\(x^2-4x+y^2-6x+15=2\)

\(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2-6x+9\right)-4-9+15-2=0\)

\(\Leftrightarrow\left(x-2\right)^2+\left(y-3\right)^2=0\)

Lại có :

\(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y-3\right)^2\ge0\end{matrix}\right.\) \(\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow x=2;y=3\)

Đặt :

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{100^2}\)

Ta có :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

.................

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Leftrightarrow A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)

a/ \(2x-3=5x+2\)

\(\Leftrightarrow5x-2x=-3-2\)

\(\Leftrightarrow3x=-5\Leftrightarrow x=-\dfrac{5}{3}\)

Vậy..

b. \(2x\left(x-1\right)=2x+2\)

\(\Leftrightarrow2x^2-4x-2=0\)

\(\Leftrightarrow x^2-2x-1=0\)

\(\Leftrightarrow\left(x-1+\sqrt{2}\right)\left(x-1-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1-\sqrt{2}\\x=1+\sqrt{2}\end{matrix}\right.\)

Vậy...

c/ ĐKXĐ : \(x\ne\pm2\)

\(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{6}{\left(x+2\right)}\)

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

\(\Leftrightarrow x^2+4x+4-x^2=6x-12\)

\(\Leftrightarrow2x-16=0\)

\(\Leftrightarrow x=8\)

Vậy..

Phương trình nhận \(x=-1\) là nghiệm 

\(\Leftrightarrow2\left(-1\right)-m=1-\left(-1\right)\)

\(\Leftrightarrow-2-m=1+1\)

\(\Leftrightarrow-2+m=2\)

\(\Leftrightarrow m=4\)

Vậy...

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{2006.2007}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{2006}-\dfrac{1}{2007}\)

\(=1-\dfrac{1}{2007}=\dfrac{2006}{2007}\)

Ta có :

Lúc đi : \(s_{AB}=50t_1\) \(\left(km\right)\)

Lúc về : \(s_{AB}=40t_2\left(km\right)\) 

\(\Leftrightarrow50t_1-40t_2=0\)

Và : \(t_1+t_2=2\left(h\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}t_1=\dfrac{8}{9}\\t_2=\dfrac{10}{9}\end{matrix}\right.\)

\(\Leftrightarrow s_{AB}=\dfrac{400}{9}\left(km\right)\)

Đổi \(s=4,5km=4500m\)

\(t=30p=1800s\)

\(A=F.s=600.4500=27.10^5\left(J\right)\)

\(P=\dfrac{A}{t}=\dfrac{27.10^5}{1800}=1500\left(W\right)\)

\(\dfrac{1}{a}-\dfrac{1}{b}=1\)

\(\Leftrightarrow\dfrac{b-a}{ab}=1\)

\(\Leftrightarrow b-a=ab\)

\(\Leftrightarrow a+ab-b=0\)

+) Với \(m=-1\) phương trình trở thành :

\(-2x+2=0\Leftrightarrow x=1\)

+) Với \(m\ne-1\) Ta có :

\(\Delta'=\left(-1\right)^2-2\left(m+1\right)=-2m\)

+ Nếu \(m=0\Leftrightarrow\) pt có 2 nghiệm kép

+ Nếu \(m>0\Leftrightarrow\) pt vô nghiệm 

+ Nếu \(m< 0\) pt có 2 nghiệm phân biệt

Vậy...

Để hàm số xác định :

\(mx^2-mx+m+3\ge0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\m>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>0\\m^2-4m\left(m+3\right)\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>0\\-3m^2-12m\ge0\end{matrix}\right.\)