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\(\left(x^2-2\right)\left(1-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)
\(=\left(x^2-x^3-2+2x\right)+\left(x^3+3^3\right)\)
\(=x^2-x^3-2+2x+x^3+27\)
\(=\left(-x^3+x^3\right)+x^2+2x+\left(-2+27\right)\)
\(=x^2+2x+25\)
=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=...=2^32-1
1.a) (2 + 1)(22 + 1)((24 + 1)(28 + 1) = (22 - 1)(22 + 1)(24 + 1)(28 + 1) = (24 - 1)(24 + 1)(28 + 1)
= (28 - 1)(28 + 1) = 216 - 1
b) 7(23 + 1)(26 + 1)(212 + 1)(224 + 1) = (23 - 1)(23 + 1)(26 + 1)(212 + 1)(224 + 1)
= (26 - 1)(26 + 1)(212 + 1)(224 + 1) = (212 - 1)(212 + 1)(224 + 1) = (224 - 1)(224 + 1) = 248 - 1
c) (x2 - x + 1)(x2 + x + 1)(x2 - 1) = [(x2 - x + 1)(x + 1)][(x2 + x + 1)(x - 1)] = (x3 + 1)(x3 - 1) = x6 - 1
2. Đặt A = 4x - x2 - 1 = -(x^2 - 4x + 4) + 3 = -(x - 2)2 + 3 \(\le\)3 \(\forall\)x
Dấu "=" xảy ra <=> x - 2 = 0 <=> x = 2
Vậy MaxA = 3 khi x = 2
\(A=\left(4x+5\right)^2-\left(3x-7\right)^2-\left(4x-1\right)\left(4x+1\right)\)
\(=\left(4x\right)^2+2.4x.5+5^2-\left[\left(3x\right)^2-2.3x.7+7^2\right]-\left[\left(4x\right)^2-1^2\right]\)
\(=16x^2+40x+25-\left(9x^2-42x+49\right)-\left(16x^2-1\right)\)
\(=16x^2+40x+25-9x^2+42x-49-16x^2+1=-9x^2+82x-23\)
A=(4x+5)2-(3x-7)2-(4x-1)(4x+1)
=16x2+40x+25-9x2+42x-49-16x2+1
=(16x2-9x2-16x2)+(40x+42x)+(25-49+1)
=-9x2+82x-23