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1) Ta có: \(3x^2-6xy+3y^2-3z^2\)
\(=3\left(x^2-2xy+y^2-z^2\right)\)
\(=3\left[\left(x^2-2xy+y^2\right)-z^2\right]\)
\(=3\left[\left(x-y\right)^2-z^2\right]\)
\(=3\left(x-y-z\right)\left(x-y+z\right)\)
2) Ta có: \(5a^2+10ab+5b^2-5c^2\)
\(=5\left(a^2+2ab+b^2-c^2\right)\)
\(=5\left[\left(a^2+2ab+b^2\right)-c^2\right]\)
\(=5\left[\left(a+b\right)^2-c^2\right]\)
\(=5\left(a+b-c\right)\left(a+b+c\right)\)
ax2-5x2-ax+5x+a-5
=x^2(a-5)-x(a-5)+(a-5)
=(a-5)(x^2-x+1)
cậu ghi sai đề rồi phải là
3ax2+3bx2+ax+bx+5a+5b
=3x^2(a+b)+x(a+b)+5(a+b)
=(a+b)(3x^2+x+5)
4a2=4b2-4a+1
=(2a)2-2*2a*1+12-4b2= (2a-1)2-(2b)2(2a-1-2b)(2a-1+2b)
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 )
=(5a2+5b2+10ab)-5c2=5.(a2+2ab+b2)-5c2=5.(a+b)2-5c2=5.[(a+b)2-c2]=5.(a+b+c).(a+b-c)
=(5a2+5b2+10ab)-5c2=5.(a2+2ab+b2)-5c2=5.(a+b)2-5c2=5.[(a+b)2-c2]=5.(a+b+c).(a+b-c)