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\(\frac{258^2-242^2}{254^2-246^2}=\frac{\left(258+242\right)\left(258-242\right)}{\left(254+246\right)\left(254-246\right)}=\frac{500.16}{500.8}=2\)
\(263^2+74.263+37^2=263^2+2.37.263+37^2=\left(263+37\right)^2=300^2=90000\)
\(136^2-92.136+46^2=136^2-46.2.136+46^2=\left(136-46\right)^2=90^2=8100\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+.....+\left(2^2-1^2\right)=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+....+\left(2+1\right)\left(2-1\right)=\left(50+49+....+1\right)=\frac{51.50}{2}=51.25=1275\)
\(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)
Tính giá trị biểu thức sau:
a, A= \(258^2-\dfrac{242^2}{254^2}-246^2\approx\) 6047,1
b, B= \(263^2+74.263+37^2=90000\)
c, C= \(136^2-92.136+46^2=8100\)
d, D = \(\left(50^2+48^2+46^2+...+2^2\right)-\left(49^2+47^2+45^2+...+1^2\right)\)
= 22100 - 20825= 1275
Bài 1:
a) \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
b) \(100^2+103^2+105^2+94^2=101^2+98^2+96^2+107^2\)
\(\Leftrightarrow100^2+103^2+105^2+94^2-101^2-98^2-96^2-107^2=0\)
\(\Leftrightarrow\left(100^2-98^2\right)+\left(103^2-101^2\right)-\left(107^2-105^2\right)-\left(96^2-94^2\right)=0\)
\(\Leftrightarrow2.198+2.204-2.212-2.190=0\)
\(\Leftrightarrow2\left(198+204-212-190\right)=0\)
\(\Leftrightarrow2.0=0\) (đúng)
Bài 2:
a) \(263^2+74.263+37^2\)
\(=263^2+2.37.263+37^2\)
\(=\left(263+37\right)^2\)
b) \(\left(50^2+48^2+46^2+...+2^2\right)-\left(49^2+47^2+45^2+...+1^2\right)\)
\(=50^2+48^2+46^2+...+2^2-49^2-47^2-45^2-...-1^2\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+\left(46^2-45^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(50+49\right)+\left(48+47\right)+\left(46+45\right)+...+\left(2+1\right)\)
\(=50+49+48+47+46+45+...+2+1\)
\(=\dfrac{\left(50+1\right).\left(50-1+1\right)}{2}=1275\)
Kết luận ...
Bài 3:
b: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)+10=0\)
\(\Leftrightarrow6x^2+12-6x^2+12x-6=0\)
hay \(x=-\dfrac{1}{2}\)
Bài 2:
a: \(x^3+3x^2+3x+1=\left(x+1\right)^3\)
b: \(m^3+9m^2n+27mn^2+27n^3=\left(m+3n\right)^3\)
\(B=263^2+74.263+37^2\)
\(=\left(263+37\right)^2\)
\(=300^2\)
\(=90000\)
\(B=263^2+74\cdot263+37^2\)
\(=263^2+2\cdot263\cdot37+37^2\)
\(=\left(263+37\right)^2\)
\(=300^2\)
\(=90000\)