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\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab-2ac+2bc\)
\(\left(a+b-c\right)^2=a^2+b^2+c^2+2ab-2bc-2ac\)
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
(a - b - c)2 = a2 + b2 + c2 - 2ab - 2ac + 2bc
(a + b - c)2 = a2 + b2 + c2 +2ab - 2ac - 2bc
a) ( x + 1 ) ( x - 1 )
= ( x2 + 1 )
b) ( x - 2y ) ( x + 2y )
= ( x2 - 4y2 )
c) 56 x 64
= ( 60 - 4 ) ( 60 + 4 )
= 3 600 - 16
= 3 584
a,12x10+18x10=10x(12+18)
=10x30
=300
b,6x15+4x15=(6+4)x15
=10x15
=150
a, 12.10+18.10
=10(12+18)
=10.30
=300
b,6.15+4.15
=15(6+4)
=15.10
=150
Có : a^2+b^2 = 4ab
<=> a^2-4ab+b^2=0
<=> (a^2-4ab+4b^2)-3b^2=0
<=> (a-2b)^2 - 3b^2 = 0
<=> (a-2b-\(\sqrt{3}b\)) . (a-2b+\(\sqrt{3}b\)) = 0
<=> \(\left[a-\left(2+\sqrt{3}\right)b\right]\). \(\left[a-\left(2-\sqrt{3}\right)b\right]\)= 0
<=> \(a-\left(2+\sqrt{3}\right)b\)= 0 hoặc \(a-\left(2-\sqrt{3}\right)b\)= 0
<=> \(a=\left(2+\sqrt{3}\right)b\)hoặc \(a=\left(2-\sqrt{3}\right)b\)
TH1 : N = \(\sqrt{3}\)
TH2 : N = \(-\sqrt{3}\)
Vậy ..............
Tk mk nha
a) = 37(37 + 126) + 622 = 9875
b) =(100-1)(100+1) = 1002 - 1 = 9999
Vì a-b=3 => (a-b)^2=9 => a^2-2ab+b^2=9 => a^2+ab+b^2=9+3ab=9+3.4=21
Ta có a^3-b^3=(a-b).(a^2+ab+b^2)=3.21=63
tick nha
a, \(2.31.12+4.6.42+8.27.3=24.31+24.42+24.27=24.\left(31+42+27\right)=31.100=3100\)
b, \(36.38+36.82+64.69+64.41=36.120+64.110=11360\)
a) \(2.31.12+4.6.42+8.27.3\)
\(=\)\(\left(2.12\right).31+\left(4.6\right).42+\left(8.3\right).27\)
\(=\)\(24.31+24.42+24.27\)
\(=\)\(24.\left(31+42+27\right)\)
\(=\)\(24.100\)
\(=\)\(2400\)