Bài 1:

a,\(127^2+146.127+73^2=127^2+2.127.73+73^2\)\(=\left(127+73\right)^2=200^2=40000\)

b,\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(18^8-\left(18^8-1\right)=1\)

\(c,100^2-99^2+98^2-97^2+...+2^2-1\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=199+195+...+3\)

áp dụng công thức Gauss ta đc đáp án là:10100

d, mk khỏi ghi đề dài dòng:

\(\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560000}{40000}=14\)Bài 2:

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

\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)Cứ tiếp tục ta đc \(A=2^{32}-1< B=2^{32}\)

\(\left(3-1\right)C=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^2+16\right)\)giải như câu a đc:\(\left(3-1\right)C=3^{32}-1\)

\(\Rightarrow C=\dfrac{3^{32}-1}{3-1}=\dfrac{3^{32}-1}{2}< D=3^{32}-1\)

1c,

\(=100^2-99^2+98^2-97^2+...+2^2-1^2\\ =\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\\ =\left(100+99\right)\cdot1+\left(98+97\right)\cdot1+...+\left(2+1\right)\cdot1\\ =100+99+98+97+...+2+1\\ =\dfrac{100\cdot101}{2}=5050\)

Bài 11:

1) Sửa lại đề là: \(A=127^2+146.127+73^2\)

\(\Rightarrow A=127^2+2.127.73+73^2\)

\(\Rightarrow A=\left(127+73\right)^2\)

\(\Rightarrow A=200^2\)

\(\Rightarrow A=40000\)

Vậy \(A=40000.\)

2) Sửa lại đề là: \(B=9^8.2^8-\left(18^4-1\right).\left(18^4+1\right)\)

\(\Rightarrow B=\left(9.2\right)^8-\left[\left(18^4\right)^2-1^2\right]\)

\(\Rightarrow B=18^8-\left(18^8-1\right)\)

\(\Rightarrow B=18^8-18^8+1\)

\(\Rightarrow B=0+1\)

\(\Rightarrow B=1\)

Vậy \(B=1.\)

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

\(\Rightarrow2D=\left(3-1\right).\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

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

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

\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(=3^{32}-1\)

\(\Rightarrow D=\frac{3^{32}-1}{2}\)

127² + 146.127 + 73²

= 127² + 2 . 73 .127 + 73²

= (127 + 73 ) ²

= 200 ²

a. 134^2 - 68.134 + 34^2 = ( 134 - 34 ) ^2 = 100^2 = 10000

b. 9^8.2^8 - ( 18^4 - 1 )(18^4 + 1 ) = 18^8 - 18^8 + 1 = 1

c. 100^2 - 99^2 + 98^2 - 97^2 + ... + 2^2 - 1 

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

= 100 + 99 + 98 + 97 + ... + 2 + 1

=( 100 + 1 ).100:2 = 5050

1.

a/ \(A=2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)

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

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

\(=2.1.\left[1^2-3xy\right]-3\left[1^2-2xy\right]\)

\(=2-6xy-3+6xy\)

\(=-1\)

Vậy...

2.

a. \(127^2+146.127+73^2\)

\(=127^2+2.73.127+73^2\)

\(=\left(127+73\right)^2=200^2=40000\)

b. \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(=18^8-18^8+1\)

\(=1\)

a)127²+146.127+73²=127²+2.127.73+73²=(127+73)²=200²=40000

b)9⁸.2⁸-(18⁴-1).(18⁴+1)=18⁸-[(18⁴)²-1²]=18⁸-(18⁸-1)=18⁸-18⁸+1=1

c)100²-99²+98²-97²+...+2²-1²=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)

a/ A = 100² - 99² + 98² -...+2² - 1²

= (100² - 99²) + (98² - 97²) +...+ (2² - 1²)

= 199 + 195 + 191 + ... + 1

= (\(\frac{199-1}{4}+1\))(\(\frac{199+1}{2}\)) = 5050

b/ Y chang câu a luôn nha

c/ \(C=\frac{780^2-220^2}{125^2+150.125+75^2}=\frac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}\)

\(=\frac{560.1000}{200^2}=14\)

d/ 2(a + 4)(a + 5) - (a + 5)² - (a² - 9a - 17) 

= 17a + 32 = 17(a + 1) + 15 = 1700 + 15 = 1715

a) Áp dụng hằng đẳng thức ta đc:

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

\(=\left(100+99\right)\left(100-99\right)+\left(98-97\right)\left(98+87\right)+...+\left(2+1\right)\left(2-1\right)\)

\(=199+195+191+...+3\)

\(=\left[\left(199-3\right):4+1\right]\cdot\left(199+3\right):2=50\cdot101=5050\)

a) Áp dụng hằng đẳng thức ta đc:

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

\(=\left(100+99\right)\left(100-99\right)+\left(98-97\right)\left(98+87\right)+...+\left(2+1\right)\left(2-1\right)\)

\(=199+195+191+...+3\)

\(=\left[\left(199-3\right):4+1\right]\cdot\left(199+3\right):2=50\cdot101=5050\)

b) mk nghĩ bước đầu tiên là phải bỏ ngoặc:

 \(=20^2+18^2+16^2+...4^2+2^2-19^2-17^2-....-3^2-1^2\)

\(=\left(20^2-19^2\right)+\left(18^2-17^2\right)+...+\left(4^2-3^2\right)-1^2\)

\(=\left(20+19\right)\left(20-19\right)+\left(18+17\right)\left(18-17\right)+...+\left(4-3\right)\left(4+3\right)-1\)

\(=\left(39+35+31+...+7\right)-1\)

\(=\left(\left[\left(39-7\right):4+1\right]\cdot\left(39+7\right):2\right)-1=207-1=206\)