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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 )
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)
3a\(x\)2 + 3b\(x\)2 + a\(x\) + b\(x\) + 5a + 5b
= (3a\(x^2\) + 3b\(x^2\)) + (a\(x\) + b\(x\)) + (5a + 5b)
= 3\(x^2\)(a + b) + \(x\)(a +b) + 5(a + b)
= (a + b)( 3\(x^2\) + \(x\) + 5)
= (a + b)(3\(x^2\) + \(x\) + 5)2
Phân tích đa thcusw thành nhân tử (Phương pháp nhân hạng tử):
1. 3a - 3b + ax - bx
= 3(a - b) + x(a - b)
= (a - b)(3 + x)
2. x3 + 3x2 + 3x + 9
= x3 + 3x2 + 3x + 1 + 8
= (x + 1)3 + 8
= (x + 1 + 2)[(x + 1)2 - 2(x + 1) + 4]
= (x + 3)(x2 + 2x + 1 - 2x - 2 + 4)
= (x + 3)(x2 + 3)
3. 10ay - 5by + 2ax - bx
= 5y(2a - b) + x(2a - b)
= (2a - b)(5y + x)
4. 5a2 - 5ax - 7a + 7x
= 5a(a - x) - 7(a - x)
= (a - x)(5a - 7)
5. x2 - 3xy + x - 3x
= x(x - 3y + 1 - 3)
= x(x - 3y - 2)
6. 7x2 - 7xy - 4x + 4y
= 7x(x - y) - 4(x - y)
= (x - y)(7x - 4)
7. 3ax - 4by - 4ay + 3bx
= (3ax + 3bx) - (4ay + 4by)
= 3x(a + b) - 4y(a + b)
= (a + b)(3x - 4y)
8. 30ax - 34bx - 15a + 17b
= (30ax - 15a) - (34bx -17b )
= 15a(x - 1) - 17b(x - 1)
= (x - 1)(15a - 17b)
9. 3x2 - 3xy + 3y2 - 3xy
= 3(x2 - xy + y2 - xy)
= 3(x2 - 2xy + y2)
= 3(x - y)2
10. 12a2 - 6ab + 3b2 - 6ab
= 3(4a2 - 2ab + b2 - 2ab)
= 3(4a2 - 4ab + b2)
= 3(2a - b)2
\(1,3ax^2+3bx^2+ax+bx+5a+5b\)
\(=3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)=\left(a+b\right)\left(3x^2+x+5\right)\)
\(2,ax-bx-2cx-2a+2b+4c=x\left(a-b-2c\right)-2\left(a-b-2c\right)=\left(x-2\right)\left(a-b-2c\right)\)
\(2ax-bx+3cx-2a+b-3c\\ =x\left(2a-b+3c\right)-\left(2a-b+3c\right)\\ =\left(x-1\right)\left(2a-b+3c\right)\)
\(ax-bx-2cx-2a+2b+4c\\ =x\left(a-b-2c\right)-2\left(a-b-2c\right)\\ =\left(x-2\right)\left(a-b-2c\right)\)
\(3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\)
\(ax^2-bx^2-2ax+2bx-3a+3b\\ =x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a+b\right)\\ =\left(x^2-2x-3\right)\left(a+b\right)\\ =\left(x+1\right)\left(x-3\right)\left(a+b\right)\)
Lời giải:
31.
\(2a^2x-5by-6a^2y+2bx=(2a^2x+2bx)-(5by+5a^2y)\)
\(=2x(a^2+b)-5y(b+a^2)=(a^2+b)(2x-5y)\)
34.
\(4acx+4bcx+4ax+4bx=4x(ac+bc+a+b)\)
\(=4x[(ac+bc)+(a+b)]=4x[c(a+b)+(a+b)]=4x(c+1)(a+b)\)
37. Sửa đề:
\(2ax^2-bx^2-2ax+bx+4a-2b\)
\(=(2ax^2-bx^2)-(2ax-bx)+(4a-2b)\)
\(=x^2(2a-b)-x(2a-b)+2(2a-b)=(2a-b)(x^2-x+2)\)
Câu 31:
\(2a^2x-5by-5a^2y+2bx\)
\(=2x\left(a^2+b\right)-5y\left(a^2+b\right)\)
\(=\left(a^2+b\right)\left(2x-5y\right)\)
Câu 34:
\(4acx+4bcx+4ax+4bx\)
\(=4cx\left(a+b\right)+4x\left(a+b\right)\)
\(=\left(a+b\right)\left(4cx+4x\right)\)
\(=4x\left(a+b\right)\left(c+1\right)\)
Câu 37:
\(2ax^2-bx^2-2ax+bx+4a-2b\)
\(=x^2\left(2a-b\right)-x\left(2a-b\right)+2\left(2x-b\right)\)
\(=\left(2a-b\right)\left(x^2-x+2\right)\)
\(=\left(2a-b\right)\left(x^2-x+2\right)\)
Ta có:
\(VT=(5a-3b+8c).(5a-3b-8c)\)
\(=\left(5a-3b\right)^2-\left(8c\right)^2\)
Mà \(a^2-b^2=4c^2\) nên:
\(VT=25^2-30ab+9b^2-16.\left(a^2-b^2\right)\)
\(=9a^2-30ab+25b^2\)
\(=\left(3a-5b\right)^2=VP\)
\(\Rightarrow\) Đpcm.