?...?

\(3/4 + 2/1 = 3/4 + 8 / 4 = 11/4\)

\(2/3 - 3/8 = 16/24 - 9 / 24 = 7 /24\)

2/9 . 3/8 = 6 /72

18/1 . 3 / 2 = 54 / 2

a)\(=\dfrac{3}{4}+\dfrac{8}{4}=\dfrac{11}{4}\)

b)\(=\dfrac{16}{24}-\dfrac{9}{24}=\dfrac{7}{24}\)

c)\(=\dfrac{2\times3}{9\times8}=\dfrac{6}{72}=\dfrac{1}{12}\)

d)\(=18\times\dfrac{3}{2}=\dfrac{54}{2}=27\)

11/4

7/24

 6 /72

 54 / 2

đề bài thiếu r, sửa lại đi bn ơi

a)  

a)  

Lời giải:
a.

$\frac{x}{-4}=\frac{3}{y}=\frac{6}{-2}=-3$

$\Rightarrow x=-4(-3)=12; y=3:(-3)=-1$

b.

$\frac{-1}{8}< \frac{x}{40}\leq \frac{1}{10}$

$\Rightarrow \frac{-5}{40}< \frac{x}{40}\leq \frac{4}{30}$

$\Rightarrow -5< x\leq 4$

Vì $x$ nguyên nên $x\in\left\{-4; -3; -2; -1; 0;1;2;3;4\right\}$

a/ 7x(2x+x) - 7x(x+3) =14

7x(2x+x-x-3)=14

x(2x-3)=2

=> x=2

Trả lời :

          Bạn kia trả lời đúng rồi !

Hok tốt nha !

Đk x>=0   

A=\(\frac{2\sqrt{x}}{\sqrt{x}+3}\)=\(\frac{2\sqrt{x}+6-6}{\sqrt{x}+3}\)=\(\frac{2\left(\sqrt{x}+3\right)-6}{\sqrt{x}+3}\)=\(2-\frac{6}{\sqrt{x}+3}\)

Để A nguyên thì \(\frac{6}{\sqrt{x}+3}\)nguyên 

=> 6\(⋮\)\(\sqrt{x}+3\)=>\(\sqrt{x}+3\in\left\{1;2;3;6\right\}\)=>\(\sqrt{x}\in\left\{0;3\right\}\)vì \(\sqrt{x}\ge0\)

vậy x\(\in\left\{0;9\right\}\)

\(ĐK:x\ge0\)

\(A=\frac{2\sqrt{x}}{\sqrt{x}+3}=\frac{2\sqrt{x}+6-6}{\sqrt{x}+3}=\frac{2\left(\sqrt{x}+3\right)-6}{\sqrt{x}+3}=2-\frac{6}{\sqrt{x}+3}\)

Để A nguyên thì \(\frac{6}{\sqrt{x}+3}\inℤ\Leftrightarrow\sqrt{x}+3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

lập bảng xét nốt nhé:)

Bài 2: 

\(\Leftrightarrow n+1\in\left\{1;2;4\right\}\)

hay \(n\in\left\{0;1;3\right\}\)

1. John has never had such an enjoyable holiday before.

2. This is the most enjoyable holiday John has ever had.

3. John has never had a more enjoyable holiday than this before.

John has never had such an enjoyable holiday before

It is the most enjoyable holiday john has ever had

John has never had a more enjoyable holiday than this before

app hay