a) \(127^2+146.127+73^2=127^2+2.73.127+73^2=\left(127+73\right)^2=40000\)b) \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)=18^8-\left(18^8-1\right)=1\)

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

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

\(=\dfrac{100\left(100+1\right)}{2}=5050\)

d) \(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\) \(=20^2-19^2+18^2-17^2+16^2-15^2+...+4^2-3^2+2^2-1^2\)

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

\(=\dfrac{20\left(20+1\right)}{2}=210\)

e) \(\dfrac{780^2-220^2}{125^2+150.125+75^2}\)

\(=\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560.1000}{200}=2800\)

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) \(\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-x^2+2xy-y^2=4xy\)

b) \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\\ =6ab^2\)

c) \(9^8\times2^8-\left(18^4-1\right)\left(18^4+1\right)=18^8-18^8+1=1\)

a) \(8^{15}.4^{13}=\left(2^3\right)^{15}.\left(2^2\right)^{13}=2^{45}.2^{26}=2^{71}\)

b) \(\left(\frac{1}{2}\right)^{18}.\left(\frac{1}{4}\right)^{28}=\left(\frac{1}{2}\right)^{18}.\left(\frac{1}{2^2}\right)^{28}=\frac{1}{2^{18}}.\frac{1}{2^{56}}=\frac{1}{2^{74}}=\left(\frac{1}{2}\right)^{74}\)

c) \(9^{12}.27^{10}=\left(3^2\right)^{12}.\left(3^3\right)^{10}=3^{24}.3^{30}=3^{54}\)

a) = (2³)¹⁵. (2²)¹³ = 2⁴⁵.2²⁶ = 2⁷¹

b) = \(\left(\frac{1}{2}\right)^{18}.\left(\left(\frac{1}{2}\right)^2\right)^{28}=\left(\frac{1}{2}\right)^{18}.\left(\frac{1}{2}\right)^{56}=\left(\frac{1}{2}\right)^{18+56}=\left(\frac{1}{2}\right)^{74}\)

c) = (3²)¹².(3³)¹⁰ = 3²⁴.3³⁰ = 3²⁴⁺³⁰ = 3⁵⁴

a: \(=2^2\cdot9\cdot\dfrac{1}{6\cdot9}\cdot\dfrac{4^2}{9^2}=\dfrac{2^2}{6}\cdot\dfrac{2^4}{3^4}=\dfrac{2^6}{2\cdot3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)

b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)

c: \(=\dfrac{\left(\dfrac{1}{2}\right)^3\cdot2^3\cdot\left(\dfrac{1}{2}\right)^2}{\left(-8\right)^2\cdot16}\cdot2^6=\dfrac{\dfrac{1}{2^2}}{64\cdot16}\cdot64=\dfrac{1}{4}:16=\dfrac{1}{64}=\left(\dfrac{1}{8}\right)^2\)

a: \(=2^2\cdot9\cdot\dfrac{1}{3^3\cdot2}\cdot\dfrac{2^4}{3^4}=\dfrac{2^4\cdot2^2}{2}\cdot\dfrac{9}{3^3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)

b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)

c: \(=\dfrac{\dfrac{1}{2^3}\cdot\dfrac{1}{2^2}\cdot8}{\left(-8\right)^2\cdot2^4}\cdot2^6=\dfrac{1}{2^2}\cdot2^6:2^{10}=\dfrac{2^4}{2^{10}}=\dfrac{1}{2^6}=\left(\dfrac{1}{8}\right)^2\)

127² + 146.127 + 73²

= 127² + 2 . 73 .127 + 73²

= (127 + 73 ) ²

= 200 ²

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

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

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

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

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

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

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

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

Ta có:

\(\begin{array}{l}{\left( {\frac{1}{4}} \right)^8} = {[{\left( {\frac{1}{2}} \right)^2}]^8} = {(\frac{1}{2})^{2.8}} = {(\frac{1}{2})^{16}};\\{\left( {\frac{1}{8}} \right)^3} = {[{(\frac{1}{2})^3}]^3} = {(\frac{1}{2})^{3.3}} = {(\frac{1}{2})^9}\end{array}\)