Question

Mik cần gấp bài 9,12 ah

Answer 1

Bài 8:

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

Answer 2

Bài 9:

\(A=\frac{a}{ab+a+1}+\frac{ba}{abc+ab+a}+\frac{c.ab}{ac.ab+c.ab+ab}\\ =\frac{a}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{1}{a+1+ab}\\ =\frac{ab+a+1}{ab+a+1}=1\)

Answer 3

Bài 10:

\((x-1)^{20}\geq 0, \forall x; (y+2)^{20}\geq 0, \forall y\)

Do đó để tổng $(x-1)^{20}+(y+2)^{20}=0$ thì $(x-1)^{20}=(y+2)^{20}=0$

$\Rightarrow x=1; y=-2$

Khi đó:

$A=2.1^5-5.(-2)^3+4=46$

 

Answer 4

Bài 11:
$|x-1|=2$

$\Rightarrow x-1=2$ hoặc $x-1=-2$

$\Rightarrow x=3$ hoặc $x=-1$

Nếu $x=3$ thì: 
$A=5.3^2+6.3-2=61$

Nếu $x=-1$ thì:

$A=5(-1)^2+6(-1)-2=-3$

 

Answer 5

Bài 12:

$(x-\frac{1}{2})^{x+1}-(x-\frac{1}{2})^{x+11}=0$

$\Rightarrow (x-\frac{1}{2})^{x+1}[1-(x-\frac{1}{2})^{10}]=0$

$\Rightarrow (x-\frac{1}{2})^{x+1}=0$ hoặc $1-(x-\frac{1}{2})^{10}=0$

Nếu $(x-\frac{1}{2})^{x+1}=0$

$\Rightarrow x-\frac{1}{2}=0$

$\Rightarrow x=\frac{1}{2}$
Nếu $1-(x-\frac{1}{2})^{10}=0$

$\Rightarrow (x-\frac{1}{2})^{10}=1=1^{10}=(-1)^{10}$

$\Rightarrow x-\frac{1}{2}=1$ hoặc $x-\frac{1}{2}=-1$
$\Rightarrow x=\frac{3}{2}$ hoặc $x=\frac{-1}{2}$