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b: \(PT\Leftrightarrow x^2+\left(m-3\right)x-m=0\)
\(\text{Δ}=\left(m-3\right)^2+4m\)
\(=m^2-6m+9+4m\)
\(=m^2-2m+1+8=\left(m-1\right)^2+8>0\)
Do đó: PT luon có hai nghiệm phân biệt
\(\dfrac{2}{x_1}+\dfrac{2}{x_2}=\dfrac{2x_1+2x_2}{x_1x_2}=\dfrac{2\cdot\left(-m+3\right)}{-m}=\dfrac{-2m+6}{-m}\)
\(\dfrac{4x_2}{x_1}+\dfrac{4x_1}{x_2}=\dfrac{4\left(x_1^2+x_2^2\right)}{x_1x_2}\)
\(=\dfrac{4\left(x_1+x_2\right)^2-8x_1x_2}{x_1x_2}=\dfrac{4\left(-m+3\right)^2-8\cdot\left(-m\right)}{-m}\)
\(=\dfrac{4\left(m-3\right)^2+8m}{-m}\)
\(=\dfrac{4m^2-24m+36+8m}{-m}=\dfrac{4m^2-16m+36}{-m}\)
c: \(A=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}+1\)
\(=\sqrt{\left(-m+3\right)^2-4\cdot\left(-m\right)}+1\)
\(=\sqrt{m^2-6m+9+4m}+1\)
\(=\sqrt{m^2-2m+1+8}+1\)
\(=\sqrt{\left(m-1\right)^2+8}+1\ge2\sqrt{2}+1\)
Dấu '=' xảy ra khi m=1
1.
\(a+b+c=0\) nên pt luôn có 2 nghiệm
\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-1\end{matrix}\right.\)
\(A=\dfrac{2x_1x_2+3}{x_1^2+x_2^2+2x_1x_2+2}=\dfrac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}=\dfrac{2\left(m-1\right)+3}{m^2+2}=\dfrac{2m+1}{m^2+2}\)
\(A=\dfrac{m^2+2-\left(m^2-2m+1\right)}{m^2+2}=1-\dfrac{\left(m-1\right)^2}{m^2+2}\le1\)
Dấu "=" xảy ra khi \(m=1\)
2.
\(\Delta=m^2-4\left(m-2\right)=\left(m-2\right)^2+4>0;\forall m\) nên pt luôn có 2 nghiệm pb
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-2\end{matrix}\right.\)
\(\dfrac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1-1\right)\left(x_2-1\right)}=4\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1^2+x_2^2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)
\(\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1+x_2\right)^2+4x_1x_2+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)
\(\Rightarrow\dfrac{\left(m-2\right)^2-2m^2+4\left(m-2\right)+4}{m-2-m+1}=4\)
\(\Rightarrow-m^2=-4\Rightarrow m=\pm2\)
Lời giải:
Để PT có 2 nghiệm phân biệt $x_1,x_2$ thì:
\(\Delta'=(m+2)^2-(m^2+m+3)>0\)
\(\Leftrightarrow 3m+1>0\Leftrightarrow m> \frac{-1}{3}\)
Áp dụng định lý Vi-et: \(\left\{\begin{matrix} x_1+x_2=2(m+2)\\ x_1x_2=m^2+m+3\end{matrix}\right.\)
\(x_1x_2=m^2+m+3=(m+\frac{1}{2})^2+\frac{11}{4}\neq 0, \forall m>\frac{-1}{3}\) nên $x_1,x_2\neq 0$ với mọi \(m> \frac{-1}{3}\).
Khi đó:
\(\frac{x_1}{x_2}+\frac{x_2}{x_1}=1\)
\(\Leftrightarrow \frac{x_1^2+x_2^2}{x_1x_2}=4\)
\(\Leftrightarrow \frac{(x_1+x_2)^2-2x_1x_2}{x_1x_2}=4\)
\(\Leftrightarrow \frac{(x_1+x_2)^2}{x_1x_2}=6\Rightarrow (x_1+x_2)^2=6x_1x_2\)
\(\Leftrightarrow 4(m+2)^2=6(m^2+m+3)\)
\(\Leftrightarrow 2m^2-10m+2=0\)
\(\Leftrightarrow m=\frac{5\pm \sqrt{21}}{2}\) (thỏa mãn)
dùng phương pháp Vi-ét ko hoàn toàn
(mình đăng lên youtube rồi đấy)
\(\Delta^'=\left(-1\right)^2-\left(m-1\right)=2-m\)
Để PT có nghiệm thì: \(m\le2\)
Khi đó theo hệ thức viet ta có: \(\hept{\begin{cases}x_1+x_2=2\\x_1x_2=m-1\end{cases}}\)
Ta có: \(x_1^4-x_1^3=x_2^4-x_2^3\)
\(\Leftrightarrow\left(x_1^4-x_2^4\right)-\left(x_1^3-x_2^3\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)\left(x_1^2+x_2^2\right)-\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left[2\left(x_1^2+x_2^2\right)-x_1^2-x_1x_2-x_2^2\right]=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1^2-x_1x_2+x_2^2\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-3x_1x_2\right]=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left[4-3\left(m-1\right)\right]=0\)
Nếu \(x_1-x_2=0\Rightarrow x_1=x_2=1\Rightarrow m=1\left(tm\right)\)
Nếu \(4-3\left(m-1\right)=0\Rightarrow m=\frac{7}{3}\left(ktm\right)\)
Vậy m = 1
pt có \(\Delta\)= (4m+1)2-4.2.(m-1) = 16m2+8m+1-8m+8=16m2+9 >0
==> pt có ngiệm với mọi m
theo hthuc vi ét ta có :\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-4m-1}{2}\\x1.x2=\dfrac{m-1}{2}\end{matrix}\right.\)(1)
mà có \(\dfrac{x1^2x2+x1x2^2}{x1^2+x2^2}=2==>\dfrac{x1.x2.\left(x1+x2\right)}{\left(x1+x2\right)^2-2x1x2}=2\) (2)
thay (1) vào (2) ta đc ........
giải ra m ( bạn tự lm nhé )
thay
\(a+b+c=1-2\left(m+3\right)+2m+5=0\)
\(\Rightarrow\) phương trình luôn có 2 nghiệm: \(\left\{{}\begin{matrix}x_1=1\\x_2=2m+5\end{matrix}\right.\)
Để 2 nghiệm của pt thỏa mãn yêu cầu của đề bài \(\Rightarrow x_2>0\Rightarrow2m+5>0\Rightarrow m>\dfrac{-5}{2}\)
\(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{4}{3}\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2m+5}}=\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{2m+5}}=\dfrac{1}{3}\Rightarrow2m+5=9\Rightarrow m=2\)
\(x^2-x+1-m=0\)
Theo Vi - ét, ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=1\\x_1x_2=\dfrac{c}{a}=1-m\end{matrix}\right.\)
Ta có :
\(5\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)-x_1x_2+4=0\)
\(\Leftrightarrow5\left(\dfrac{x_2+x_1}{x_1x_2}\right)-x_1x_2+4=0\)
\(\Leftrightarrow5\left(\dfrac{1}{1-m}\right)-\left(1-m\right)+4=0\)
\(\Leftrightarrow\dfrac{5}{1-m}-1+m+4=0\)
\(\Leftrightarrow\dfrac{5}{1-m}+m+3=0\)
\(\Leftrightarrow\dfrac{5+m\left(1-m\right)+3\left(1-m\right)}{1-m}=0\)
\(\Leftrightarrow5+m-m^2+3-3m=0\)
\(\Leftrightarrow-m^2-2m+8=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=2\\m=-4\end{matrix}\right.\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-4\right)\\x_1x_2=-m^2+4\end{matrix}\right.\)
\(\dfrac{x_1+x_2}{x_1x_2}+\dfrac{4}{x_1x_2}=1\)
Thay vào ta được : \(\dfrac{2\left(m-4\right)+4}{-m^2+4}=1\Leftrightarrow\dfrac{2m-4}{\left(2-m\right)\left(m+2\right)}=1\Leftrightarrow\dfrac{-2}{m+2}=1\Rightarrow-2=m+2\Leftrightarrow m=-4\)