Cách 2 : Tách : \(\sqrt{5-x^3}-2-\left(\sqrt[3]{x^2+7}-2\right)\) -> Dùng liên hợp 

Đặt \(AB=a;AC=b;BC=a\) . Ta có : \(p=\dfrac{a+b+c}{2}=18\)

S = \(\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}=54\) \(=pr=18r\Rightarrow r=3\)  (cm) 

Ta có : (...) = \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\left(x+1\right)-\left[\sqrt[3]{x^2+7}-\left(x+1\right)\right]}{x^2-1}\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\left(x+1\right)}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{5-x^3-\left(x+1\right)^2}{\left(\sqrt{5-x^3}+x+1\right)\left(x^2-1\right)}\)  

\(=\lim\limits_{x\rightarrow1}\dfrac{-x^3-x^2-2x+4}{...}\)  \(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+2x+4\right)\left(x-1\right)}{...}\)   

= \(\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+2x+4\right)}{\left(x+1\right)\left(\sqrt{5-x^3}+x+1\right)}=\dfrac{-7}{8}\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\left(x+1\right)}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{x^2+7-x^3-3x^2-3x-1}{\left(x^2-1\right)\left[\sqrt[3]{\left(x+7\right)^2}+\left(x+1\right)\sqrt[3]{x^2+7}+\left(x+1\right)^2\right]}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+3x+6\right)\left(x-1\right)}{...}\)  

\(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+3x+6\right)}{\left(x+1\right)\left[\sqrt[3]{\left(x^2+7\right)^2}+\sqrt[3]{x^2+7}\left(x+1\right)+\left(x+1\right)^2\right]}\)

\(=\dfrac{-\left(1+3+6\right)}{\left(1+1\right)\left(4+2.2+4\right)}=\dfrac{-5}{12}\)

Suy ra : \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}=\dfrac{-7}{8}+\dfrac{5}{12}=\dfrac{-11}{24}\)

\(5y-3x=2xy-11\) \(\Leftrightarrow\left(2y+3\right)\left(2x-5\right)=7\)

x ; y \(\in Z\Rightarrow2x-5;2y+3\in Z\) \(\Rightarrow2y+3;2x-5\inƯ\left(7\right)\) 

-> Đến đây bạn lập bảng ... 

Uk 

1. 2 tam giác cân có một cặp góc = nhau thì đồng dạng 

3. \(3x-6=0\Leftrightarrow x=2\) ; \(x^2-4=0\Leftrightarrow x=\pm2\)

=> P/t x^2 - 4 = 0 là p/t hệ quả của p/t 3x - 6 = 0

=> 2 p/t không tương đương 

Ta có : \(f\left(2\right)=2a+b-6\)

\(\lim\limits_{x\rightarrow2^+}\dfrac{x-\sqrt{x+2}}{x^2-4}=\lim\limits_{x\rightarrow2^+}\dfrac{x^2-x-2}{\left(x-2\right)\left(x+2\right)\left(x+\sqrt{x+2}\right)}\)  

\(=\lim\limits_{x\rightarrow2^+}\dfrac{x+1}{\left(x+2\right)\left(x+\sqrt{x+2}\right)}=\dfrac{3}{16}\)

\(\lim\limits_{x\rightarrow2^-}x^2+ax+3b=4+2a+3b\) 

H/s liên tục tại điểm x = 2 \(\Leftrightarrow\dfrac{3}{16}=2a+3b+4=2a+b-6\)

Suy ra : \(a=\dfrac{179}{32};b=-5\) => t = a + b = 19/32 . Chọn C 

1.S 2.Đ 3.S 4.Đ

MN // BC => \(\dfrac{AM}{AB}=\dfrac{AN}{AC}\Rightarrow\dfrac{3}{AB}=\dfrac{4}{4+8}\Rightarrow AB=9\left(cm\right)\)  . Chọn B

37 . If you were on a new planet ; what would you do ?

38 . I prefer reading books than watching movies

39 .  It is the computer that hasn't been used for over a year . 

40 .  Because the subway is fast ; people like using it .