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26 tháng 8 2019
A=9b^2c-3bc^2-9ac^2-3a^2c-9a^2b-3a^2+28abc
A=9.(b^2c-ac^2-a^2.b)-3.(bc^2+a^2.c+3a^2)+28abc
A=9.(b.(bc-a^2)-ac^2)-3.(c.(bc+a^2)+3a^2)+28abc
k dung mik nhe!!!!!
NL
Nguyễn Lê Phước Thịnh
CTVHS
4 tháng 6
a: \(\left(a-2b\right)^2-4b^2\)
\(=\left(a-2b\right)^2-\left(2b\right)^2\)
=(a-2b-2b)(a-2b+2b)
=a(a-4b)
b: \(\left(a-b\right)^2-c^2=\left(a-b-c\right)\left(a-b+c\right)\)
c: \(\left(a+b\right)^2-4=\left(a+b\right)^2-2^2=\left(a+b+2\right)\left(a+b-2\right)\)
d: \(\left(a+3b\right)^2-9b^2\)
\(=\left(a+3b\right)^2-\left(3b\right)^2\)
=(a+3b-3b)(a+3b+3b)=a(a+6b)
e: \(\left(x-3\right)^3-27\)
\(=\left(x-3-3\right)\left\lbrack\left(x-3\right)^2+3\left(x-3\right)+9\right\rbrack\)
\(=\left(x-6\right)\left(x^2-6x+9+3x-9+9\right)=\left(x-6\right)\left(x^2-3x+9\right)\)
f: \(\left(x+1\right)^3-125\)
\(=\left(x+1-5\right)\left\lbrack\left(x+1\right)^2+5\left(x+1\right)+25\right\rbrack\)
\(=\left(x-4\right)\left(x^2+2x+1+5x+5+25\right)=\left(x-4\right)\left(x^2+7x+31\right)\)
8 tháng 9 2020
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 )
HH
30 tháng 12 2016
\(a,3a+3b-a^2-ab\)
\(=\left(3a-a^2\right)+\left(3b-ab\right)\)
\(=a\left(3-a\right)+b\left(3-a\right)\)
\(=\left(a+b\right)\left(3-a\right)\)
\(b,8y^2-8yz-13y+13z\)
\(=\left(8y^2-8yz\right)-\left(13y-13z\right)\)
\(=8y\left(y-z\right)-13\left(y-z\right)\)
\(=\left(y-z\right)\left(8y-13\right)\)
\(c,3b^2+3c^2-ab^2-ac^2+2a-6\)
\(=\left(3b^2-ab^2\right)+\left(3c^2-ac^2\right)+\left(2a-6\right)\)
\(=b^2\left(3-a\right)+c^2\left(3-a\right)-2\left(3-a\right)\)
\(=\left(3-a\right)\left(b^2+c^2-2\right)\)
A(B+C)+3(B+C)=(B+C)(A+3)
\(a\left(b+c\right)+3b+3c\)
\(\Rightarrow a\left(b+c\right)+3\left(b+c\right)\)
\(\Rightarrow\left(a+3\right)\left(b+c\right)\)
Ta có : a(b+c)+3b+3c=ab+ac+3b+3c=(ab+3b)+(ac+3c)=b(a+3)+c(a+3)=(a+3)(a+c)
\(a.\left(b+c\right)+3b+3c\)
\(=a.\left(b+c\right)+3.\left(b+c\right)\)
\(=\left(b+c\right).\left(a+3\right)\)