Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100

4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4

4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100

4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101

4A=98.99.100.101

=>A=98.99.100.101/4

=> A=24497550

a,\(\frac{31}{1000}\)

b,\(\frac{8}{108}\)

c,0

a,\(\frac{49}{97}\)

b,\(\frac{-1}{4751}\)

Bài 9:

a) Ta có: \(A=\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)

\(=4x^2+4xy+y^2-4x^2+y^2-xy-y^2\)

\(=3xy-y^2\)

\(=3\cdot\left(-2\right)\cdot3-3^2=-18-9=-27\)

b) Ta có: \(B=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)

\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)

\(=-13ab+2a+b-2\)

\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)

\(=\dfrac{31}{2}\)

Bài 7: 

a) \(498^2=\left(500-2\right)^2=250000-2000+4=248004\)

b) \(93\cdot107=100^2-7^2=10000-49=9951\)

c) \(163^2+74\cdot163+37^2=\left(163+37\right)^2=200^2=40000\)

d) \(1995^2-1994\cdot1996=1995^2-1995^2+1=1\)

e) \(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(=18^8-18^8+1=1\)

f) \(125^2-2\cdot125\cdot25+25^2=\left(125-25\right)^2=100^2=10000\)

\(a,=\left(3x-5\right)\left(3x+3\right)=3\left(x+1\right)\left(3x-5\right)\\ b,=\left(5x-4-7x\right)\left(5x-4+7x\right)=\left(-2x-4\right)\left(12x-4\right)\\ =-8\left(x+2\right)\left(x-3\right)\\ c,=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\\ =\left(x+14\right)\left(3x-4\right)\\ d,=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\\ =\left(x+5\right)\left(5x-3\right)\\ e,=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\\ =\left(4x+7\right)\left(8x+11\right)\\ f,=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\\ =\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\\ =\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\\ g,=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\\ =\left(a-b\right)\left(x-y\right)\left(a+b\right)\left(x+y\right)\)

\(h,=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\\ =\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\\ =\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)

a: \(\left(3x-1\right)^2-16\)

\(=\left(3x-1-4\right)\left(3x-1+4\right)\)

\(=\left(3x+3\right)\left(3x-5\right)\)

\(=3\left(x+1\right)\left(3x-5\right)\)

b: \(\left(5x-4\right)^2-49x^2\)

\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)

\(=\left(-2x-4\right)\left(12x-4\right)\)

\(=-8\left(x+2\right)\left(3x-1\right)\)

a: =(6x)^2-(3x-2)^2

=(6x-3x+2)(6x+3x-2)

=(9x-2)(3x+2)

d: \(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)

\(=4x\cdot\left[x^2+2x+1+x^2-2x+1\right]\)

=8x(x^2+1)

e: =(4x)^2-2*4x*3y+(3y)^2

=(4x-3y)^2

f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)

\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)

g: =(4x)^3+1^3

=(4x+1)(16x^2-4x+1)

k: =x^3(27x^3-8)

=x^3(3x-2)(9x^2+6x+4)

l: =(x^3-y^3)(x^3+y^3)

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

$36x^2-(3x-2)^2$

$=[6x-(3x-2)][6x+(3x-2)]$

$=(6x-3x+2)(6x+3x-2)$

$=(3x+2)(9x-2)$

$16(4x+5)^5-25(2x+2)^2$

Câu này không có dạng hằng đẳng thức hiệu hai bình phương vì số mũ của hai phần là $5$ và $2$.

Có thể đặt nhân tử chung $1$ nhưng không phân tích tiếp được bằng các phương pháp thông thường.

$(x-y+4)^2$

Đây đã là một tích:

$(x-y+4)^2=(x-y+4)(x-y+4)$

$(x+1)^4-(x-1)^4$

$=[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]$

$=[x^2+2x+1-x^2+2x-1][x^2+2x+1+x^2-2x+1]$

$=4x(2x^2+2)$

$=8x(x^2+1)$

$16x^2-24xy+9y^2$

$=(4x)^2-2\cdot4x\cdot3y+(3y)^2$

$=(4x-3y)^2$

$-\dfrac{x^4}{4}+2x^2y^3-4y^6$

$=-\dfrac14(x^4-8x^2y^3+16y^6)$

$=-\dfrac14(x^2-4y^3)^2$

$64x^3+1$

$=(4x)^3+1^3$

$=(4x+1)(16x^2-4x+1)$

$x^3y^6z^9-125$

$=(xy^2z^3)^3-5^3$

$=(xy^2z^3-5)(x^2y^4z^6+5xy^2z^3+25)$

$27x^6-8x^3$

$=x^3(27x^3-8)$

$=x^3[(3x)^3-2^3]$

$=x^3(3x-2)(9x^2+6x+4)$

$x^6-y^6$

$=(x^3-y^3)(x^3+y^3)$

$=(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)$

$27x^3-54x^2y+36xy^2-8y^3$

$=(3x)^3-3(3x)^2(2y)+3(3x)(2y)^2-(2y)^3$

$=(3x-2y)^3$

$y^9-9x^2y^6+27x^4y^3-27x^6$

$=(y^3)^3-3(y^3)^2(3x^2)+3(y^3)(3x^2)^2-(3x^2)^3$

$=(y^3-3x^2)^3$

đăng ít thôi bạn đăng nhiều thế mới nhìn vào đã thấy chóng mặt huống chi là giải

2: \(=16\cdot38-16\cdot2-38\cdot16+38=-32+38=6\)

3: \(=6-7+4-3=10-10=0\)

4: \(=-125\left(26-36\right)=-125\cdot\left(-10\right)=1250\)

5: \(=5^3-5^3=0\)

6: \(=24\left(-125+225\right)=100\cdot24=2400\)

7: =21(135+65)-200

=200*21-200

=200*20=4000

8: =42+58-19+19=100

mh biết làm bài này rùi bn có cần mi2h đang cho bn ko?