Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 )
Từ \(4a^2+b^2=5ab\), ta có: \(4a^2-4ab-ab+b^2\)=0
Hay: (a-b) (4a-b)=0
Vì: 2a>b>0 nên 4a-b \(\ne\)0 .
Từ: (.) \(\Rightarrow\)
Từ: a-b=0 . Tức là: a=b
Thay a=b vào C ta được :
C= \(\frac{ab}{4a^2-b^2}=\frac{a^2}{4a^2-a^2}=\frac{1}{3}\)(do a\(\ne\)0)
a)\(2a^2-3ab+b^2\)
=\(a^2+a^2-2ab-ab+b^2\)
=\(\left(a-b\right)^2+a\left(a-b\right)\)
=\(\left(a-b\right)\left(2a-b\right)\)
b)\(x^2-7x-30\)
=\(x^2-10x+3x-30\)
=\(x\left(x-10\right)+3\left(x-10\right)\)
=\(\left(x-10\right)\left(x+3\right)\)
c)\(6a^2-5ab-6b^2\)
=\(6a^2-9ab+4ab-6b^2\)
=\(3a\left(2a-3b\right)+2b\left(2a-3b\right)\)
=\(\left(2a-3b\right)\left(3a+2b\right)\)
d)\(a^4+a^2+1\)
=\(a^4+2a^2-a^2+1\)
=\(\left(a^2+1\right)^2-a^2\)
=\(\left(a^2+1-a\right)\left(a^2+1+a\right)\)
e)\(x^3+6x^2+11x+6\)
=\(x\left(x^2+6x+9+2\right)+6\)
\(=x\left(\left(x+3\right)^2+2\right)+6\)
=\(x\left(x+3\right)^2+2x+6\)
=\(x\left(x+3\right)^2+2\left(x+3\right)\)
=\(\left(x+3\right)\left(x^2+3x+2\right)\)
a. \(2a^2+5ab-3b^2-7b-2\)
\(=\left(2a^2+6ab+2a\right)-\left(ab+3b^2+b\right)-\left(2a+6b+2\right)\)
\(=2a\left(a+3b+1\right)-b\left(a+3b+1\right)-2\left(a+3b+1\right)\)
\(=\left(2a-b-2\right)\left(a+3b+1\right)\)
b. \(2x^2-7xy+x+3y^2-3y\)
\(=\left(2x^2-xy\right)-\left(6xy-3y^2\right)+\left(x-3y\right)\)
\(=x\left(2x-y\right)-3y\left(2x-y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y+1\right)\)
c. \(6x^2-xy-2y^2+3x-2y\)
\(=\left(6x^2+3xy\right)-\left(4xy-2y^2\right)+\left(3x-2y\right)\)
\(=3x\left(2x+y\right)-2y\left(2x+y\right)+\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(2x+y\right)+\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(2x+y+1\right)\)
4a2=4b2-4a+1
=(2a)2-2*2a*1+12-4b2= (2a-1)2-(2b)2(2a-1-2b)(2a-1+2b)
a) = (xyz+xy) +(z+1) +(yz+zx)+(x+y)
= xy(z+1) +(z+1)+z(x+y)+(x+y)
= (z+1)(xy+1)+(x+y)(Z+1)
=(z+1)(xy+1+x+y)
\(3y^2\left(a-3x\right)-a\left(a-3x\right)=\left(3y^2-a\right)\left(a-3x\right)\)
sai rồi
thì mik nói là mik lm sai òi mà