8ax - 4ay - 2x + y

= (8ax - 4ay) - (2x - y)

= 4a(2x - y) - (2x - y)

= (2x - y)(4a - 1)

8ax-4ay-2x+y

=4a(2x-y)-(2x-y)

=(4a-1)(2x-y)

a) a^2+b^2+c^2=ab+ac+bc. cm a=b=c

Ta có: a^2 + b^2 + c^2 = ab + bc + ca 

<=> 2.a^2 + 2.b^2 + 2.c^2 = 2.ab + 2.bc + 2.ca 

<=> ( a^2 - 2ab + b^2 ) + ( b^2 - 2bc +c^2 ) + ( c^2 - 2ac + a^2 ) =0 

<=> (a-b)^2 + (b-c)^2 + (c -a)^2 =0 (1) 

Vì (a-b)^2 ; (b-c)^2 ; (c -a)^2 ≧ 0 với mọi a,b,c. 

=> (a-b)^2 + (b-c)^2 + (c -a)^2 ≧ 0 (2) 

Từ (1) và (2) khẳng định dấu "=" khi: 

a - b = 0; b - c = 0 ; c - a = 0 => a=b=c 

Vậy a=b=c  ( đpcm )

b)a+b+c=o cm a^3+b^3+c^3=3abc

thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có :

a^3+b^3+c^3-3abc=0

<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc=0

<=>[(a+b)^3 +c^3] -3ab.(a+b+c)=0

<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c)=0

<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab)...

<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)=0

luôn đúng do a+b+c=0

c)tìm x 8x^3-12x^2+6x=28

8x^3 - 12x^2 + 6x - 28 = 0

8x^2(x-2)+4x(x-2)+14(x-2)=0

<=>2(x-2)(4x^2+2x+7) = 0

Ta có 4x^2+2x+7=(2x)^2+2.2x1/2 +1/4 -1/4+28/4=(2x+1/2)^2+27/4 >0 V x

=>x-2=0 <=>x=2

Vậy PT có tập ngiệm là  S ={2}

\(5\left(x-2\right)+2>1-2\left(x+1\right)\)

\(\Leftrightarrow5x-10+2>1-2x-2\)

\(\Leftrightarrow7x>7\)

\(\Leftrightarrow x=1\)

\(5+3x\left(x+3\right)< \left(3x-1\right)\left(x+2\right)\)

\(\Leftrightarrow5+3x^2+9x< 3x^2-x+6x-2\)

\(\Leftrightarrow4x< -7\)

\(\Leftrightarrow x< \frac{-7}{4}\)

\(\frac{x+3}{3x-2}>1\)

\(x+3>3x-2\)
 

\(2x< 5\)

\(x< \frac{5}{2}\)

5(3-2x)+5(x+4)6-4x 
= 15-10x++30x+120-4x
= (15+120)+(-10x+30x-4x)
= 135+16x

5(3-2x)+5(x+4)6-4x 

= 15-10x++30x+120-4x

= (15+120)+(-10x+30x-4x)

= 135+16x

\(ax-b>bx+a\)

\(\Leftrightarrow ax-a=bx+b\)

\(\Leftrightarrow a\left(x-1\right)-b\left(x-1\right)>0\)

\(\left(x-1\right)\left(a-b\right)>0\)

\(a^3+b^3-c^3+3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)-c^3+3abc\)

\(=\left(a+b-c\right)^3+3c\left(a+b\right)\left(a+b-c\right)-3ab\left(a+b-c\right)\)

\(=\left(a+b-c\right)\left(a^2+b^2+c^2-ab+bc+ca\right)=0\)

\(\Leftrightarrow a+b-c=0\)

\(\Leftrightarrow a+b=c\)

\(P=\left(\frac{a+b}{c}\right)^{10}=1^{10}=1\)

Ta có

 \(x\left(1+2x\right)-2\left(2x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)

Trả lời:

\(x\left(1+2x\right)-2\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=2\end{cases}}}\)

Vậy x = - 1/2; x = 2 là nghiệm của pt.