ĐK : \(y\ne2x,a\ne-b\)

\(A=\frac{ac+bx+ax+bc}{ay+2bx+2ax+by}\)

\(=\frac{\left(ac+ax\right)+\left(bx+bc\right)}{\left(ay+by\right)+\left(2ax+2bx\right)}\)

\(=\frac{a\left(c+x\right)+b\left(c+x\right)}{y\left(a+b\right)+2x\left(a+b\right)}\)

\(=\frac{\left(c+x\right)\left(a+b\right)}{\left(a+b\right)\left(y+2x\right)}\)

\(=\frac{c+x}{y+2x}\) không phụ thuộc vào \(a,b\) ( đpcm )

\(2ax-bx+3cx-2a+b-3c\\ =x\left(2a-b+3c\right)-\left(2a-b+3c\right)\\ =\left(x-1\right)\left(2a-b+3c\right)\)

\(ax-bx-2cx-2a+2b+4c\\ =x\left(a-b-2c\right)-2\left(a-b-2c\right)\\ =\left(x-2\right)\left(a-b-2c\right)\)

\(3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\)

\(ax^2-bx^2-2ax+2bx-3a+3b\\ =x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a+b\right)\\ =\left(x^2-2x-3\right)\left(a+b\right)\\ =\left(x+1\right)\left(x-3\right)\left(a+b\right)\)

\(A=\frac{ac+bx+ax+bc}{ay+2bx+2ax+by}=\frac{a\left(c+x\right)+b\left(c+x\right)}{a\left(y+2x\right)+b\left(y+2x\right)}=\frac{\left(c+x\right)\left(a+b\right)}{\left(y+2x\right)\left(a+b\right)}\)

Do \(a\ne-b\Rightarrow a+b\ne0\Rightarrow\)\(A=\frac{c+x}{y+2x}\), giá trị không phụ thuộc vào a; b (đpcm)

a) A = ax - ay - bx + by 

A = a(x - y) - b(x - y)

A = (a - b)(x - y)

A = 1. (-2) = -2

B = 2ax + 2ay - 2bx - 2by

B = 2a(x + y) - 2b(x + y)

B = 2(x + y)(a - b)

B = 2. (-5).(-2) = 20

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 )

\(\text{a) x}^4+2x^2+1=\left(x^2+1\right)^2\)

\(\text{b) 3}ax^2+3bx^2+ãx+bx+5a+5b=\left(3ax^2+3bx^2\right) +\left(ax+bx\right)+\left(5a+5b\right)=3x^2+x\left(a+b\right)+5\left(a+b\right)=\left(a+b\right)\left(3x^2+x+5\right)\)

\(\text{c) a}x^2-bx^2-2ax+2bx-3a+3b=\left(\text{a}x^2-bx^2\right)-\left(2ax-2bx\right)-\left(3a-3b\right)=x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a-b\right)=\left(x^2-2x-3\right)\left(a-b\right)\)

a) \(x^4+2x^2+1\)

\(=x^2\left(x^2+1\right)+\left(x^2+1\right)\)

\(=\left(x^2+1\right)^2\)

b) \(3ax^2+3bx^2+ax+bx+5a+5b\)

\(=3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\)

\(=\left(a+b\right)\left(3x^2+x+5\right)\)

c) \(ax^2-bx^2-2ax+2bx-3a+3b\)

\(=x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a-b\right)\)

\(=\left(a-b\right)\left(x^2-2x-3\right)\)

\(=\left(a-b\right)\left(x-3\right)\left(x+1\right)\)

Câu d mình chịu

Gợi ý: ab + cx + ax + bc = (a + c)(x + b)

Và ay + 2cx + 2ax + cy = (a + c)(2x + y).

\(\Delta_1'=b^2-ac\) ; \(\Delta_2'=c^2-ab\) ; \(\Delta_3'=a^2-bc\)

\(\Rightarrow\Delta_1'+\Delta_2'+\Delta_3'=a^2+b^2+c^2-ab-bc-ca\)

\(=\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c-a\right)^2\ge0\) ; \(\forall a;b;c\)

\(\Rightarrow\) Tồn tại ít nhất 1 trong 3 giá trị \(\Delta_1';\Delta_2';\Delta_3'\) không âm

\(\Rightarrow\) Ít nhất 1 trong 3 pt nói trên có nghiệm