\(A=\frac{\sqrt{x}-2+3}{\sqrt{x}-2}=1+\frac{3}{\sqrt{x}-2}\)

A nguyên khi \(\frac{3}{\sqrt{x}-2}\) nguyên

<=> \(\sqrt{x}+2\) là ước của 3

\(\Leftrightarrow\sqrt{x}+2\in\left\{1;3;-1;-3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{-1;1;-3;-5\right\}\)

\(\Leftrightarrow x\in\left\{1;2;9;25\right\}\)

Vậy x=1 ; x= 9 ; x=25

My stomachache _____disappeared______ after I took the tablets.      APPEAR

It was _____unkindly______ of you not to help him when he was in trouble.       KIND

When my father has free time he often tells us____funny___ stories. FUN

Look! the children are eating_____hungrily_____.        HUNGRY​

It's a nice shop and the assistants are all polite and very __helpfull___. HELP

Catching the common cold is ____unpleasant___ for everybody. PLEASE

Do you know a good ____decorater__ to decorate my house?   DECORATE

TT

cô on mà dc có 1 cái

Số 809 chia 11 dư 6 chia 4 dư 1 va 19 dư 11

\(\Rightarrow\left|x+\frac{11}{2}\right|>\frac{11}{2}\)

\(\Rightarrow\left[\begin{array}{nghiempt}-\left(x+\frac{11}{2}\right)>\frac{11}{2}\\x+\frac{11}{2}>\frac{11}{2}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}-x-\frac{11}{2}>\frac{11}{2}\\x>0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x< 11\\x>0\end{array}\right.\)

Vaay x<11 ; x>0

\(\Rightarrow\left[\begin{array}{nghiempt}x-\frac{5}{3}< \frac{1}{3}\\-\left(x-\frac{5}{3}\right)< \frac{1}{3}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x< 2\\-x+\frac{5}{3}< \frac{1}{3}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x< 2\\-x< -\frac{4}{3}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x< 2\\x>\frac{4}{3}\end{array}\right.\)

=> \(\frac{4}{3}< x< 2\)

Nửa chu vi hình chữ nhật là :

60 : 2 = 30 ( cm )

Chiều dài là :

(30 - 8 ) : 2 = 12 ( cm )

chiều rộng là :

30-12=18 ( cm )

Diện tích hình chữ nhật là :

18 x 12 = 216 ( cm2)

Đ/S 216 cm2

\(\left|3\right|< \left|5\right|\)

\(\left|-1\right|>\left|0\right|\)

\(\left|-3\right|< \left|-5\right|\)

\(\left|2\right|=\left|-2\right|\)

\(\Rightarrow\left(x+1\right)\left(x-3\right)=\left(x+3\right)\left(2x-3\right)\)

=> \(x^2-3x+x-3=2x^2-3x+6x-9\)

=> \(-3+9=2x^2-3x+6x-x^2+3x-x\)

=> \(x^2+5x=6\)

=> \(x\left(x+5\right)=6\)

THử với các ước của 6 ta được x=1

Vậy x=1