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\(1,ax+ay+bx+by\)
\(=\left(ax+ay\right)+\left(bx+by\right)\)
\(=a\left(x+y\right)+b\left(x+y\right)\)
\(=\left(x+y\right)\left(a+b\right)\)
\(2,ax+ay+2x+2y\)
\(=\left(ax+ay\right)+\left(2x+2y\right)\)
\(=a\left(x+y\right)+2\left(x+y\right)\)
\(=\left(x+y\right)\left(a+2\right)\)
\(3,ax+ay-bx-by\)
\(=\left(ax+ay\right)-\left(bx+by\right)\)
\(=a\left(x+y\right)-b\left(x+y\right)\)
\(=\left(x+y\right)\left(a-b\right)\)
\(4,ax+ay-2x-2y\)
\(=\left(ax+ay\right)-\left(2x+2y\right)\)
\(=a\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(a-2\right)\)
ax + ay + bx + by
= a.(x+y) + b.(x+y)
= (x+y).(a+b)
ax + ay + 2x + 2y
= a.(x+y) + 2.(x+y)
= (x+y).(a+2)
ax + ay - bx - by
= a.(x+y) - b.(x+y)
= (x+y).(a-b)
ax + ay - 2x - 2y
= a.(x+y) - 2.(x+y)
= (x+y).(a-2)
A = 4acx + 4bcx + 4ax + 4bx ( đã sửa '-' )
= 4x( ac + bc + a + b )
= 4x[ c( a + b ) + ( a + b ) ]
= 4x( a + b )( c + 1 )
B = ax - bx + cx - 3a + 3b - 3c
= x( a - b + c ) - 3( a - b + c )
= ( a - b + c )( x - 3 )
C = 2ax - bx + 3cx - 2a + b - 3c
= x( 2a - b + 3c ) - ( 2a - b + 3c )
= ( 2a - b + 3c )( x - 1 )
D = ax - bx - 2cx - 2a + 2b + 4c
= x( a - b - 2c ) - 2( a - b - 2c )
= ( a - b - 2c )( x - 2 )
E = 3ax2 + 3bx2 + ax + bx + 5a + 5b
= 3x2( a + b ) + x( a + b ) + 5( a + b )
= ( a + b )( 3x2 + x + 5 )
F = ax2 - bx2 - 2ax + 2bx - 3a + 3b
= x2( a - b ) - 2x( a - b ) - 3( a - b )
= ( a - b )( x2 - 2x - 3 )
= ( a - b )( x2 + x - 3x - 3 )
= ( a - b )[ x( x + 1 ) - 3( x + 1 ) ]
= ( a - b )( x + 1 )( x - 3 )
\(\left(ax+by\right)^2-\left(ay+bx\right)^2\)
\(=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\)
\(=\left(ax-ay-bx+by\right)\left(ax+ay+bx+by\right)\)
\(=\left[a\left(x-y\right)-b\left(x-y\right)\right]\left[a\left(x+y\right)+b\left(x+y\right)\right]\)
\(=\left(a-b\right)\left(x-y\right)\left(a+b\right)\left(x+y\right)\)
1, \(a.\left(b+c\right)+3b+3c=a.\left(b+c\right)+3.\left(b+c\right)\)= \(\left(b+c\right).\left(3+a\right)\)
2, \(a.\left(m-n\right)+\left(m-n\right)=\left(m-n\right).\left(a+1\right)\)
3, \(7a^2-7ax-9a+9x=7a.\left(a-x\right)-9.\left(a-x\right)\)= \(\left(a-x\right).\left(7a-9\right)\)
4, \(4x+by+4y+bx=4.\left(x+y\right)+b.\left(x+y\right)\)= \(\left(x+y\right).\left(4+b\right)\)
5, \(ay-ax-2x+2y=a.\left(y-x\right)+2.\left(y-x\right)\)= \(\left(y-x\right).\left(a+2\right)\)
Chúc bạn học tốt. Có gì không hiểu thì chat hỏi mik nhé. ^^
\(2ax-bx+3cx-2a+b-3c\\ =x\left(2a-b+3c\right)-\left(2a-b+3c\right)\\ =\left(x-1\right)\left(2a-b+3c\right)\)
\(ax-bx-2cx-2a+2b+4c\\ =x\left(a-b-2c\right)-2\left(a-b-2c\right)\\ =\left(x-2\right)\left(a-b-2c\right)\)
\(3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\)
\(ax^2-bx^2-2ax+2bx-3a+3b\\ =x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a+b\right)\\ =\left(x^2-2x-3\right)\left(a+b\right)\\ =\left(x+1\right)\left(x-3\right)\left(a+b\right)\)
\(\left(ax+by\right)^2-\left(ay+bx\right)^2\)
(ax+ay+bx+by)(ax−ay+by−bx) \(=\left(ax+ay+bx+by\right)\left(ax-ay+by-bx\right)\)
\(=\left(a+b\right)\left(x+y\right)\left(a-b\right)\left(x-y\right)\)
\(\left(ax+by\right)^2-\left(ay+bx\right)^2=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\)
\(=\left[a\left(x-y\right)-b\left(x-y\right)\right].\left[a\left(x+y\right)+b\left(x+y\right)\right]\)
\(=\left(a-b\right)\left(x-y\right)\left(a+b\right)\left(x+y\right)\)
a) \(\dfrac{ax+ay-bx-by}{ax-ay-bx+by}=\dfrac{a\left(x+y\right)-b\left(x+y\right)}{a\left(x-y\right)-b\left(x-y\right)}=\dfrac{\left(a-b\right)\left(x+y\right)}{\left(a-b\right)\left(x-y\right)}=\dfrac{x+y}{x-y}\)
b) \(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\dfrac{a+b-c}{a+c-b}\)
1) \(=5\left(x+y\right)-\left(x-y\right)\left(x+y\right)=\left(x+y\right)\left(5-x+y\right)\)
2) \(=3\left(x^2-4x+4\right)=3\left(x-2\right)^2\)
3) \(=\left(x^2-1\right)\left(x^2+1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
4) \(=\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)\)
5) \(=3\left(a^2-10a+25-b^2\right)=3\left(\left(a-5\right)^2-b^2\right)=3\left(a-5-b\right)\left(a-5+b\right)\)
6) \(=a\left(x-y\right)\left(x+y\right)+b\left(x+y\right)=\left(x+y\right)\left(ax-ay+b\right)\)
\(ax^2+by^2-ay^2-bx^2=x^2\left(a-b\right)-y^2\left(a-b\right)=\left(a-b\right)\left(x^2-y^2\right)=\left(a-b\right)\left(x-y\right)\left(x+y\right)\)
nnnvnvvnvnvnnvneech
\(ax-bx+ay-by-3a+3b\)
\(=x\left(a-b\right)+y\left(a-b\right)-3\left(a-b\right)\)
\(=\left(a-b\right)\left(x+y-3\right)\)