hệ số ta có :\(\hept{\begin{cases}4b-a^2=0\\b^2=81\end{cases}}\)

b² = 81 => b = ±9

Với b = 9 => 36 - a² = 0 => a = 36 => a = ±6

Với b = -9 => -36 - a² = 0 => a² = -36 ( vô lí )

=> Với b = 9 ; a = 6 => a + b = 9 + 6 = 15

=> Với b = 9 ; a = -6 => a + b = -6 + 9 = 3

\(D=-x^2+2x+4=-\left(x^2-2x+1\right)+5=-\left(x-1\right)^2+5\)

Ta có : \(-\left(x-1\right)^2\le0\forall x;-\left(x-1\right)^2+5\le5\)

Vậy GTLN D = 5 <=> x = 1 

a) \(C=\left(\frac{a+2}{3a}+\frac{2}{a+1}-3\right).\frac{a+1}{2-4a}+\frac{a^2-3a+1}{3a}\)

\(C=\left(\frac{\left(a+2\right)\left(a+1\right)}{3a\left(a+1\right)}+\frac{2.3a}{3a\left(a+1\right)}-\frac{3.3a.\left(a+1\right)}{3a\left(a+1\right)}\right).\frac{a+1}{2-4a}+\frac{a^2-3a+1}{3a}\)

\(C=\frac{a^2+3a+2+6a-9a^2-9a}{3a\left(a+1\right)}.\frac{a+1}{2-4a}+\frac{a^2-3a+1}{3a}\)

\(C=\frac{-8a^2+2}{3a\left(a+1\right)}.\frac{a+1}{2\left(1-2a\right)}+\frac{a^2-3a+1}{3a}\)

\(C=\frac{2\left(1-2a\right)\left(1+2a\right)}{3a\left(a+1\right)}.\frac{a+1}{2\left(1-2a\right)}+\frac{a^2-3a+1}{3a}\)

\(C=\frac{1+2a}{3a}+\frac{a^2-3a+1}{3a}\)

\(C=\frac{a^2-a+2}{3a}\)

b) \(C=\frac{a^2-a+2}{3a}\)

Để \(C\inℤ\)thì \(a^2-a+2=a\left(a-1\right)+2⋮3\)

- \(a=3k\left(k\inℤ\right)\): \(a\left(a-1\right)+2=3k\left(3k-1\right)+2⋮̸3\)

- \(a=3k+1\left(k\inℤ\right)\): \(a\left(a-1\right)+2=\left(3k+1\right).3k+2⋮̸3\)

- \(a=3k+2\left(k\inℤ\right)\): \(a\left(a-1\right)+2=\left(3k+2\right)\left(3k+1\right)+2=9k^2+9k+4⋮̸3\)

Vậy không có giá trị nào của \(a\inℤ\)để \(C\inℤ\).

Bạn tham khảo ạ

Nguồn: H

Nguồn: H.o.c.24