Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bn lên mạng hoặc vào câu hỏi tương tự nha!
chúc bn hok tốt!
hahaha!
#conmeo#
2) Đẳng thức điều kiện tương đương với \(\left(1+a\right)\left(1+b\right)\left(1+c\right)=1\Rightarrow1+a,1+b,1+c\ne0\)
Ta có: \(S=\frac{1}{1+\left(1+a\right)+\left(1+a\right)\left(1+b\right)}+\frac{1}{1+\left(1+b\right)+\left(1+b\right)\left(1+c\right)}\)\(+\frac{1}{1+\left(1+c\right)+\left(1+c\right)\left(1+a\right)}\)
\(=\frac{1}{1+\left(1+a\right)+\left(1+a\right)\left(1+b\right)}+\frac{1+a}{\left(1+a\right)\left[1+\left(1+b\right)+\left(1+b\right)\left(1+c\right)\right]}\)\(+\frac{\left(1+a\right)\left(1+b\right)}{\left(1+a\right)\left(1+b\right)\text{[}1+\left(1+c\right)+\left(1+c\right)\left(1+a\right)\text{]}}=\frac{1+\left(1+a\right)+\left(1+a\right)\left(1+b\right)}{1+\left(1+a\right)+\left(1+a\right)\left(1+b\right)}=1\)
\(A\left(x_0;2x_0\right)\Rightarrow OA=\sqrt{x_0^2+4x_0^2}=3\sqrt{5}\)
\(\Leftrightarrow5x_0^2=45\Rightarrow x_0^2=9\)
\(\Rightarrow x_0=-3\Rightarrow y_0=-6\)
Sửa: \(\left(d\right):y=\left(m-2\right)x+m+1\)
PT giao (d) với Ox \(y=0\Leftrightarrow x\left(m-2\right)=-m-1\Leftrightarrow x=\dfrac{m+1}{2-m}\Leftrightarrow A\left(\dfrac{m+1}{2-m};0\right)\Leftrightarrow OA=\left|\dfrac{m+1}{2-m}\right|\)
PT giao (d) với Oy \(x=0\Leftrightarrow y=m+1\Leftrightarrow B\left(0;m+1\right)\Leftrightarrow OB=\left|m+1\right|\)
Áp dụng HTL: \(\dfrac{1}{OA^2}+\dfrac{1}{OB^2}=\dfrac{1}{\left(\sqrt{2}\right)^2}=\dfrac{1}{2}\)
\(\Leftrightarrow\left|\dfrac{2-m}{m+1}\right|^2+\dfrac{1}{\left|m+1\right|^2}=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{\left(2-m\right)^2}{\left(m+1\right)^2}+\dfrac{1}{\left(m+1\right)^2}=\dfrac{1}{2}\\ \Leftrightarrow2\left(2-m\right)^2+2=\left(m+1\right)^2\\ \Leftrightarrow8-8m+2m^2+2=m^2+2m+1\\ \Leftrightarrow m^2-10m+9=0\\ \Leftrightarrow\left[{}\begin{matrix}m=-1\\m=-9\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}m=-1\\m=-9\end{matrix}\right.\) thỏa mãn đề bài
a: \(d\left(O;d\right)=\dfrac{\left|-\sqrt{3}\cdot0+\left(-1\right)\cdot0+\sqrt{3}m\right|}{\sqrt{\left(-\sqrt{3}\right)^2+\left(-1\right)^2}}=\dfrac{m\sqrt{3}}{2}\)
b: Để d=3 thì \(m\sqrt{3}=6\)
=>\(m=2\sqrt{3}\)