1.2 với \(x\ge0,x\in Z\)

A=\(\dfrac{2\sqrt{x}+7}{\sqrt{x}+2}=2+\dfrac{3}{\sqrt{x}+2}\in Z< =>\sqrt{x}+2\inƯ\left(3\right)=\left(\pm1;\pm3\right)\)

*\(\sqrt{x}+2=1=>\sqrt{x}=-1\)(vô lí)

*\(\sqrt{x}+2=-1=>\sqrt{x}=-3\)(vô lí
*\(\sqrt{x}+2=3=>x=1\)(TM)

*\(\sqrt{x}+2=-3=\sqrt{x}=-5\)(vô lí)

vậy x=1 thì A\(\in Z\)

Ta có:

\(\dfrac{\sqrt{x}+5}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}+2\right)+3}{\sqrt{x}+2}=1+\dfrac{3}{\sqrt{x}+2}\)

Để \(A\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\in Z\)

\(\Rightarrow\left(\sqrt{x}+2\right)\inƯ_{\left(3\right)}=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow\sqrt{x}=\left\{-1;-3;1;-5\right\}\)

Mà \(\sqrt{x}\ge0\)

Nên \(\sqrt{x}=1\Rightarrow x=1\)

Vậy x=1 thì \(A\in Z\)

A = \(\dfrac{4\sqrt{x}+9}{2\sqrt{x}+1}\)

Mà \(4\sqrt{x}+9>0\)

\(2\sqrt{x}+1>0\)

=> A > 0

A = \(\dfrac{2\left(2\sqrt{x}+1\right)+7}{2\sqrt{x}+1}\) = \(2+\dfrac{7}{2\sqrt{x}+1}\)

Mà \(2\sqrt{x}+1\ge1< =>\dfrac{7}{2\sqrt{x}+1}\le7\)

<=> \(A\le9\)

<=> 0 < A \(\le9\)

Mà A thuộc Z

<=> A \(\in\){1;2;3;4;5;6;7;8;9}

Đến đây bn thay A vào để tìm x nhé

A = \(\dfrac{2\left(2\sqrt{x}+1\right)+7}{2\sqrt{x}+1}=2+\dfrac{7}{2\sqrt{x}+1}\)

Mà \(2\sqrt{x}+1>0< =>\dfrac{7}{2\sqrt{x}+1}>0\)

<=> A > 2

Có \(2\sqrt{x}+1\ge1< =>\dfrac{7}{2\sqrt{x}+1}\le7\)

<=> \(A\le9\)

<=> 2 < A \(\le9\)

Mà A thuộc Z

<=> \(A\in\left\{3;4;5;6;7;8;9\right\}\)

Đến đây bn thay A vào để tìm x nhé

A = \(\dfrac{6\sqrt{x}+8}{3\sqrt{x}+2}=2+\dfrac{4}{3\sqrt{x}+2}\)

Có \(3\sqrt{x}+2>0< =>\dfrac{4}{3\sqrt{x}+2}>0\) <=> A > 2

Có: \(3\sqrt{x}+2\ge2< =>\dfrac{4}{3\sqrt{x}+2}\le2\) <=> A \(\le4\)

<=> 2 < A \(\le4\)

Mà A nguyên

<=> \(\left[{}\begin{matrix}A=3\\A=4\end{matrix}\right.\)

TH1: A = 3

<=> \(\dfrac{4}{3\sqrt{x}+2}=1\)

<=> \(3\sqrt{x}+2=4< =>x=\dfrac{4}{9}\)

TH2: A = 4

<=> \(\dfrac{4}{3\sqrt{x}+2}=2< =>3\sqrt{x}+2=2< =>x=0\)

`A=(2sqrtx+17)/(sqrtx+5)`

`=(2sqrtx+10+7)/(sqrtx+5)`

`=(2(sqrtx+5)+7)/(sqrtx+5)`

`=2+7/(sqrtx+5)`

`A in ZZ`

`=>7/(sqrtx+5) in ZZ`

`=>sqrtx+5 in Ư(7)={+-1,+-7}`

Mà `sqrtx+5>=5`

`=>sqrtx+5=7`

`=>sqrtx=2`

`=>x=4`

Vậy `x=4` thì `A in ZZ`

Hì nhìn lộn đề bài =="

`A=(2\sqrtx+17)/(sqrtx+5)`

`A=(2sqrtx+10+7)/(sqrtx+5)`

`=(2(sqrtx+5)+7)/(sqrtx+5)`

`=2+7/(sqrtx+5)>2`

`A=2+7/(sqrtx+5)<=2+7/5=17/5`

`=>2<A<=17/5`

Mà `A in ZZ`

`=>A=3`

`=>2sqrtx+17=3sqrtx+15`

`=>sqrtx=2`

`=>x=4`

\(3.A=\dfrac{2\sqrt{x}+17}{\sqrt{x}+5}=\dfrac{2\left(\sqrt{x}+5\right)+7}{\sqrt{x}+5}\)\(=2+\dfrac{7}{\sqrt{x}+5}\)

\(\sqrt{x}+5\ge5=>2+\dfrac{7}{\sqrt{x}+5}\le2+\dfrac{7}{5}=3,4\)

dấu'=' xảy ra<=>x=0=>MaxA=3,4

3.\(A=\dfrac{2\sqrt{x}+17}{\sqrt{x}+5}=2+\dfrac{7}{\sqrt{x}+5}\)

\(A\in Z< =>\sqrt{x}+5\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(=>x\in\left\{4\right\}\)

1 interested in travelling by plane

2 15 minutes riding his bike to school

3 is made by my mother everymorning

4 take care of her little brother

5 able to speak 2 languages when he was young

6 able to play the piano

7 is played all over VN

8 to play voleyball very well when he was young

9 to go fishing when he was young

10 way to the shopping mall

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camping

was

couldn't

percussion

takes

look after

go around 

go through

wind

how long

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do