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17 tháng 8 2018
a)\(4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(=\dfrac{1}{2}.\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(=\dfrac{1}{2}.\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(=\dfrac{1}{2}.\left(3^8-1\right)\left(3^8+1\right)\)
\(=\dfrac{1}{2}.\left(3^{16}-1\right)\)
\(=\dfrac{1}{2}3^{16}-\dfrac{1}{2}\)
b) \(48\left(5^2+1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)
\(=2.\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)
\(=2.\left(5^4-1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)
\(=2.\left(5^8+1\right).....\left(5^{32}+1\right)\)
\(=2.\left(5^{32}-1\right)\)
\(=2.5^{32}-2\)
Tham khảo nhé~
23 tháng 10 2021
\(a,=\left(a+5+\dfrac{1}{2}-a\right)^2=\left(\dfrac{11}{2}\right)^2=\dfrac{121}{4}\\ b,=\dfrac{\left(x+y\right)^2-16}{3x\left(x-4+y\right)}=\dfrac{\left(x+y-4\right)\left(x+y+4\right)}{3x\left(x+y-4\right)}=\dfrac{x+y+4}{3x}\)
23 tháng 10 2021
a, \(\left(a+5\right)^2+2\left(a+5\right)\left(\dfrac{1}{2}-a\right)+\left(\dfrac{1}{2}-a\right)^2=\left(a+5+\dfrac{1}{2}-a\right)^2=\left(\dfrac{11}{2}\right)^2=\dfrac{121}{4}\)
b,\(\dfrac{x^2-16+2xy+y^2}{3x^2-12x+3xy}=\dfrac{\left(x^2+2xy+y^2\right)-4^2}{3x\left(x-4+y\right)}=\dfrac{\left(x+y-4\right)\left(x+y+4\right)}{3x\left(x+y-4\right)}=\dfrac{x+y+4}{3x}\)
5 tháng 7 2020
a) \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x\left(x^2-5x+1\right)-2\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-5x^2+x-2x^2+10x-2-x^3-11x\)
\(=-7x^2-2\)
b) \(\left(x-1\right)\left(x^2+x+1\right)+x^3-2\)
\(=x\left(x^2+x+1\right)-1\left(x^2+x+1\right)+x^3-2\)
\(=x^3+x^2+x-x^2-x-1+x^3-2\)
\(=2x^3-3\)
c) \(\left(x-y\right)\left(x+y\right)-2x\left(x-y\right)\)
\(=x\left(x+y\right)-y\left(x+y\right)-2x\left(x-y\right)\)
\(=x^2+xy-yx-y^2-2x^2+2xy\)
\(=-x^2-y^2+2xy\)
5 tháng 7 2020
a, \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-7x^2+11x-2-x^3-11x=-7x^2-2\)
b, \(\left(x-1\right)\left(x^2+x+1\right)+\left(x^3-2\right)\)
\(=x^3-1+x^3-2=2x^3-3\)
c, \(\left(x-y\right)\left(x+y\right)-2x\left(x-y\right)\)
\(=x^2-y^2-2x^2+2xy=-x^2-y^2+2xy\)
22 tháng 6 2017
\(\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
\(=\frac{x^3\left(x-1\right)-\left(x-1\right)}{x^4+x^3+x^2+2x^2+2x+2}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}\)
\(=\frac{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2+2\right)}\)
\(=\frac{\left(x-1\right)^2}{\left(x^2+2\right)}\)
17 tháng 6 2017
mình làm bài 2 trước nha:
a) y.(a-b)+a.(y-b)=a.y-b.y+a.y-b.y
=(a.y+a.y)-(b.y+b.y)
=2.a.y-2.b.y
=2.y.(a-b)
b)x2.(x+y)-y.(x2-y2)=x3+x2.y-x2y+y3=x3+y3
TD
1
18 tháng 10 2021
\(a,=x^2-3x-10-x^2+3x=-10\\ b,=\left(x+1\right)\left(x+1-x+1\right)=2\left(x+1\right)=2x+2\)
Ta có:
\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(P=\frac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(P=\frac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(P=\frac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)
\(P=\frac{5^{32}-1}{2}\)