d: Xét ΔABK và ΔCKB có 

AB=CK

KB chung

AK=CB

Do đó: ΔABK=ΔCKB

Em cảm ơn ạ!!
 

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

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

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

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

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

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

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

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

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

Em cảm ơn nhìu ạ ❤️😍

a: Xét ΔBAD vuông tại A và ΔBHI vuông tại H có

\(\hat{ABD}=\hat{HBI}\)

Do đó: ΔBAD~ΔBHI

=>\(\frac{BA}{BH}=\frac{BD}{BI}\)

=>\(BA\cdot BI=BH\cdot BD\)

b: Xét ΔBAH có BI là phân giác

nên \(\frac{IA}{IH}=\frac{BA}{BH}\) (1)

Xét ΔBAC có BD là phân giác

nên \(\frac{BC}{BA}=\frac{DC}{DA}\) (2)

Xét ΔBHA vuông tại H và ΔBAC vuông tại A có

góc HBA chung

Do đó: ΔBHA~ΔBAC

=>\(\frac{BH}{BA}=\frac{BA}{BC}\)

=>\(\frac{BA}{BH}=\frac{BC}{BA}\left(3\right)\)

Từ (1),(2),(3) suy ra \(\frac{IA}{IH}=\frac{DC}{DA}\)

=>\(AI\cdot AD=IH\cdot DC\)

c: Ta có: ED⊥AC

AB⊥CA

Do đó: ED//AB

=>\(\hat{IAB}=\hat{IFD}\)

Xét ΔIAB và ΔIFD có

\(\hat{IAB}=\hat{IFD}\)

\(\hat{AIB}=\hat{FID}\) (hai góc đối đỉnh)

Do đó: ΔIAB~ΔIFD

=>\(\frac{IA}{FI}=\frac{AB}{FD}\)

=>\(\frac{AI}{AB}=\frac{FI}{FD}\)

=>\(AI\cdot FD=FI\cdot AB\)

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 \(x\in Z\)

\(\Rightarrow x=1\)