Chứng minh đẳng thức sau:
1002 + 1032 + 1052 + 942 = 1012 + 982 + 962 + 1072
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DV
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 7 2023
a, A = 1002 - 992 + 982 - 972 +...+ 22 - 12
A = (1002 - 992) + (982 - 972) +...+ (22 - 1)2
A = (100 - 99)(100+99) + (98-97)(98+97)+..+(2-1)(2+1)
A = 1.199 + 1.195 + 1.191 +...+1.3
A = 3 + ...+191+ 195 + 199
Dãy số trên là dãy số cách đều với khoảng cách là: 199 -195=4
Dãy số trên có số hạng là: (199 - 3): 4 + 1 = 50 (số )
A = (199 +3) \(\times\) 50 : 2 = 5050
NL
Nguyễn Lê Phước Thịnh
CTVHS
6 tháng 5
Bài 3:
a: \(A=4x^2+4x+11\)
\(=4x^2+4x+1+10\)
\(=\left(2x+1\right)^2+10\ge10\forall x\)
Dấu '=' xảy ra khi 2x+1=0
=>2x=-1
=>\(x=-\frac12\)
b: \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)
Dấu '=' xảy ra khi \(x^2+5x=0\)
=>x(x+5)=0
=>x=0 hoặc x=-5
c: \(C=x^2-2x+y^2-4y+7\)
\(=x^2-2x+1+y^2-4y+4+2\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)
Dấu '=' xảy ra khi x-1=0 và y-2=0
=>x=1 và y=2
Bài 4:
a: \(A=5-8x-x^2\)
\(=-x^2-8x-16+21\)
\(=-\left(x+4\right)^2+21\le21\forall x\)
Dấu '=' xảy ra khi x+4=0
=>x=-4
b: \(B=5-x^2+2x-4y^2-4y\)
\(=-x^2+2x-1-4y^2-4y-1+7\)
\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\forall x,y\)
Dấu '=' xảy ra khi x-1=0 và 2y+1=0
=>x=1 và y=-1/2
Bài 5:
a: \(a^2+b^2+c^2=ab+ac+bc\)
=>\(2\left(a^2+b^2+c^2\right)=2\left(ab+ac+bc\right)\)
=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
=>\(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
=>\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
=>a=b=c
b: \(a^2-2a+b^2+4b+4c^2-4c+6=0\)
=>\(a^2-2a+1+b^2+4b+4+4c^2-4c+1=0\)
=>\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)
=>a-1=0 và b+2=0 và 2c-1=0
=>a=1 và b=-2 và c=1/2
Bài 1:
a: \(A=100^2-99^2+98^2-97^2+\cdots+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\cdots+\left(2-1\right)\left(2+1\right)\)
=100+99+98+87+...+2+1
\(=100\cdot\frac{\left(100+1\right)}{2}=5050\)
b: \(B=3\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)
c: \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)
\(=2c^2\)
Bài 2:
a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3ab^2-3a^2b\)
\(=a^3+b^3\)
b: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
=(a+b+c)\(\left(a^2+b^2+c^2-ab-ac-bc\right)\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
6 tháng 5
Bài 3:
a: \(A=4x^2+4x+11\)
\(=4x^2+4x+1+10\)
\(=\left(2x+1\right)^2+10\ge10\forall x\)
Dấu '=' xảy ra khi 2x+1=0
=>2x=-1
=>\(x=-\frac12\)
b: \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)
Dấu '=' xảy ra khi \(x^2+5x=0\)
=>x(x+5)=0
=>x=0 hoặc x=-5
c: \(C=x^2-2x+y^2-4y+7\)
\(=x^2-2x+1+y^2-4y+4+2\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)
Dấu '=' xảy ra khi x-1=0 và y-2=0
=>x=1 và y=2
Bài 4:
a: \(A=5-8x-x^2\)
\(=-x^2-8x-16+21\)
\(=-\left(x+4\right)^2+21\le21\forall x\)
Dấu '=' xảy ra khi x+4=0
=>x=-4
b: \(B=5-x^2+2x-4y^2-4y\)
\(=-x^2+2x-1-4y^2-4y-1+7\)
\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\forall x,y\)
Dấu '=' xảy ra khi x-1=0 và 2y+1=0
=>x=1 và y=-1/2
Bài 5:
a: \(a^2+b^2+c^2=ab+ac+bc\)
=>\(2\left(a^2+b^2+c^2\right)=2\left(ab+ac+bc\right)\)
=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
=>\(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
=>\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
=>a=b=c
b: \(a^2-2a+b^2+4b+4c^2-4c+6=0\)
=>\(a^2-2a+1+b^2+4b+4+4c^2-4c+1=0\)
=>\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)
=>a-1=0 và b+2=0 và 2c-1=0
=>a=1 và b=-2 và c=1/2
Bài 1:
a: \(A=100^2-99^2+98^2-97^2+\cdots+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\cdots+\left(2-1\right)\left(2+1\right)\)
=100+99+98+87+...+2+1
\(=100\cdot\frac{\left(100+1\right)}{2}=5050\)
b: \(B=3\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)
c: \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)
\(=2c^2\)
Bài 2:
a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3ab^2-3a^2b\)
\(=a^3+b^3\)
b: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
=(a+b+c)\(\left(a^2+b^2+c^2-ab-ac-bc\right)\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
6 tháng 5
Bài 1:
a: \(A=100^2-99^2+98^2-97^2+\cdots+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\cdots+\left(2-1\right)\left(2+1\right)\)
=100+99+98+87+...+2+1
\(=100\cdot\frac{\left(100+1\right)}{2}=5050\)
b: \(B=3\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)
c: \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)
\(=2c^2\)
Bài 2:
a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3ab^2-3a^2b\)
\(=a^3+b^3\)
b: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
=(a+b+c)\(\left(a^2+b^2+c^2-ab-ac-bc\right)\)
13 tháng 10 2021
\(2,\\ a,a^3+b^3=a^3=3a^2b+3ab^2+b^3-3a^2b-3ab^2\\ =\left(a+b\right)^3-3ab\left(a+b\right)\\ b,a^3+b^3+c^3-3abc\\ =\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\\ =\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ac-ab-bc\right)\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
5 tháng 5
Bài 4:
a: \(A=5-8x-x^2\)
\(=-\left(x^2+8x-5\right)\)
\(=-\left(x^2+8x+16-21\right)\)
\(=-\left(x+4\right)^2+21\le21\forall x\)
Dấu '=' xảy ra khi x+4=0
=>x=-4
b: \(B=5-x^2+2x-4y^2-4y\)
\(=-x^2+2x-1-4y^2-4y-1+7\)
\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\forall x,y\)
Dấu '=' xảy ra khi x-1=0 và 2y+1=0
=>x=1 và y=-1/2
Bài 3:
a: \(A=4x^2+4x+11\)
\(=4x^2+4x+1+10=\left(2x+1\right)^2+10\ge10\forall x\)
Dấu '=' xảy ra khi 2x+1=0
=>2x=-1
=>\(x=-\frac12\)
b: \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\forall x\)
Dấu '=' xảy ra khi \(x^2+5x=0\)
=>x(x+5)=0
=>x=0 hoặc x=-5
c: \(C=x^2-2x+y^2-4y+7\)
\(=x^2-2x+1+y^2-4y+4+2\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)
Dấu '=' xảy ra khi x-1=0 và y-2=0
=>x=1 và y=2
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.