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\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+...+\left(100+99\right)\left(100-99\right)\)
\(=1+2+3+4+...+100=\frac{\left(100+1\right).100}{2}=5050\)
Bài làm :
Ta có :
\(-1^2+2^2-3^2+4^2-5^2+....+100^2\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+....+\left(100^2-99^2\right)\)
\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+....+\left(100+99\right)\left(100-99\right)\)
\(=1+2+3+4+....+100=\frac{\left(100+1\right).100}{2}=5050\)
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
như thế này chứ:
A=1002-992+982-972+...+22-12
B=12-22+32-42+...-20082-20092
C=3.(22+1)(24+1)(28+1)(216+1)-232
a) \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+....+\left(2-1\right)\left(2+1\right)\)
\(=199+195+....+3\)
\(=\frac{\left(199+3\right)\left[\left(199-3\right):4+1\right]}{2}\)
\(=5050\)
\(100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+....+\left(2-1\right)\left(2+1\right)\)
\(=1.199+1.195+...+1.3\)
\(=199+195+....+3\)
\(=\left[\left(\dfrac{199-3}{4}\right)+1\right]:2.\left(199+3\right)=5050\)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
\(3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{64}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)\)
\(=\left(2^8-1\right)......\left(2^{64}+1\right)=2^{128}-1\)
2) 100^2-99^2+98^2-97^2+...+2^2-1^2
=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)
=1.199+1.195+...+1.3
= 199+195+191+...+7+3
= 5050
cho tam giác ABC có Â =100 ,M là trung điểm của BC tren tia doi cua tia MA lay diem K sao cho KM=MA
a )tính số đo gocABK
b) Về phía ngoài của tam giác ABC vẽ các đoạn thẳng ADvuong góc và bằng AB ,AE vuông góc và bằng AC, chứng minh tam giác ABk bang tam giác DAE
C/M :MA vuong goc DE
dùng hằng đẳng thức A^2 - B^2 = (A - B)(A + B) nhé phần b chuyển vế sang rồi dùng hđt là Okay
a) \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+...+2+1\)
\(=\frac{100.\left(100+1\right)}{2}=5050\)
b) \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=...=\left(2^{64}-1\right)\left(2^{64}+1\right)+1^2=2^{128}-1^2+1^2=2^{128}\)
c) \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)
\(=2c^2\)
a/Có A=100^2+99^2+98^2+...+1^2 -2(99^2+97^2+..+1)
= Sigma(100)(x=1)(x^2) -2((1^2+2^2+3^2+..+99^2)-(2^2+4^2+...+98^2)
=Sigma(100)(x=1)(x^2)-2.Sigma(99)(x=1)(x^2)+4sigma(49)(x=1)(x^2)
=5050
b/bạn lấy 3=2^2-1 rồi dùng hiệu 2 bình nhé
c/tách ra được thôi
1: Ta có: \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
2: Ta có: \(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+...+2+1\)
=5050