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Bài 2.
a) 1013 = (100+1)3 = 1003+3.1002.1+3.100.12+13
= 1000000+30000+300+1 = 1030301
b) 2993 = (300-1)3 = 3003-3.3002.1+3.300.12-13
= 27000000 - 270000 + 900 -1 = 26730899
c) 993 = (100-1)3 = 1003-3.1002.1+3.100.12-1
= 1000000 - 30000 + 300 -1 = 970299
\(1,\\ b,A=\left(u-v\right)^3+3uv\left(u+v\right)\\ A=u^3-3u^2v+3uv^2-v^3+3u^2v+3uv^2=u^3-v^3\\ c,6\left(c-d\right)\left(c+d\right)+2\left(c-d\right)^2-\left(c-d\right)^3\\ =6c^2-6d^2+2c^2-4cd+2d^2-c^3+3c^2d-3cd^2+d^3\\ =8c^2-c^3-4d^2-4cd+3c^2d-3cd^2+d^3\)
\(2,\\ a,101^3=\left(100+1\right)^3\\ =100^3+3\cdot10000\cdot1+3\cdot100\cdot1+1\\ =1000000+30000+300+1=1030301\\ b,299^3=\left(300-1\right)^3\\ =300^3-3\cdot90000\cdot1+3\cdot300\cdot1-1\\ =27000000-270000+900-1\\ =26730899\\ c,99^3=\left(100-1\right)^3\\ =100^3-3\cdot10000\cdot1+3\cdot100\cdot1-1\\ =1000000-30000+300-1=970299\)
a) A = u 3 + 6 uv 2 – v 3 .
b) B = ( c + 2 d ) + ( c − 2 d 3 = 8 c 3 .
Câu 3:
a: \(49^2=2401\)
b: \(51^2=2601\)
c: \(99\cdot100=9900\)
Lời giải:
a.
$A=(u-v)^3+3uv(u+v)=u^3-3u^2v+3uv^2-v^3+3u^2v+3uv^2$
$=u^3-v^3+6uv^2$
b.
$3(c-2d)^2+3(c+2d)^2+(c+2d)^3+(c-2d)^3$
$3[(c-2d)^2+(c+2d)^2]+[(c+2d)+(c-2d)][(c+2d)^2-(c+2d)(c-2d)+(c-2d)^2]$
$=3(2c^2+8d^2)+2c[2c^2+8d^2-(c^2-4d^2)]$
$=6(c^2+4d^2)+2c(c^2+12d^2)$
$=2c^3+24cd^2+6c^2+24d^2$
\(a.\left(3x-1\right)^2+\left(x+3\right)\left(2x-1\right)\)
\(=9x^2-6x+1-2x^2+x-6x+3\)
\(=7x^2-11x+4\)
a) ( 100 + 1 ) 3 = 100 3 + 3 . 100 2 + 3.100 + 1 = 1030301.
b) ( 47 + 3 ) 3 = 50 3 = 125000.
c) ( 300 – 1 ) 3 = 26730899.
d) ( 1008 – 23 ) 3 = 1000 3 = 10 9 .
a) C = c + d + 2 ( c − d ) 3 = ( 3 c − d ) 3 .
b) D = m − n ( n + p ) 3 = ( m − 2 n − p ) 3 .
a. A = (a + b)3 - (a - b)3
A = \(\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
A = (a + b - a + b)\(\left[a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right]\)
A = 2b(a2 + a2 + a2 + 2ab - 2ab + b2 - b2 + b2)
A = 2b(3a2 + b2)
A = 6a2b + 2b3