Xét hiệu :

\(100^2+103^2+105^2+94^2-\left(101^2+98^2+96^2+107^2\right)\)

\(=100^2+103^2+105^2+94^2-101^2-98^2-96^2-107^2\)

\(=\left(100^2-98^2\right)+\left(103^2+101^2\right)-\left(107^2-105^2\right)-\left(96^2-94^2\right)\)

\(=\left(100-98\right)\left(100+98\right)+\left(103-101\right)\left(103+101\right)-\left(96-94\right)\left(96+94\right)\)\(-\left(107-105\right)\left(107+105\right)\)

\(=2.198+2.204-2.212-2.190\)

\(=2.\left(198+204-212-190\right)\)

\(=2.0\)

\(=0\)

VẬY dpcm

Ta có:  

100²+103²+105²+94²=101²+98²+96²+107²

=>100²+103²+105²+94²-(101²+98²+96²+107²)=0

=>100²+103²+105²+94²-101²-98²-96²-107²=0

=>(100²-98²) + (103²-101²) + (105²-107²) + (94²-96²) = 0

=>(100-98)(100+98) + (103-101)(103+101) + (105-107)(105+107) + (94-96)(94+96) = 0

=>2.(100+98) + 2.(103+101) - 2.(105+107) - 2.(94+96) = 0

=>2.[(100+98)+(103+101)-(105+107)-(94+96)] = 0

=>2.(198+204-212-190)=0

=>2.0=0

                     Chứng tỏ 100²+103²+105²+94²=101²+98²+96²+107²

cám ơn các bn nhiều ạ!!!

Xét hiệu , ta có :

100² + 103² + 105² + 94² - ( 101² + 98² + 96² + 107² )

= 100² + 103² + 105² + 94² - 101² - 98² - 96² - 107²

= ( 100² - 98² ) + ( 103² - 101² ) - ( 107² - 105² ) - ( 96² - 94² )

= ( 100 - 98 ).( 100 + 98 ) + ( 103 - 101 ).( 103 + 101 ) - ( 107 - 105 ). ( 107 + 105 ) - ( 96 - 94 ).( 96 + 94 )

= 2.198 + 2.204 - 2.212 - 2.190 = 2.( 198 + 204 - 212 - 190)

= 2.0 = 0

Vậy 100² + 103² + 105² + 94² = 101² + 98² + 96² + 107².

dùng hàng đẳng thức A^2-B^2=(A-B)(A+B) nhé còn phần b chuyển vế sang rồi dùng HĐT là được

a) \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)

b) \(100^2+103^2+105^2+94^2=101^2+98^2+96^2+107^2\)

\(\Leftrightarrow\left(100^2-98^2\right)+\left(103^2-101^2\right)+\left(105-107^2\right)+\left(94^2-96^2\right)=0\)

\(\Leftrightarrow2\left(100+98+103+101-105-107-94-96\right)=0\)

\(\Leftrightarrow2\times0=0\)(ĐPCM)

Bạn xem lại đề

a) \(\left(a+b\right)^2=[-\left(a+b\right)]^2=\left(-a-b\right)^2\)

b)\(\left(a-b\right)^2=[-\left(a-b\right)]^2=\left(b-a\right)^2\)

c)\(\left(a-b\right)^3=-[-\left(a-b\right)]^3=-\left(b-a\right)^3\)

dùng hằng đẳng thức A^2 - B^2 = (A - B)(A + B) nhé phần b chuyển vế sang rồi dùng hđt là Okay

Giải thích rõ hơn được hong

\(VT-VP=\left(100^2-96^2\right)+\left(105^2-101^2\right)-\left(107^2-103^2\right)-\left(98^2-94^2\right)\)

\(=\left(100-96\right)\left(100+96\right)+\left(105-101\right)\left(105+101\right)-\left(107-103\right)\left(107+103\right)-\left(98-94\right)\left(98+94\right)\)

\(=4\left(196+206-210-192\right)=0\)

=> VT=VP

a) Đặt A = (2 + 1)(2² + 1)(2⁴ + 1 )(2⁸ +1)( 2¹⁶ +1 )

=> A = ( 2² - 1 ) (2² + 1)(2⁴ + 1 )(2⁸ +1)( 2¹⁶ +1 )

=> A = (2⁴ - 1)(2⁴ + 1 )(2⁸ +1)( 2¹⁶ +1 )

=> A = (2⁸ - 1)(2⁸ +1)( 2¹⁶ +1 )

=> A= (2¹⁶ -1 ) (2¹⁶ + 1) = 2³² - 1 => đpcm

b)

<=> \(\left(100^2-98^2\right)+\left(103^2-101^2\right)+\left(105^2-107^2\right)+\left(94^2-96^2\right)\) = 0

<=> \(\left(100-98\right)\left(100+98\right)+\left(103-101\right)\left(103+101\right)\)+ (105 -107)(105+107) + (94 - 96)(96 + 94) = 0

<=> \(2.198+2.204-2.212-2.190\) = 0

<=> \(2\left(198+204-212-190\right)=0\)

<=> \(\left(198-190\right)+\left(204-212\right)=0\)

<=> \(-8+8=0\) (luôn đúng) => đpcm

P/s: đây ko phải bài lớp 10 đâu!

bn kiểm tra lại đề giúp mk vs