15:

a: (a-b)^2

=a^2-2ab+b^2

=a^2+2ab+b^2-4ab

=(a+b)^2-4ab

b: (x+y)^2+(x-y)^2

=x^2+2xy+y^2+x^2-2xy+y^2

=2x^2+2y^2

=2(x^2+y^2)

14:

P=9x^2-12x+4

=(3x)^2-2*3x*2+2^2

=(3x-2)^2

a: Khi x=14 thì P=(3*14-2)^2=40^2=1600

b: Khi x=2/3 thì P=(3*2/3-2)^2=0

c: Khi x=-8/3 thì P=(-8/3*3-2)^2=(-10)^2=100

12:

a: 101^2=(100+1)^2

=100^2+2*100+1

=10000+200+1

=10201

b: 75^2-50*75+25^2

=(75-25)^2=50^2=2500

c: 103*97

=(100+3)(100-3)

=100^2-9

=9991

Bài 1: b) B = (a - b - c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc Bài 2: a) (x - 1/3)^2 = x^2 - 2/3x + 1/9 b) (x + y^2/3)^3 = x^3 + 3x^2y^2/3 + 3xy^4/9 + y^6/27 Bài 3: a) (2x + 1)(4x^2 - 2x + 1) = 8x^3 + 4x^2 - 2x + 4x^2 - 2x + 1 = 8x^3 + 8x^2 - 4x + 1 b) (1 - x^2)(1 + x^2 + x^4) = 1 - x^4 + x^2 + x^2 - x^4 + x^6 = x^6 - 2x^4 + 2x^2 + 1 c) (y - x/y)(y^2 + x + x^2/y^2) = y^3 + xy - x^2 + y^2 + xy + x^2 + x/y^2 - x - x^2/y^2 = y^3 + 2xy + y^2 - x Bài 4: a) M = (x + 3)(x^2 - 3x + 9) = x^3 - 3x^2 + 9x + 3x^2 - 9x + 27 = x^3 + 27 b) N = (1 - 3x)(1 + 3x + 9^2) = 1 - 9x + 3x - 27x^2 + 9 + 27x + 81 = -27x^2 + 27 c) P = (x - 1/2)(x^2 + x/2 + 1/4) = x^3 + x^2/2 + x^2/4 - x/2 - x/4 + 1/8 = x^3 + 3x^2/4 - 3x/4 + 1/8 d) Q = (2x + 3y)(4x^2 - 6xy + 9y^2) = 8x^3 - 12x^2y + 18xy^2 + 12x^2y - 18xy^2 + 27y^3 = 8x^3 + 27y^3 Bài 5: a) x^2 + 6x + 9 = (x + 3)^2 b) 9x^2 - 6x + 1 = (3x - 1)^2 c) x^2y^2 + xy + 1/4 = (xy + 1/2)^2 d) (x - y)^2 + 6(x - y) + 9 = (x - y + 3)^2

Bài 6: a) x^2 + 6x + 9 = (x + 3)^2 b) 4x^2 - 4x + 1 = (2x - 1)^2 c) 9x^2 - 12xy + 4y^2 = (3x - 2y)^2 d) (x - y)(x + y/3) = x^2 - y^2/9 Bài 7: a) -x^3 + 3x^2 - 3x + 1 = -(x - 1)^3 b) x^3 + x^2 + 1/3x + 1/27 = (x + 1/3)^3 c) x^6 - 3^4y + 3^2y^2 - y^3 = (x^2 - 3y)^3 d) (x - y)^3 + (x - y)^2 + 1/3(x - y) + 1/27 = (x - y + 1/3)^3 Bài 8: a) x^3 + 27 = (x + 3)(x^2 - 3x + 9) b) x^3 - 1/8 = (x - 1/2)(x^2 + 1/2x + 1/4) c) 8x^3 + y^3 = (2x)^3 + y^3 = (2x + y)(4x^2 - 2xy + y^2) d) 8^3 - 27y^3 = 512 - 27y^3 = (8 - 3y)(64 + 24y + 9y^2) Bài 9: a) A = -(-28)^3 + 6(-28)^2 - 12(-28) + 8 = -21952 b) B = 8(1/2)^3 + 12(1/2)^2 + 6(1/2) + 1 = 2 c) C = (20 + 2(1))^3 - 6(20 + 2(1))^2 + 12(20 + 2(1)) - 8 = 0 Bài 10: a) 11^3 - 1 = 1330 b) x^3 - y^3 = (x - y)(x^2 + xy + y^2) = 6(6^2 + 9) = 450Bài 11: a) M = (x + 3)(x^2 - 3x + 9) - (3 - 2x)(4x^2 + 6x + 9) = x^3 - 3x^2 + 9x + 3x^2 - 9x + 27 - (12x^2 + 18x + 27 - 8x^2 - 12x - 18) = x^3 - 12x^2 + 9x + 27 - 4x^2 - 12x + 9 = x^3 - 16x^2 - 3x + 36 b) N = (x - 2y)(x^2 + 2xy + 4y^2) + 16y^3 = (x - 2y)(x^2 + 2xy + 4y^2) + 16y^3 = x^3 - 2x^2y + 2xy^2 - 4y^3 + 16y^3 = x^3 - 2x^2y + 2xy^2 + 12y^3 Bài 12: a) 101^2 = 10201 b) 75^2 - 50.75 + 25^2 = 5625 - 3750 + 625 = 2500 c) 103.97 = 10091 Bài 13: a) 101^3 = 1030301 b) 98^3 + 6.98^2 + 12.98 + 8 = 941192 c) 99^3 = 970299 Bài 14: a) P = 9(34)^2 - 12(34) + 4 = 8296 b) P = 9(2/3)^2 - 12(2/3) + 4 = 0 c) P = 9(-8/3)^2 - 12(-8/3) + 4 = 128 Bài 15: a) (a - b)^2 = (a + b)^2 - 4ab = a^2 - 2ab + b^2 = a^2 + 2ab + b^2 - 4ab = (a + b)^2 - 4ab b) (x + y)^2 + (x - y)^2 = 2(x^2 + y^2) = x^2 + 2xy + y^2 + x^2 - 2xy + y^2 = 2x^2 + 2y^2 

a) đáp án A=1

b) B=0

c) C=1

Biến đổi vế trái ta có 

(a+b+c)^2 = (a+b + c)( a+b+c) = a(a+b + c) + b(a+b+c ) + c (a+b+c )

                                              = a^2 + ab +ac + ab + b^2 + bc + ac + bc + c^2 

                                               = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac => ĐPCM

Ta có:

(a + b + c)2 = (a + b + c)(a + b + c)

= a2 + ab + ac + ab + b2 + bc + ac + bc + c2

= a2 + b2 + c2 + 2ab + 2bc + 2ac (đpcm)

Vậy (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac.

Lời giải:

Xét tử :

\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}=\frac{a^2}{a^2+bc+(-ab-ac)}+\frac{b^2}{b^2+ac+(-ab-bc)}+\frac{c^2}{c^2+ab+(-bc-ac)}\)

\(=\frac{a^2}{a(a-b)-c(a-b)}+\frac{b^2}{b(b-c)-a(b-c)}+\frac{c^2}{c(c-a)-b(c-a)}\)

\(=\frac{a^2}{(a-c)(a-b)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)}\)

\(=\frac{a^2(c-b)+b^2(a-c)+c^2(b-a)}{(a-b)(b-c)(c-a)}\)

\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)

Xét mẫu (tương tự bên tử)

\(\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}=\frac{bc}{(a-c)(a-b)}+\frac{ac}{(b-a)(b-c)}+\frac{ab}{(c-a)(c-b)}\)

\(=\frac{bc(c-b)+ac(a-c)+ab(b-a)}{(a-b)(b-c)(c-a)}=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(a-b)(b-c)(c-a)}\)

\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)

Do đó:

\(A=\frac{1}{1}=1\)