a) 3a + 4b - 5c - 2a - 3b + 5c

= ( 3a - 2a ) + ( 4b - 3b ) - ( 5c - 5c )

= a + b

b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

= ( 7a - 3a - 4a ) + ( 3b + 2b + b ) - ( 4c + 2c + 2c ) 

= 6b - 8c

a) 3a + 4b - 5c - 2a - 3b + 5c

= (3a - 2a) + (4b - 3b) - (5c - 5c)

= a + b - 0 = a + b

b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

= (7a - 3a - 4a) + (3b + 2b + b) - ( 4c + 2c + 2c)

= 0 + 6b - 8c = 6b - 8c

4 a - 9 a - 1 2 a 1 2 - 3 a - 1 2 + a - 4 + 3 a - 1 a 1 2 - a - 1 2

Chọn B

1.

Cho $a+b+c=0$.

Ta có:

$a+b+c=0\Rightarrow a=-(b+c)$

$\Rightarrow a^2=(b+c)^2=b^2+c^2+2bc$

$\Rightarrow a^2-b^2-c^2=2bc$

Tương tự:

$b^2-c^2-a^2=2ca$

$c^2-a^2-b^2=2ab$

Do đó: $A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}$

$=\dfrac{a^3+b^3+c^3}{2abc}$

Mà $a+b+c=0$ nên:

$a^3+b^3+c^3=3abc$

Suy ra $A=\dfrac{3abc}{2abc}$

$A=\dfrac32$

2.

Cho $abc=2$.

Ta có:

$A=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}$

Vì $abc=2$ nên:

$ab=\dfrac2c,\quad bc=\dfrac2a,\quad ac=\dfrac2b$

Suy ra:

$A=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{2bc}{2+2bc+2b}$

$=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{bc}{1+bc+b}$

Quy đồng và sử dụng $abc=2$:

$A=1$

a: \(A=\left(2x-1\right)\left(4x^2+2x+1\right)-7\left(x^3+1\right)\)

\(=\left(2x\right)^3-1^3-7x^3-7\)

\(=8x^3-1-7x^3-7=x^3-8\)

b: Thay x=-1/2 vào A, ta được:

\(A=\left(-\dfrac{1}{2}\right)^3-8=-\dfrac{1}{8}-8=-\dfrac{65}{8}\)

c: \(A=x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

Để A là số nguyên tố thì x-2=1

=>x=3

Đáp án B

P = a − 3 − 4 a − 1 a 1 2 − 4 a − 1 2 − 1 a − 1 2 = a 1 2 − 3 a − 1 2 − 4 a − 3 2 − a 1 2 + 4 a − 1 2 a − 1 2 ( a 1 2 − 4 a − 1 2 )       = a − 1 2 − 4 a − 3 2 1 − 4 a − 1 = a − 1 2

A = 2a-3b+5c-5b+4a-2c-3c+4b+7a = 13a-4b

k mk nha

A=-(-2a+3b-5c)+(-5b+4a-2c)-(3c-4b-7a)

   =2a-3b+5c-5b+4a-2c-3c+4b+7a

   =13a-4b.

\(\frac{a}{ab+a+2}\)+ \(\frac{b}{bc+b+1}\)+ \(\frac{2c}{ac+2c+2}\)

= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{a\left(bc+b+1\right)}\)+ \(\frac{2abc}{ab\left(ac+2c+2\right)}\)

= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{abc+ab+a}\)+ \(\frac{2abc}{a^2bc+2abc+2ab}\)

= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{ab+a+2}\)+ \(\frac{2}{ab+a+2}\)   (vì  abc = 2  )

= \(\frac{ab+a+2}{ab+a+2}\)= 1

tại sao lại nhân vs a và ab z bn

M\(=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2bc}{b\left(ac+2c+2\right)}\)

M = \(\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{b+1+bc}+\dfrac{2bc}{abc+2bc+2b}\)

M=\(\dfrac{1}{b+1+bc}+\dfrac{b}{b+1+bc}+\dfrac{2bc}{2+2bc+2b}\)

M = \(\dfrac{1+b}{b+1+bc}+\dfrac{2bc}{2\left(1+bc+b\right)}\)

M = \(\dfrac{1+b}{b+1+bc}+\dfrac{bc}{b+1+bc}=\dfrac{1+b+bc}{b+1+bc}=1\)