Ta có : \(y=\dfrac{x}{x-1}=1+\dfrac{1}{x-1}\Rightarrow y'=\dfrac{-1}{\left(x-1\right)^2}\)

Giả sử M(xo ; yo) là tiếp điểm của tiếp tuyến d với đths trên \(\). Ta có : 

 PT d : \(y=\dfrac{-1}{\left(x_0-1\right)^2}\left(x-x_0\right)+\dfrac{x_0}{x_{0-1}}=\dfrac{-x}{\left(x_0-1\right)^2}+\dfrac{x_0^2}{\left(x_0-1\right)^2}\) 

K/C từ B(1;1) đến d : d(B;d) = \(\left|\dfrac{\dfrac{1}{\left(x_0-1\right)^2}+1-\dfrac{x_0^2}{\left(x_0-1\right)^2}}{\sqrt{\dfrac{1}{\left(x_0-1\right)^4}+1}}\right|\)  

= \(\left|\dfrac{2\left(1-x_0\right)}{\left(x_0-1\right)^2}\right|:\dfrac{\sqrt{\left(x_0-1\right)^4+1}}{\left(x_0-1\right)^2}=\dfrac{2\left|1-x_0\right|}{\sqrt{\left(1-x_0\right)^4+1}}\)   \(\le\dfrac{2\left|1-x_0\right|}{\sqrt{2\left(1-x_0\right)^2}}=\sqrt{2}\)

" = " \(\Leftrightarrow\left[{}\begin{matrix}x_0=0\\x_0=2\end{matrix}\right.\)

Suy ra : y = -x hoặc y = -x + 4 

13C-14C-15C-16B-17D-18C-19A-20B

\(1\dfrac{1}{5}-60\%+0,4:\dfrac{5}{8}=\dfrac{6}{5}-\dfrac{3}{5}+\dfrac{2}{5}:\dfrac{5}{8}=\dfrac{3}{5}+\dfrac{16}{25}=\dfrac{31}{25}\)

\(a^4+1\ge a\left(a^2+1\right)\Leftrightarrow a^3\left(a-1\right)-\left(a-1\right)\ge0\) \(\Leftrightarrow\left(a^3-1\right)\left(a-1\right)\ge0\) 

\(\Leftrightarrow\left(a-1\right)^2\left(a^2+a+1\right)\ge0\) (Luôn đúng)

Ta có : \(f'\left(1\right)=f'\left(2\right)=0\) ; \(g\left(x\right)=f\left(x^2+4x-m\right)\) \(\Rightarrow g'\left(x\right)=\left(2x+4\right)f'\left(x^2+4x-m\right)\)

g'(x) = 0 \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\f'\left(x^2+4x-m\right)=0\left(1\right)\end{matrix}\right.\) 

g'(x) có nhiều no nhất \(\Leftrightarrow\left(1\right)\) có nhiều no nhất \(\Leftrightarrow x^2+4x-m=1\) và \(x^2+4x-m=2\) đều có 2 no 

\(x^2+4x-m=1\) có 2 no \(\Leftrightarrow\Delta'=m+5>0\Leftrightarrow m>-5\)

\(x^2+4x-m=2\) có 2 no \(\Leftrightarrow m>-6\)

Vậy m > -5 

Mà m \(\in\left[-2021;2022\right]\) nên m \(\in\left[-4;2022\right]\)

=> Có : 2023 + 4 = 2027 giá trị nguyên của m t/m

6h30' - 2h = 4h30' = 4,5h 

Số lớn : \(\dfrac{500+50}{2}=275\)  . Số bé : \(275-50=225\) . Đ/s : ... 

Bài có trên Web 

Do \(\dfrac{\pi}{2}< \alpha< \pi\Rightarrow sin\alpha>0\) . Suy ra : \(sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(-\dfrac{5}{13}\right)^2}=\sqrt{1-\dfrac{25}{169}}=\dfrac{12}{13}\) 

\(sin2\alpha=2sin\alpha cos\alpha=2.\dfrac{-5}{13}.\dfrac{12}{13}=-\dfrac{120}{169}\)

tan \(\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{12}{13}:-\dfrac{5}{13}=-\dfrac{12}{5}\)

\(sin\left(\dfrac{\pi}{6}-\alpha\right)=\dfrac{1}{2}.cos\alpha-\dfrac{\sqrt{3}}{2}.sin\alpha=\dfrac{1}{2}.\dfrac{-5}{13}-\dfrac{\sqrt{3}}{2}.\dfrac{12}{13}=-\dfrac{12\sqrt{3}+5}{26}\)

\(cos\left(\dfrac{\pi}{3}+\alpha\right)=\dfrac{1}{2}cos\alpha-\dfrac{\sqrt{3}}{2}.sin\alpha=\dfrac{-12\sqrt{3}-5}{26}\)

Dễ dàng c/m :  \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\forall x;y>0\) (*)

a ; b ; c là độ dài 3 cạnh của \(\Delta\Rightarrow\left\{{}\begin{matrix}a;b;c>0\\a+b-c;b+c-a;c+a-b>0\end{matrix}\right.\)

AD (*) ta được : \(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{2b}=\dfrac{2}{b}\) 

CMTT : \(\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{2}{c};\dfrac{1}{c+a-b}+\dfrac{1}{a+b-c}\ge\dfrac{2}{a}\)

Suy ra : \(2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\Rightarrow\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)