Dễ ợt

ax + by + by + ay

= ( ax + ay ) + ( by + bx )

= a( x + y ) + b ( y + x )

= ( a + b ) ( x + y ) 

Ta nhóm lại với nhau rồi đặt thừa số chung ra ngoài:

ax + by + bx + ay

= ( ax + ay ) + ( by + bx )

= a( x + y ) + b( y + x )

= ( a + b )( x + y )

a(y+x)+b(x+y)

tk nha

Ta nhóm lại với nhau rồi đặt thừa số chung ra ngoài

ax + by + bx + ay

= ( ax + ay ) + ( by + bx )

= a(x + y) + b(y + x)

= (a + b)(x + y )

a( b + c ) - d( b + c )

= a . b + a . c - d . b + d . c

= a . b + a . c - d . b + d . c

=axb+axc-dxb+dxc

=axb+axc-dxb+dxc

=axb+axc-dxb+dxc

a: 6*6*6*6*6=6^5

b: 25*25*25*5=5^2*5^2*5^2*5=5^7

c: 8*2*2*16*4

=2^3*2^2*2^4*2^2

=2^11

d: \(3\cdot3\cdot3\cdot15\cdot5\cdot5\cdot5=15\cdot15\cdot15\cdot15=15^4\)

\(2\cdot2\cdot2\cdot3\cdot3=2^3\cdot3^2\)

a,\(\left(3x-1\right)\left(3x+1\right)=3x\left(3x+1\right)-\left(3x+1\right)=3x^2+3x-3x-1=3x^2-1\)

b,\(\left(x-3\right)\left(x+5\right)=x\left(x+5\right)-3\left(x+5\right)=x^2+5x-3x-15=x^2+2x-15\)

a) ab+ac= a(b+c)

b) ab-ac+ad= a(b-c+d)

c) ax-bx-cx+dx= x(a-b-c+d)

a.(b+c)

a.(b-c+d)

x.(a-b-c+d)

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\(a\left(b+c\right)-d\left(b+c\right)\)

\(=\left(b+c\right)\left(a-d\right)\)

\(ac-ad+bc-bd\)

\(=a\left(c-d\right)+b\left(c-d\right)\)

\(=\left(c-d\right)\left(a+b\right)\)

\(ax+by+bx+ay\)

\(=\left(ax+ay\right)+\left(by+bx\right)\)

\(=a\left(x+y\right)+b\left(y+x\right)\)

\(=\left(x+y\right)\left(a+b\right)\)

ab+ac-do+dc

a(c-d)+b(c-d)

a(x+y)+b(y+x)

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