d: Xét ΔABK và ΔCKB có 

AB=CK

KB chung

AK=CB

Do đó: ΔABK=ΔCKB

d: Xét ΔABK và ΔCKB có 

AB=CK

KB chung

AK=CB

Do đó: ΔABK=ΔCKB

\(\Leftrightarrow\dfrac{90}{x}-\dfrac{90}{x+5}=\dfrac{1}{5}\)

\(\Leftrightarrow x\left(x+5\right)=450x+2250-450x\)

\(\Leftrightarrow x^2+5x-2250=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=45\\x=-50\end{matrix}\right.\)

ĐKXĐ: \(x\ne0;x\ne-5\)

\(\dfrac{180}{x}-\dfrac{180}{x+5}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{180.5.\left(x+5\right)}{5x\left(x+5\right)}-\dfrac{180.5.x}{5x\left(x+5\right)}=\dfrac{2.x.\left(x+5\right)}{5x\left(x+5\right)}\)

\(\Leftrightarrow900x+4500-900x=2x^2+10x\)

\(\Leftrightarrow2x^2+10x-4500=0\)

\(\Leftrightarrow x^2+5x-2250=0\)

\(\Leftrightarrow x^2+50x-45x-2250=0\)

\(\Leftrightarrow x\left(x+50\right)-45\left(x+50\right)=0\)

\(\Leftrightarrow\left(x+50\right)\left(x-45\right)=0\)

=>   x+50=0         hoặc                 x-45=0

         x=-50                                    x=45

Vậy....

1. Ha went to doctor's yesterday because he was sick

2. He ate some spinach last night

3. The doctor said that the vegetable could be dirty so he must wash carefully

4. The doctor gave him some medicine

5. He felt better after take medicine

c: \(\dfrac{2x}{x+5}+\dfrac{10x}{x^2+5x}\)

\(=\dfrac{2x}{x+5}+\dfrac{10x}{x\left(x+5\right)}\)

\(=\dfrac{2x}{x+5}+\dfrac{10}{x+5}=\dfrac{2x+10}{x+5}=\dfrac{2\left(x+5\right)}{x+5}=2\)

d: \(\dfrac{x}{x^2-36}+\dfrac{x-6}{x^2+6x}+\dfrac{-36}{\left(x^2-6x\right)\left(x+6\right)}\)

\(=\dfrac{x}{\left(x-6\right)\left(x+6\right)}+\dfrac{x-6}{x\left(x+6\right)}+\dfrac{-36}{x\left(x-6\right)\left(x+6\right)}\)

\(=\dfrac{x^2+\left(x-6\right)^2-36}{x\left(x-6\right)\left(x+6\right)}\)

\(=\dfrac{x^2+x^2-12x+36-36}{x\left(x-6\right)\left(x+6\right)}=\dfrac{2x^2-12x}{x\left(x-6\right)\left(x+6\right)}\)

\(=\dfrac{2\left(x^2-6x\right)}{\left(x^2-6x\right)\left(x+6\right)}=\dfrac{2}{x+6}\)

hearing => to hear (mệnh đề quan hệ rút gọn) 

c) \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}\le x\le\dfrac{7}{10}+\dfrac{27}{6}\)

\(\Leftrightarrow\dfrac{2}{3}\le x\le\dfrac{26}{5}=5,2\), mà \(x\in Z\)

\(\Rightarrow x\in\left\{1;2;3;4;5\right\}\)

d) \(-\dfrac{31}{14}+\dfrac{115}{131}+\dfrac{111}{74}\le x\le\dfrac{6}{36}+\dfrac{9}{27}+\dfrac{48}{96}\)

\(\Leftrightarrow\dfrac{150}{917}\le x\le1\) , mà \(x\in Z\)

\(\Rightarrow x=1\)