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18 tháng 8 2016

a) ( x2 - 2x + 2 )( x- 2 )( x2 + 2x + 2 )( x2 + 2 )

= [ ( x+ 2 )2 - 4x2 ] ( x4 - 4 )

= ( x4 + 4 ) ( x4 - 4 )

= x8 - 16

b) ( a + b + c )2 + ( a + b - c )2 + ( 2a -b )2

= 2 ( a2 + b2 + c2 ) + 2 ( ab + bc + ac ) + 2 ( ab - bc - ac ) + ( 4a2 - 4ab + b2 )

= 2 ( a+ b2 + c2 ) + 4ab - 4ab + 4a2 + b2

= 6a2 + 3b2 + 2c2

c) 1002 - 992 + 982 - 972 + ..... + 22 - 12

= ( 100 - 99 )( 100 + 99 ) + ( 98 - 97 )( 98 + 97 ) + ..... + ( 2 - 1 )( 2 + 1 )

= 199 + 197 + 195 + ..... + 5 + 3

\(\frac{\left(199+3\right)\left(\left(199-3\right)\frac{1}{2}+1\right)}{2}\)

= 9999

d) 3 ( 22 + 1 )( 24 +1 )......( 264 + 1 ) + 1

= ( 22 -1 )( 22 + 1 )(24 + 1 ).....( 264 + 1 ) + 1

= ( 24 -1 )( 24 + 1 )( 28 + 1 )......( 264 + 1 ) +1

= ( 28 -1 )( 28 + 1).....( 264 + 1) +1

............

= ( 264 - 1)( 264 +1 ) + 1

= 2128

4 tháng 10 2017

a,b,c,f tìm cách áp dụng HĐT vào nhé! động não tí xem :)

d) Sửa đề :\(100^2-99^2+98^2-97^2+...+2^2-1^2\)

\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)

\(=199+195+...+3\)

Khi đó tổng sẽ là:

\(\dfrac{\left(199+3\right)\left[\dfrac{\left(199-3\right)}{4}+1\right]}{2}=5050.\)

e) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)

\(=2^{128}-1+1\)

\(=2^{128}.\)

4 tháng 10 2017

lười quá, ko mún tính^^

21 tháng 10 2016

A = 1002 - 992 + 982 - 972 + . . . + 22 - 12

= (100 - 99)(100 + 99) + (98 - 97)(98 + 97) + . . . (2 - 1)(2 + 1)

= 199 + 195 + . . . + 3

= 5050

B = 3(22 + 1)(24 + 1) . . . (264 + 1) + 1

= (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1)(264 + 1) + 1

= (24 - 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1) + 1

= (28 - 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1) + 1

= (216 - 1)(216 + 1)(232 + 1)(264 + 1) + 1

= (232 - 1)(232 + 1)(264 + 1) + 1

= (264 - 1)(264 + 1) + 1

= 2128 - 1 + 1

= 2128

22 tháng 10 2016

Câu C mk chép nhầm đề đó

9 tháng 8 2020

a. \(\left(a^2+a-1\right)\left(a^2-a+1\right)=a^4+a^2+1\)

b. \(\left(a+2\right)\left(a-2\right)\left(a^2+2a+4\right)\left(a^2-2x+4\right)=a^6-64\)

c. \(\left(2+3y\right)^2-\left(2x-3y\right)^2-12xy=4+12y-4x^2\)

d. \(\left(x+1\right)^3-\left(x-1\right)^3-\left(x^3-1\right)-\left(x-1\right)\left(x^2+x+1\right)=-2x^3+6x^2+4\)

9 tháng 8 2020

\(A=\left(a^2+\left(a-1\right)\right)\left(a^2-\left(a-1\right)\right)=a^4-\left(a-1\right)^2=a^4-\left(a^2-2a+1\right)=a^4-a^2+2a-1\)

\(B=\left(a+2\right)\left(a^2-2a+4\right)\left(a-2\right)\left(a^2+2a+4\right)=\left(a^3+8\right)\left(a^3-8\right)=a^6-64\)

\(C=9y^2+12y+4-\left(4x^2-12xy+9y^2\right)-12xy=12y+4-4x^2\)

\(D=x^3+3x^2+3x+1-x^3+3x^2-3x+1-x^3+1-x+1=-x^3+6x^2-x+4\)

18 tháng 7 2017

a) \(\left(x^2-2x+2\right)\left(x-2\right)\left(x^2-2x+2\right)\left(x+2\right)\)

\(=\left(x^3-2x^2-2x^2+4x+2x-4\right)\left(x^3+2^3\right)\)

\(=\left(x^3-4x^2+6x-4\right)\left(x^3+8\right)\)

\(=x^6+8x^3-4x^5-32x^2+6x^4+48x-4x^3-32\)

\(=x^6-4x^5+4x^3-32x^2+48x-32\)

b) \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)

\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]+x^3-3x\left(x^2-1\right)\)

\(=2x\left[\left(x^2+2x+1\right)-\left(x^2-1\right)+\left(x^2-2x+1\right)\right]+x^3-\left(3x^3-3x\right)\)

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

\(=2x\left(x^2+3\right)+x^3-3x^3+3x\)

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

\(=9x\)

2 câu kia đợi tí đã nhé!

18 tháng 7 2017

c) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+\left(2a-b\right)^2\)

\(=\left(a^2+b^2+c^2+2ab+2bc+2ca\right)+\left(a^2+b^2+c^2+2ab-2bc-2ca\right)+\left(4a^2-4ab+b^2\right)\)

\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca+4a^2-4ab+b^2\)

\(=6a^2+3b^2+2c^2\)

d) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+2\left(a+b\right)^2\)

\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca+2a^2+2ab+b^2\)

\(=4a^2+4b^2+2c^2+6ab.\)

8 tháng 7 2018

1272 + 146.127 + 732

= 1272 + 2 . 73 .127 + 732

= (127 + 73 ) 2

= 200 2

15 tháng 10 2023

a: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

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

\(=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=16x^2\)

b: \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3+2x^2-x-2-x^3+8\)

\(=2x^2-x+6\)

15 tháng 10 2023

a) \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

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

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

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

\(=\left(4x\right)^2\)

\(=16x^2\)

b) \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

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

\(=x^3+2x^2-x-2-x^3+8\)

\(=2x^2-x+6\)