Bài 13:

ĐKXĐ: x∉{0;2;-2;1/2}

a: \(B=\left(\frac{x+2}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\frac{2x^2-x}{x^2-2x}\)

\(=\left(\frac{-\left(x+2\right)}{x-2}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right):\frac{x\left(2x-1\right)}{x\left(x-2\right)}\)

\(=\frac{-\left(x+2\right)^2-4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{2x-1}\)

\(=\frac{-x^2-4x-4-4x^2+x^2-4x+4}{x+2}\cdot\frac{1}{2x-1}=\frac{-4x^2-8x}{\left(x+2\right)\left(2x-1\right)}\)

\(=\frac{-4x\left(x+2\right)}{\left.\left(x+2\right)\left(2x-1\right)\right.}=\frac{-4x}{2x-1}\)
b: |x|=3

=>x=3 hoặc x=-3

Khi x=3 thì \(B=\frac{-4\cdot3}{2\cdot3-1}=\frac{-12}{5}\)

Khi x=-3 thì \(B=\frac{-4\cdot\left(-3\right)}{2\cdot\left(-3\right)-1}=\frac{12}{-6-1}=\frac{-12}{7}\)

c: Để B nguyên thì -4x⋮2x-1

=>-4x+2-2⋮2x-1

=>-2⋮2x-1

mà 2x-1 lẻ

nên 2x-1∈{1;-1}

=>2x∈{2;0}

=>x∈{1;0}

Kết hợp ĐKXĐ, ta được: x=1

Bài 12:

a: ĐKXĐ: a∉{1;-1;-2}

b: \(P=\left(\frac{a+1}{2a-2}+\frac{1}{2-2a^2}\right)\cdot\frac{2a+2}{a+2}\)

\(=\left(\frac{a+1}{2\left(a-1\right)}-\frac{1}{2\left(a-1\right)\left(a+1\right)}\right)\cdot\frac{2\left(a+1\right)}{a+2}\)

\(=\frac{\left(a+1\right)^2-1}{2\left(a-1\right)\left(a+1\right)}\cdot\frac{2\left(a+1\right)}{a+2}=\frac{a\left(a+2\right)}{\left(a-1\right)\left(a+2\right)}=\frac{a}{a-1}\)

c: |a|=2

=>a=2(nhận) hoặc a=-2(loại)

Khi a=2 thì \(P=\frac{2}{2-1}=\frac21=2\)

Bài 11:

a: ĐKXĐ: x∉{2;-3}

b: \(P=\frac{x+2}{x+3}-\frac{5}{x^2+3x-2x-6}+\frac{1}{2-x}\)

\(=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)

\(=\frac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)

c: \(P=\frac{-3}{4}\)

=>\(\frac{x-4}{x-2}=\frac{-3}{4}\)

=>4(x-4)=-3(x-2)

=>4x-16=-3x+6

=>7x=22

=>\(x=\frac{22}{7}\) (nhận)

d: Để P nguyên thì x-4⋮x-2

=>x-2-2⋮x-2

=>-2⋮x-2

=>x-2∈{1;-1;2;-2}

=>x∈{3;1;4;0}

e: \(x^2-9=0\)

=>\(x^2=9\)

=>x=3(nhận) hoặc x=-3(loại)

Khi x=3 thì \(P=\frac{3-4}{3-2}=-1\)

Gọi chiều rộng là x

Chiều dài là x+15

Theo đề, ta có phương trình:

\(\left(x+5\right)\left(x+12\right)=x\left(x+15\right)+80\)

\(\Leftrightarrow x^2+17x+60-x^2-15x=80\)

=>2x+60=80

=>x=10

Vậy: Chiều rộng là 10m

Chiều dài là 25m

Gọi độ dài quãng đường là x

Theo đề, ta có: 

\(\dfrac{x}{42}-\dfrac{x}{46}=\dfrac{3}{4}\)

hay x=362,25(km)

Gọi quãng đường AB là x (x>0)

Vận tốc xe máy đi từ A đến B là \(\dfrac{x}{30}\)

Vận tốc xe máy lúc về là \(\dfrac{x}{35}\)

Theo đề bài, ta có:

\(\dfrac{x}{30}-\dfrac{x}{35}=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{7x-6x}{210}=\dfrac{70}{210}\)

\(\Leftrightarrow x=70\left(km\right)\left(tm\right)\)

Vậy quãng đường AB là 70km

a: \(P=\dfrac{8}{x\left(x+4\right)}+\dfrac{5x}{x\left(x+4\right)}-\dfrac{2x+8}{x\left(x+4\right)}=\dfrac{8+5x-2x-8}{x\left(x+4\right)}=\dfrac{3}{x+4}\)

b: Thay x=1/2 vào P, ta được:

P=3:9/2=3x2/9=6/9=2/3

Với khác 0 ; x khác 4 

\(P=\dfrac{8+5x-2x-8}{x\left(x+4\right)}=\dfrac{3x}{x\left(x+4\right)}=\dfrac{3}{x+4}\)

Thay x = 1/2 vào P ta được \(\dfrac{3}{\dfrac{1}{2}+4}=\dfrac{3}{\dfrac{9}{2}}=3:\dfrac{9}{2}=\dfrac{2}{3}\)

1 were you doing

2 were having - rang

3 takes - is celebrated 

4 was formed

5 have lost - haven't found

6 is held to worship

7 skating

8 getting up

9 reading - doing

10 has been built

11 swimming - feel

12 were watching - failed

13 has worked - graduated

14 have been invited

15 will be discussing

16 decided not to stay

17 to pass - testing

18 not to phone

19 doing

20 to stay - do

Bài 6

\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow\dfrac{1}{x+5}=-3\Leftrightarrow-3\left(x+5\right)=1\Leftrightarrow x=-\dfrac{16}{3}\\ \Leftrightarrow Q=\left(3x-7\right)^2=\left[3\cdot\left(-\dfrac{16}{3}\right)-7\right]^2=529\)

Bài 7:

\(a,ĐK:x\ne\pm3\\ b,P=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\\ b,P=4\Leftrightarrow4\left(x-3\right)=4\Leftrightarrow x=4\)

\(d,=\dfrac{3y}{5x\left(x-y\right)}\\ e,=\dfrac{5x\left(x+2\right)\left(2-x\right)}{4\left(x-2\right)\left(x+2\right)}=\dfrac{-5x}{4}\\ f,=\dfrac{3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(6-x\right)}=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\\ g,=\dfrac{3xy\left(x-3y\right)\left(x+3y\right)}{2x^2y^2\left(x-3y\right)}=\dfrac{3\left(x+3y\right)}{2xy}\\ h,=\dfrac{45x^2y\left(x-y\right)\left(x+y\right)}{10xy\left(y-x\right)}=\dfrac{-9x\left(x+y\right)}{2}\\ i,=\dfrac{12\left(a-b\right)\left(a+b\right)\left(a^2+ab+b^2\right)}{3\left(a+b\right)\left(a-b\right)^2}=\dfrac{4\left(a^2+ab+b^2\right)}{a-b}\)

e: \(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=-\dfrac{5}{2}\)