rút gọn biểu thức: A=a(2b-c)-b(a+c)-a(c+b)
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a) (3a-2b+c)-(2a+b)-(c-a)
= 3a - 2b + c - 2a - b - c + a
= 3a - 2b - b + c - c - 2a + a
= 3a - (2b + b) + (c - c) - (2a - a)
= 3a - 3b + 0 - a
= 3a - a - 3b
= 2a - 3b
b) (a-b)-(b+c)-(c+a)
= a - b - b - c - c - a
= a - a - b - b - c - c
= ( a - a) - ( b + b) - ( c + c)
= 0 - 2b - 2c
b) (a-b)-(b+c)-(c+a)
= a - b - b - c -c - a
= (a - a ) - ( b + b ) - ( c + c )
= 0 - 2b - 2c
= -2b - 2c
= -(2b + 2c)
= -2(b+c)
Ta có:
$M=\dfrac{ab+bc+ca-a-b-c}{a^2b-a^2-b+1}$
Vì $abc=1$ nên:
$c=\dfrac1{ab}$
Suy ra: $M=\dfrac{ab+\dfrac1a+\dfrac1b-a-b-\dfrac1{ab}}{a^2b-a^2-b+1}$
$=\dfrac{(a-1)(b-1)(ab-1)}{ab(a^2-1)(b-1)}$
$=\dfrac{(a-1)(b-1)(ab-1)}{ab(a-1)(a+1)(b-1)}$
$=\dfrac{ab-1}{ab(a+1)}$
Mà $ab=\dfrac1c$ nên:
$M=\dfrac{\frac1c-1}{\frac1c(a+1)}$
$=\dfrac{1-c}{a+1}$
Vậy: $M=\dfrac{1-c}{a+1}$
(2a-3b)-(5b-d)+(8b-d)
\(=2a-3b-5b+d+8b-d\)
\(=2a+\left(8b-3b-5b\right)+\left(d-d\right)\)
=2a
\(\left(a+b\right)-\left(d-4b\right)-\left(c-10d\right)\)
\(=a+b-d+4b-c+10d\)
\(=a+5b+11d-c\)
a) (2x+3)2-2(2x+3)(2x+5)+(2x+5)2
=4x2+12x+9-(4x+6)(2x+5)+4x2+20x+25
=4x2+12x+9-(8x2+12x+20x+30)+4x2+20x+25
=4x2+12x+9-8x2-12x-20x-30+4x2+20x+25
=4
b) (x2+x+1)(x2-x+1)(x2-1)
=((x2+1)2-x2)(x2-1)
=(x4+x2+1)(x2-1)
=x6+x4+x2-x4-x2-1
=x6-1
c)(a+b-c)2+(a-b+c)2-2(b-c)2
=a2+b2+c2+2ab-2ac-2bc+a2+b2+c2-2ab+2ac-2bc-2(b2-2bc+c2)
=2a2+2b2+2c2-4bc-2b2+4bc-2c2
=2a2
d) (a+b+c)2+(a-b-c)2+(b-c-a)2+(c-a-b)2
= a2+b2+c2+2ab+2ac+2bc+a2+b2+c2-2ab-2ac+2bc+a2+b2+c2+2bc-2ab+2ac+a2+b2+c2-2ac-2bc+2ab
=4a2+4b2+4c2+4ab+4bc
a) Ta có: \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left(2x+3-2x-5\right)^2\)
=4
b) Ta có: \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6-1\)
E=(-a-b+c+d)-(d+c-b-2a)
E=-a-b+c+d-d-c+b+2a
E=-a+(-)b+c+d+(-d)+(-c)+b+2a
E=-a+(-b)+c+d+(-d)+(-c)+b+2a
E=(2a-a)+(-b+b)+(-d+d)+(-c+c)=a+0+0+0=a
A=a(2b-c-c-b)-b(a+c)
=a(b-2c)-b(a+c)
=ab+2ac-ab-bc
=2ac-bc