18. iron-doing
19. wakes
còn lời giải chi tiết thì... mk ko biết .-.

18. help somebody do sth: giúp ai làm gì

hate doing sth: ghét làm gì

19. wakes - Đặt điều kiện trong tương lai thì mệnh đề điều kiện dùng thì hiện tại: KHi Harry thức giấc, chúng tôi sẽ nói cho anh ấy biết tin tức tốt lành này.

\(A=\dfrac{2}{1x3}+\dfrac{2}{3x5}+\dfrac{2}{5x7}+...+\dfrac{2}{21x23}\)

\(A=2x\left(\dfrac{1}{1x3}+\dfrac{1}{3x5}+\dfrac{1}{5x7}+...+\dfrac{1}{21x23}\right)\)

\(A=2x\dfrac{1}{2}x\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{21}-\dfrac{1}{23}\right)\)

\(A=1-\dfrac{1}{23}\)

\(A=\dfrac{22}{23}\)

\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(B=\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+\dfrac{1}{5x6}+\dfrac{1}{6x7}+\dfrac{1}{7x8}+\dfrac{1}{8x9}+\dfrac{1}{9x10}\)

\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(B=\dfrac{1}{2}-\dfrac{1}{10}\)

\(B=\dfrac{5}{10}-\dfrac{1}{10}\)

\(B=\dfrac{4}{10}\)

\(B=\dfrac{2}{5}\)

\(C=\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+\dfrac{1}{208}\)

\(C=\dfrac{1}{4x7}+\dfrac{1}{7x10}+\dfrac{1}{10x13}+\dfrac{1}{13x16}\)

\(C=\dfrac{1}{3}x\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\left(\dfrac{4}{16}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\dfrac{3}{16}\)

\(C=\dfrac{1}{16}\)

Câu 1

A ∪ B = 1; 3; 5; 7; 8; 9}

Câu 2:

A \ B = {0; 1}

Câu 3

Ta có:

A \ B = {0; 1}

B \ A = {5; 6}

⇒ X = (A \ B) ∩ (B \ A) = ∅

c) Ta có: \(P=2x+\dfrac{1}{x+1}\)

\(\Leftrightarrow\dfrac{-x}{x+1}=2x+\dfrac{1}{x+1}\)

\(\Leftrightarrow\dfrac{-x}{x+1}=\dfrac{2x\left(x+1\right)+1}{x+1}\)

Suy ra: \(2x^2+2x+1=-x\)

\(\Leftrightarrow2x^2+3x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: Để \(P=2x+\dfrac{1}{x+1}\) thì \(x=-\dfrac{1}{2}\)

`a)F=((x+1)/(1-x)-(1-x)/(x+1)-(4x^2)/(x^2-1)):(4x^2-4)/(x^2-2x+1)`

`đk:x ne +-1`

`F=((-(x+1)^2+(x-1)^2-4x^2)/(x^2-1)):(4(x-1)(x+1))/(x-1)^2`

`=(-x^2-2x-1+x^2-2x+1-4x^2)/(x^2-1):(4(x+1))/(x-1)`

`=(-4x^2-4x)/((x-1)(x+1)).(x-1)/(4(x+1))`

`=(-4(x-1))/((x-1)(x+1)).(x-1)/(4(x+1))`

`=-4/(x+1).(x-1)/(4(x+1)`

`=(1-x)/(x+1)^2`

`F<-1`

`<=>(1-x-(x+1)^2)/(x+1)^2<0`

Vì `(x+1)^2>0`

`=>1-x-(x+1)^2<0`

`<=>(x+1)^2+x-1>0`

`<=>x^2+2x+1+x-1>0`

`<=>x^2+3x>0`

`<=>x(x+3)>0`

`<=>` $\left[ \begin{array}{l}x>0\\x<-3\end{array} \right.$

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

a) Ta có: \(F=\left(\dfrac{x+1}{1-x}-\dfrac{1-x}{x+1}-\dfrac{4x^2}{x^2-1}\right):\dfrac{4x^2-4}{x^2-2x+1}\)

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

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

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

\(=\dfrac{-4x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{4\left(x+1\right)}\)

\(=\dfrac{-4x}{x-1}\cdot\dfrac{x-1}{4\left(x+1\right)}\)

\(=\dfrac{-4x}{4\left(x+1\right)}\)

\(=\dfrac{-x}{x+1}\)

b) Để F<-1 thì F+1<0

\(\Leftrightarrow\dfrac{-x}{x+1}+1< 0\)

\(\Leftrightarrow\dfrac{-x+x+1}{x+1}< 0\)

\(\Leftrightarrow\dfrac{1}{x+1}< 0\)

mà 1>0

nên x+1<0

hay x<-1

Kết hợp ĐKXĐ, ta được: x<-1

a: \(A=1+3+3^2+...+3^{11}\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\)

\(=\left(1+3\right)+3^2\left(1+3\right)+...+3^{10}\left(1+3\right)\)

\(=4\left(1+3^2+...+3^{10}\right)⋮4\)

b: \(B=16^5+2^{15}\)

\(=\left(2^4\right)^5+2^{15}\)

\(=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33⋮33\)

c: \(C=5+5^2+5^3+...+5^8\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^7+5^8\right)\)

\(=30+5^2\left(5+5^2\right)+...+5^6\left(5+5^2\right)\)

\(=30\left(1+5^2+...+5^6\right)⋮30\)

d: \(45⋮9;99⋮9;180⋮9\)

=>\(45+99+180⋮9\)

=>D chia hết cho 9

Bài 1:

câu a: 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)

        = \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)

        = \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)

         = \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)

          = \(\dfrac{29}{6}\)

b, (15,25 + 3,75) \(\times\) 4 + ( 20,71 + 5,29)\(\times\) 5

     = 19 \(\times\) 4 + 26 \(\times\) 5

     = 76 + 130

     = 206

c, \(\dfrac{4}{5}\) \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{4}{5}\) \(\times\) \(\dfrac{1}{3}\) - \(\dfrac{4}{5}\) \(\times\) \(\dfrac{1}{4}\)

= \(\dfrac{2}{5}\)    + \(\dfrac{4}{15}\) - \(\dfrac{1}{5}\)

= \(\dfrac{6}{15}\) + \(\dfrac{4}{15}\) - \(\dfrac{3}{15}\)

= \(\dfrac{7}{15}\) 

d, 1\(\dfrac{5}{7}\) + 7\(\dfrac{3}{6}\) + 2\(\dfrac{2}{7}\) - 4\(\dfrac{3}{6}\)

= (1 + 2 + \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\)) + ( 7 + \(\dfrac{3}{6}\) - 4 - \(\dfrac{3}{6}\))

=  3 + 1 + 3 

= 7

e, 8,4 \(\times\) \(x\) + 1,6 \(\times\) \(x\) = 10

   (8,4 + 1,6) \(\times\) \(x\)      = 10

     10 \(\times\) \(x\)                = 10

               \(x\)                = 1