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a) \(A=1999\cdot2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1\)
=> \(A< B\)
b) \(A=12^6\)
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)
=> \(A>B\)
c) \(A=2011\cdot2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1\)
\(B=2012^2\)
=> \(A< B\)
d) \(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)}{2}\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)..\left(3^{64}+1\right)}{2}\)
\(=\frac{\left(3^8-1\right).....\left(3^{64}+1\right)}{2}\)
\(=\frac{3^{128}-1}{2}\)
\(B=3^{128}-1\)
=> \(A< B\)
Bài 8:
b. 1+8x6y3 = 13+23(x2)3y3 = 13+(2x2y)3
= (1+2x2y)(1-2x2y+4x4y2)
e. 27x3+\(\dfrac{y^3}{8}\)\(=\left(3x\right)^3+\left(\dfrac{y}{2}\right)^3\)
= (3x+\(\dfrac{y}{2}\))(9x2-\(\dfrac{3xy}{2}\)+\(\dfrac{y^2}{4}\))
Bài 9:
c. 1- 9x +27x2 -27x3 = 13-3.12.3x+3.(3x)2-(3x)3
= (1-3x)3
d. x3+\(\dfrac{3}{2}x^2\)+\(\dfrac{3}{4}x+\dfrac{1}{8}\) = x3+\(3x^2.\dfrac{1}{2}\)+\(3x.\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3\)
= (x+\(\dfrac{1}{2}\))3
f. x2 - 2xy +y2 -4m2 +4m.n - n2 = (x2 - 2xy +y2)-((2m)2 -2.2m.n + n2)
= (x-y)2-(2m-n)2 = (x-y-2m+n)(x-y+2m-n)
a. \(1-2y+y^2=\left(1-y\right)^2\)
b. \(\left(x+1\right)^2-25=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)
c. \(1-4x^2=\left(1+2x\right)\left(1-2x\right)\)
d. \(8-27x^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
e. \(27+27x+9x^2+x^3=\left(x+3\right)^3\)
f, \(8x^3-12x^2y+6xy^2-y^3=\left(2x-y\right)^3\)
g, \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(\left(a\right)1-2y+y^2\)
\(\Leftrightarrow y^2-2y+1\)
\(\Leftrightarrow\left(y-1\right)^2\)
\(\left(b\right)\left(x+1\right)^2-25\)
\(\Leftrightarrow\left(x+1\right)^2-5^2\)
\(\Leftrightarrow\left(x-4\right)\left(x+6\right)\)
\(\left(c\right)1-4x^2\)
\(\Leftrightarrow1-\left(2x\right)^2\)
\(\Leftrightarrow\left(1-2x\right)\left(1+2x\right)\)
\(\left(d\right)8-27x^3\)
\(\Leftrightarrow2^3-\left(3x\right)^3\)
\(\Leftrightarrow\left(2-3x\right)\left(4+6x+9x^2\right)\)
\(\left(e\right)27+27x+9x^2+x^3\)
\(\Leftrightarrow\left(x+3\right)^3\)
\(\left(f\right)8x^3-12x^2y+6xy^2-y^3\)
\(\Leftrightarrow\left(2x\right)^3-12x^2y+6xy^2-y^3\)
\(\Leftrightarrow\left(2x-y\right)^3\)
\(\left(g\right)x^3+8y^3\)
\(\Leftrightarrow\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
c) n3 - 2 = (n3 - 8) + 6 = (n -2)(n2 + 2n + 4) + 6
Để n3 - 2 chia hết cho n - 2 <=> 6 chia hết cho n - 2 <=> n - 2 \(\in\) Ư(6) = {-6;-3;-2;-1;1;2;3;6}
Tương ứng n \(\in\) {-4; -1; 0; 1; 3; 4; 5; 8}
Vậy.....
d) n3 - 3n2 - 3n - 1 = (n3 - 1) - (3n2 + 3n + 3) + 3 = (n -1).(n2 + n + 1) - 3.(n2 + n + 1) + 3 = (n - 4)(n2 + n + 1) + 3
Để n3 - 3n2 - 3n - 1 chia hết cho n2 + n + 1 thì (n - 4)(n2 + n + 1) + 3 chia hết cho n2 + n + 1
<=> 3 chia hết cho n2 + n + 1 <=> n2 + n + 1 \(\in\) Ư(3) = {-3;-1;1;3}
Mà n2 + n + 1 = (n + \(\frac{1}{2}\))2 + \(\frac{3}{4}\) > 0 với mọi n nên n2 + n + 1 = 1 hoặc = 3
n2 + n + 1 = 1 <=> n = 0 hoặc n = -1
n2 + n + 1 = 3 <=> n2 + n - 2 = 0 <=> (n -1)(n +2) = 0 <=> n = 1 hoặc n = -2
Vậy ...
e) n4 - 2n3 + 2n2 - 2n + 1 = (n4 - 2n3 + n2) + (n2 - 2n + 1) = (n2 - n)2 + (n -1)2 = n2(n -1)2 + (n -1)2 = (n-1)2.(n2 + 1)
n4 - 1 = (n2 - 1).(n2 + 1) = (n -1)(n +1)(n2 + 1)
=> \(\frac{n^4-2n^3+2n^2-2n+1}{n^4-1}=\frac{\left(n-1\right)^2\left(n^2+1\right)}{\left(n-1\right)\left(n+1\right)\left(n^2+1\right)}=\frac{n-1}{n+1}\)( Điều kiện: n- 1 ; n + 1 khác 0 => n khác 1;-1)
Để n4 - 2n3 + 2n2 - 2n + 1 chia hết cho n4 - 1 thì \(\frac{n-1}{n+1}\) nguyên <=> n - 1 chia hết cho n + 1
<=> (n + 1) - 2 chia hết cho n +1
<=> 2 chia hết cho n + 1 <=> n + 1 \(\in\) Ư(2) = {-2;-1;1;2} <=> n \(\in\){-3; -2; 0; 1}
n = 1 Loại
Vậy n = -3 hoặc -2; 0 thì...
a) n2 + 2n - 4 = n2 + 2n - 15 + 11 = (n2 + 5n - 3n -15) + 11 = (n - 3)(n + 5) + 11
để n2 + 2n - 4 chia hết cho 11 <=> (n - 3).(n +5) chia hết cho 11 <=> n - 3 chia hết cho 11 hoặc n + 5 chia hết cho 11 ( Vì 11 là số nguyên tố)
n- 3 chia hết cho 11 <=> n = 11k + 3 ( k nguyên)
n + 5 chia hết cho 11 <=> n = 11k' - 5 ( k' nguyên)
Vậy với n = 11k + 3 hoặc n = 11k' - 5 thì.....
b) 2n3 + n2 + 7n + 1 = n2. (2n - 1) + 2n2 + 7n + 1 = n2. (2n -1) + n.(2n -1) + 8n + 1
= (n2 + n)(2n -1) + 4.(2n -1) + 5 = (n2 + n + 4)(2n -1) + 5
Để 2n3 + n2 + 7n + 1 chia hết cho 2n - 1 <=> (n2 + n + 4)(2n -1) + 5 chia hết cho 2n -1
<=> 5 chia hết cho 2n -1 <=> 2n - 1 \(\in\)Ư(5) = {-5;-1;1;5}
2n -1 = -5 => n = -2
2n -1 = -1 => n = 0
2n -1 = 1 => n = 1
2n -1 = 5 => n = 3
Vậy....
a) Ta có: \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1< 2000^2\)
Vậy A < 20002
c) \(E=26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50\)
\(F=27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52\)
Vì 50 < 52 => 2.50 < 2.52
=> E < F