Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x_1^2-2}{x_1+1}.\frac{x_2^2-2}{x_2+1}=4\)
\(\frac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1+x\right)\left(x_2+1\right)}=4\)
\(\frac{\left(x_1.x_2\right)^2-2x_1^2-2x_2^2+4}{x_1.x_2+x_1+x_2+1}=4\)
\(\frac{\left(x_1.x_2\right)^2-2\left(x^2_1+x_2^2\right)+4}{x_1.x_2+\left(x_1+x_2\right)+1}=4\)
\(\frac{\left(m-2\right)^2-2.\left[\left(x_1+x_2\right)-2x_1x_2\right]+4}{m-2+\left(-m\right)+1}=4\)
\(\frac{m^2-4m+4-2.\left[m^2-2\left(m-2\right)\right]+4}{-1}=4\)
\(\Leftrightarrow m^2-4m+4-2\left(m^2-2m+4\right)+4=-4\)
\(\Leftrightarrow m^2-4m+4-2m^2+4m-8+4+4=0\)
\(\Leftrightarrow-m^2+4=0\)
\(\Leftrightarrow m^2-4=0\)
\(\Leftrightarrow m^2=4\)
\(\Leftrightarrow m=\pm2\)
vậy \(m=\pm2\) là các giá trị cần tìm
Ta có :
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}+1}-\frac{2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để P đạt GTNN thì \(1-\frac{2}{\sqrt{x}+1}\) phải đạt GTNN hay \(\frac{2}{\sqrt{x}+1}>0\) và đạt GTLN \(\Rightarrow\)\(\sqrt{x}+1>0\) và đạt GTNN
\(\Rightarrow\)\(\sqrt{x}+1=1\)
\(\Rightarrow\)\(\sqrt{x}=0\)
\(\Rightarrow\)\(x=0\)
Suy ra :
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{0}-1}{\sqrt{0}+1}=\frac{-1}{1}=-1\)
Vậy \(P_{min}=-1\) khi \(x=0\)
\(\Delta'=m^2-m^2+m>0\Rightarrow m>0\)
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m^2-m\end{matrix}\right.\)
a/ Kết hợp Viet và đề bài ta có hệ:
\(\left\{{}\begin{matrix}x_1+x_2=2m\\2x_1+3x_2=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x_1+2x_2=4m\\2x_1+3x_2=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_2=-4m+6\\x_1=6m-6\end{matrix}\right.\)
\(x_1x_2=m^2-m\Leftrightarrow\left(-4m+6\right)\left(6m-6\right)=m^2-m\)
\(\Leftrightarrow25m^2-61m+36=0\Rightarrow\left[{}\begin{matrix}m=1\\m=\frac{36}{25}\end{matrix}\right.\)
b/ \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1=3x_2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x_2=2m\\x_1=3x_2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_2=\frac{m}{2}\\x_1=\frac{3m}{2}\end{matrix}\right.\)
\(\Rightarrow\frac{3m^2}{4}=m^2-m\Leftrightarrow\frac{m^2}{4}-m=0\Rightarrow\left[{}\begin{matrix}m=0\left(l\right)\\m=4\end{matrix}\right.\)
\(\Rightarrow\)\(x_1^2+\left(x_1+x_2\right)x_2+4m^2-6=0\)
\(\Rightarrow x_1^2+x_1x_2+x_2^2+4m^2-6=0\)
\(\Rightarrow\left(x_1+x_2\right)^2-2x_1x_2+x_1x_2+4m^2-6=0\)
\(\Rightarrow\left(4m\right)^2-x_1x_2+4m^2-6=0\)
\(\Rightarrow16m^2-\left(4m^2-6\right)+4m^2-6=0\)
\(\Rightarrow16m^2-4m^2+6+4m^2-6=0\)
\(\Rightarrow16m^2=0\Rightarrow m=0\)
c/
\(\left|x_1\right|+\left|x_2\right|=3\)
\(\Leftrightarrow\left(\left|x_1\right|+\left|x_2\right|\right)^2=9\)
\(\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=9\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=9\)
\(\Leftrightarrow4\left(m+1\right)^2-2\left(m^2-3m\right)+2\left|m^2-3m\right|=9\)
- Với \(m^2-3m\ge0\Leftrightarrow\left[{}\begin{matrix}m\ge3\\-\frac{1}{5}< m\le0\end{matrix}\right.\)
\(\Rightarrow4\left(m+1\right)^2-2\left(m^2-3m\right)+2\left(m^2-3m\right)=9\)
\(\Leftrightarrow4\left(m+1\right)^2=9\Rightarrow\left(m+1\right)^2=\frac{9}{4}\)
\(\Rightarrow\left[{}\begin{matrix}m+1=\frac{3}{2}\\m+1=-\frac{3}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=\frac{1}{3}\left(l\right)\\m=-\frac{5}{2}< -\frac{1}{5}\left(l\right)\end{matrix}\right.\)
- Với \(m^2-3m< 0\Rightarrow0< m< 3\)
\(\Rightarrow4\left(m+1\right)^2-2\left(m^2-3m\right)-2\left(m^2-3m\right)=9\)
\(\Leftrightarrow20m-5=0\Rightarrow m=\frac{1}{4}\) (thỏa mãn)
\(\Delta'=\left(m+1\right)^2-m^2+3m=5m+1>0\Rightarrow m>-\frac{1}{5}\)
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2-3m\end{matrix}\right.\)
a/ \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\2x_1-3x_2=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x_1+3x_2=6\left(m+1\right)\\2x_1-3x_2=8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{6m+14}{5}\\x_2=\frac{4m-4}{5}\end{matrix}\right.\)
\(x_1x_2=m^2-3m\)
\(\Leftrightarrow\left(\frac{6m+14}{5}\right)\left(\frac{4m-4}{5}\right)=m^2-3m\)
Bạn tự khai triển và giải pt bậc 2 này
b/ \(\left|x_1-x_2\right|=4\)
\(\Leftrightarrow\left(x_1-x_2\right)^2=16\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=16\)
\(\Leftrightarrow4\left(m+1\right)^2-4\left(m^2-3m\right)=16\)
\(\Leftrightarrow5m+1=4\)
Do \(x_1;x_2\) là hai nghiệm của pt nên ta có những điều sau:
\(x_1+x_2=5\) ; \(x_1x_2=-1\); \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=27\)
\(x_1^2-5x_1-1=0\Rightarrow x_1^2+3x_1-2=8x_1-1\)
Tương tự: \(x_2^2+3x_2-2=8x_2-1\)
\(x_1^2+2x_1=7x_1+1\Rightarrow x_1^3+2x_1^2=7x_1^2+x_1\)
Tương tự: \(x_2^3+2x_2^2=7x_2^2+x_2\)
Thay vào:
\(M=\left(8x_1-1\right)\left(8x_2-1\right)=64\left(x_1x_2\right)-8\left(x_1+x_2\right)+1=...\)
\(N=\left(7x_1^2+x_1-1\right)\left(7x_2^2+x_2-1\right)\)
\(N=49\left(x_1x_2\right)^2+7x_1x_2\left(x_1+x_2\right)-7\left(x_1^2+x_2^2\right)-\left(x_1+x_2\right)+1\)
Bạn tự thay số