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Đặt A = \(100^2-99^2+98^2-97^2+96^2-95^2+2^2-1^2\)
A\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\left(96-95\right)\left(96+95\right)+...+\left(2-1\right)\left(2+1\right)\)
A \(=199+195+191+...+3\)
A gồm \(\dfrac{\left(199-3\right)}{4}+1=50\) ( số hạng )
Vậy A = \(\dfrac{\left(199+3\right).50}{2}=5050\)
Đặt \(A=100^2-99^2+98^2-97^2+96^2-95^2+...+2^2-1^2\)
\(A=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(A=2.100-1+2.98-1+2.96-1+...+2.2-1\)
\(A=2.\left(100+98+...+2\right)-50\)
\(A=\dfrac{2.\left[\left(100-2\right):2+1\right].\left(100+2\right)}{2}-50\)
\(A=50.102-50\)
\(A=50.\left(201-1\right)\)
\(A=50.101\)
\(A=5050\)
Gíup được câu b thoy.
\(B=\left(5x-2\right)\left(5x+2\right)-\left(5x-1\right)^2\\ =25x^2-4-\left(25x^2-10x+1\right)\\ =25x^2-4-25x^2+10x-1\\ =10x-5=5\left(x-1\right)\)
A=\(x^2+y^2+9+2xy+6x+6y+9-x^2-y^2-9\)
A=2xy+6x+6y
B=\(25x^2-4-25x^2+10x-1\)
B=10x-5
C=\(100^2+98^2+...+2^2-\left(99^2+97^2+...+1\right)\)
C=\(100^2-99^2+98^2-97^2+...+4^2-3^2+2^2-1\)
C=\(\left(100-99\right)\left(100+99\right)+...+\left(2-1\right)\left(2+1\right)\)
C=\(199+195+191+...+7+3\)
C=\(\left(199+3\right)\left(\left(199-3\right):4+1\right)=10100\)
a) 1272 + 146.127 + 732
= 1272 + 2.73.127 + 732
= (127 + 73)2 = 2002 = 40000
b) 98 . 28 - (184 - 1)(184 + 1)
= (9.2)8 - 188 + 1
= 188 - 188 + 1 = 1
c) \(\frac{780^2-220^2}{125^2+150.125+75^2}=\frac{\left(780-220\right)\left(780+220\right)}{125^2+2.75.125+75^2}=\frac{560.1000}{\left(125+75\right)^2}=\frac{560000}{200^2}\)
\(=\frac{560000}{40000}=14\)
a) 1272 + 146.127 + 732
= 1272 + 2.73.127 + 732
= ( 127 + 73 )2
= 2002 = 40 000
b) 98.28 - ( 184 - 1 )( 184 + 1 )
= ( 9.2 )8 - [ ( 184 )2 - 12 ]
= 188 - 188 + 1
= 1
c) \(\frac{780^2-220^2}{125^2+150\cdot125+75^2}\)
\(=\frac{\left(780-220\right)\left(780+220\right)}{125^2+2\cdot75\cdot125+75^2}\)
\(=\frac{560\cdot1000}{\left(125+75\right)^2}\)
\(=\frac{560000}{200^2}\)
\(=\frac{560000}{40000}=14\)
a/ \(127^2+146\cdot127+73^2=127^2+2\cdot73\cdot127+73^2=\left(127+73\right)^2=200^2=40000\)
b/ \(100^2-99^2+98^2-97^2+...+2^2-1^2=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=100+99+98+97+...+2+1=\frac{\left(100\right)\left(100+1\right)}{2}=5050\)
c/ \(34^2+66^2+68\cdot66=34^2+66^2+2\cdot34\cdot66=\left(34+66\right)^2=100^2=10000\)