1B

2:

2x-1=x+3

⇔2x-x=3+1

⇔x=4

→D

3A

Câu 2:

a, (x+3)(x-1/2)=0

⇒x+3=0 hoặc x-1/2=0

⇔x=-3 hoặc x=1/2

Vậy S={-3;1/2}

b, $\frac{x+1}{99}$+ $\frac{x+2}{98}=$ $\frac{x+3}{97}+$ $\frac{x+4}{96}$

$⇔(\frac{x+1}{99}+1)+($ $\frac{x+2}{98}+1)=($ $\frac{x+3}{97}+1)+($ $\frac{x+4}{96}+1)$

$⇔\frac{x+100}{99}+$ $\frac{x+100}{98}=$ $\frac{x+100}{97}+$ $\frac{x+100}{96}$

$⇔(x+100).($$\frac{1}{99}+$ $\frac{1}{98}-$ $\frac{1}{97}-$ $\frac{1}{96})=0$

$⇒x+100=0$ (vì: $\frac{1}{99}+$ $\frac{1}{98}-$ $\frac{1}{97}-$ $\frac{1}{96}$ $\neq0$)

$⇔x=-100$

Vậy S={-100} 

$(x+100)$($\frac{1}{99}$+ $\frac{1}{98}$- $\frac{1}{97}$-$\frac{1}{96}$)=0 Đáp án:

 

Giải thích các bước giải:

1B

2D

3A

Câu 2

A Đkxđ x thuộc R

pt ⇔ \(\left[ \begin{array}{l}x+3=0\\x-1/2=0\end{array} \right.\) 

    ⇔\(\left[ \begin{array}{l}x=-3\\x=1/2\end{array} \right.\) 

Vây...

B ĐKXD x thuộc R

pt ⇔$\frac{x+1}{99}$ +$1$+$\frac{x+2}{98}$ +$1$=$\frac{x+3}{97}$+$1$+ $\frac{x+4}{96}$ +$1$

    ⇔$\frac{x+100}{99}$ +$\frac{x+100}{98}$ =$\frac{x+100}{97}$+ $\frac{x+100}{96}$      

    ⇔(x+100)(1/99+1/98-1/97-1/96)=0

Mà 1/99+1/98-1/97-1/96<0

⇒x+100=0

⇔x=-100