Câu 3: 

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x+y}{3+2}=\dfrac{90}{5}=18\)

Do đó: x=54; y=36

Đề bài thấy hơi sao sao ấy

Dễ nhưng dài lắm :)) cô cho lm đầy 

1,\(\sqrt{\left(x-1\right)^2}=\left|x-1\right|=-\left(x-1\right)=1-x\)

2,\(\sqrt{\left(a-2b\right)^2}=\left|a-2b\right|=-\left(a-2b\right)=2b-a\)

3,\(\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)

Bài 5:

Gọi số sách 7A,7B,7C,7D lần lượt là \(a,b,c,d\in \mathbb{N^*},sách\)

Áp dụng tc dtsbn:

\(\dfrac{a}{37}=\dfrac{b}{37}=\dfrac{c}{40}=\dfrac{d}{36}=\dfrac{c-d}{40-36}=\dfrac{12}{4}=3\\ \Rightarrow\left\{{}\begin{matrix}a=111\\b=111\\c=120\\d=108\end{matrix}\right.\)

Vậy ...

Bài 6:

Gọi cd, cr theo thứ tự là \(a,b>0;m\)

\(\Rightarrow a:b=5:4\Rightarrow\dfrac{a}{5}=\dfrac{b}{4}\)

Đặt \(\dfrac{a}{5}=\dfrac{b}{4}=k\Rightarrow a=5k;b=4k\)

Mà \(ab=500\Rightarrow20k^2=500\Rightarrow k^2=25\Rightarrow k=5\left(k>0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a=5\cdot5=25\\b=5\cdot4=20\end{matrix}\right.\\ \Rightarrow\text{Chu vi là }2\left(a+b\right)=2\left(25+20\right)=90\left(m\right)\)

25. amusement.

26. relaxed.

27. diving.

28. operate

29. environmentally

30. largest.

- Bạn tham khảo đáp án nhé! Chúc bạn học tốt.

1. D => bỏ

2. C

3. C

4. A => the

5. A

6. C =>  next

7. C => taught

8. C => the most

9. D => from

10. B => would

II

1. I wish I would become a singer.

2.  She used to walk to school when she was a child.

3. If I were you, I would learn Chinese

4. The man who I met yesterday was my uncle

5. The trees are watered by me everyday

Bài 1: 

a: 4/9=4/9

5/3=15/9

b: 3/14=-18/84

-5/6=-70/84

c: -2/25=-14/175

-3/35=-15/175

d: -7/20=-35/100

-11/25=-44/100

e: -25/75=-1/3=-5/15

-12/30=-2/5=-6/15

-12 . x - 7 . x = -76

Mẫu: x = 1 => -19 

x         Kết quả

1              -19

2              -38

3              -57

4              -76

thì kết quả chính xác là x = 4

 -12x-7x=-76

    -19x  =-76

         x  =-76:-19

         x  =    4

\(a,\dfrac{11x}{2x-5}+\dfrac{x-30}{2x-5}=\dfrac{11x+x-30}{2x-5}=\dfrac{12x-30}{2x-5}=\dfrac{6\left(2x-5\right)}{2x-5}=6\)

\(b,\dfrac{3x^2-1}{2x}+\dfrac{x^2+1}{2x}=\dfrac{3x^2-1+x^2+1}{2x}=\dfrac{4x^2}{2x}=2x\)

\(c,\dfrac{3}{2x-5}+\dfrac{-2}{2x+5}+\dfrac{-20}{4x^2-25}=\dfrac{3\left(2x+5\right)}{\left(2x-5\right)\left(2x+5\right)}-\dfrac{2\left(2x-5\right)}{\left(2x-5\right)\left(2x+5\right)}-\dfrac{20}{\left(2x-5\right)\left(2x+5\right)}=\dfrac{6x+15-4x+10-20}{\left(2x-5\right)\left(2x+5\right)}=\dfrac{2x+5}{\left(2x-5\right)\left(2x+5\right)}=\dfrac{1}{2x-5}\)

\(d,\dfrac{x-2}{x-1}+\dfrac{x-3}{x+1}+\dfrac{4-2x^2}{x^2-1}=\dfrac{\left(x-2\right)\left(x+1\right)+\left(x-3\right)\left(x-1\right)+4-2x^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+x-2+x^2-3x-x+3+4-2x^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{-5x+5}{\left(x-1\right)\left(x+1\right)}=\dfrac{-5\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{-5}{x-1}\)

\(e,\dfrac{x+1}{x-1}+\dfrac{1-x}{x+1}+\dfrac{4}{x^2-1}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2x+1-x^2+2x-1+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\)