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\(a+b+c=1-2\left(m+3\right)+2m+5=0\)
\(\Rightarrow\) 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 \(\Rightarrow x_2>0\Rightarrow2m+5>0\Rightarrow m>\dfrac{-5}{2}\)
\(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{4}{3}\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2m+5}}=\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{2m+5}}=\dfrac{1}{3}\Rightarrow2m+5=9\Rightarrow m=2\)
\(\Delta=\left(-5\right)^2-4\cdot1\cdot\left(2m-1\right)\)
\(=25-8m+4\\ =29-8m\)
Để pt có nghiệm \(\Leftrightarrow\Delta\ge0\)
\(\Leftrightarrow29-8m\ge0\\ \Leftrightarrow-8m\ge-29\\ \Leftrightarrow m\le\dfrac{29}{8}\)
Với \(m\le\dfrac{29}{8}\) theo vi-ét ta có
\(\left\{{}\begin{matrix}x_1+x_2=5\\x_1\cdot x_2=2m-1\end{matrix}\right.\)
Có \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=\dfrac{19}{3}\)
\(\Leftrightarrow\dfrac{x_1^2+x^2_2}{x_1x_2}=\dfrac{19}{3}\)
\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=\dfrac{19}{3}\)
\(\Leftrightarrow\dfrac{5^2-2\left(2m-1\right)}{2m-1}=\dfrac{19}{3}\) (đkxđ \(m\ne\dfrac{1}{2}\) )
\(\Leftrightarrow\dfrac{25-4m+2}{2m-1}=\dfrac{19}{3}\\ \Leftrightarrow\dfrac{27-4m}{2m-1}=\dfrac{19}{3}\\ \Leftrightarrow3\left(27-4m\right)=19\left(2m-1\right)\)
\(\Leftrightarrow81-12m=38m-19\\ \Leftrightarrow81+19=38m+12m\\ \Leftrightarrow100=50m\)
\(\Leftrightarrow m=2\) ( Thỏa mãn \(m\le\dfrac{29}{8};m\ne\dfrac{1}{2}\) )
Vậy......................................
-theo vi-ét ta có:
\(x_1x_2=\dfrac{c}{a}=2m-1\left(1\right)\)
\(x_1+x_2=\dfrac{-b}{a}=5\left(2\right)\)
- theo đề bài ta lại có:
\(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=\dfrac{19}{3}\)
<=>\(\dfrac{x_1^2+x_2^2}{x_1x_2}=\dfrac{19}{3}\)
<=>\(\dfrac{x_1^2+2x_1x_2+x_2^2-2x_1x_2}{x_1x_2}=\dfrac{19}{3}\)
<=>\(\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=\dfrac{19}{3}\)(3)
-thay (1),(2) vào (3) ta được:
\(\dfrac{5^2-2\left(2m-1\right)}{2m-1}=\dfrac{19}{3}\)
=) m=2
vậy m=2
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
Lời giải:
Để pt có hai nghiệm phân biệt thì \(\Delta'=1+2m>0\Leftrightarrow m> \frac{-1}{2}\)
a)
Áp dụng hệ thức Viete, với $x_1,x_2$ là hai nghiệm của pt:
\(\left\{\begin{matrix} x_1+x_2=2\\ x_1x_2=-2m\end{matrix}\right.\)
Khi đó: \((x_1^2+1)(x_2^2+1)=5\)
\(\Leftrightarrow (x_1x_2)^2+x_1^2+x_2^2=4\)
\(\Leftrightarrow (x_1x_2)^2+(x_1+x_2)^2-2x_1x_2=4\)
\(\Leftrightarrow 4m^2+4+4m=4\)
\(\Leftrightarrow m(m+1)=0\Rightarrow m=0\) do \(m> \frac{-1}{2}\)
b)
Ta có:
\(u=\frac{1}{x_1+1}+\frac{1}{x_2+1}=\frac{x_1+x_2+2}{(x_1+1)(x_2+1)}\)
\(=\frac{x_1+x_2+2}{x_1x_2+(x_1+x_2)+1}=\frac{2+2}{-2m+2+1}=\frac{4}{3-2m}\)
\(v=\frac{1}{x_1+1}.\frac{1}{x_2+1}=\frac{1}{(x_1+1)(x_2+1)}=\frac{1}{x_1+x_2+x_1x_2+1}=\frac{1}{2-2m+1}=\frac{1}{3-2m}\)
Do đó pt nhận \(\frac{1}{x_1+1}; \frac{1}{x_2+1}\) làm nghiệm theo định lý Viete đảo là:
\(X^2-\frac{4}{3-2m}X+\frac{1}{3-2m}=0\)
\(\Leftrightarrow (3-2m)X^2-4X+1=0\)
f(x) =x^2 -2x -2m
a) f(x) có hai nghiệm pb <=> 1 +2m > 0 => m>-1/2
P=\(\left(x_1^2+1\right)\left(x_2^2+1\right)=\left(x_1.x_2\right)^2+\left(x_1+x_2\right)^2-2x_1x_2+1\)
\(P=\left(x_1x_2-1\right)^2+\left(x_1+x_2\right)^2=\left(2m+1\right)^2+4\)
\(P=5\Leftrightarrow\left(2m+1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}2m+1=-1;m=-1\left(l\right)\\2m+1=1;m=0\left(n\right)\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}m\ge\dfrac{1}{2}\\1+2-2m\ne0\end{matrix}\right.\) <=> \(m\in[\dfrac{-1}{2};\dfrac{3}{2})U\left(\dfrac{3}{2};\infty\right)\)
\(\left\{{}\begin{matrix}\dfrac{1}{x_1+1}+\dfrac{1}{x_2+1}=\dfrac{x_1+x_2+2}{x_1x_2+\left(x_1+x_2\right)+1}=\dfrac{4}{3-2m}\\\dfrac{1}{x_1+1}.\dfrac{1}{x_2+1}=\dfrac{1}{3-2m}\end{matrix}\right.\)
phương trình cần tìm
\(g\left(x\right)=x^2-\dfrac{4}{3-2m}+\dfrac{1}{3-2m}\) \(\Leftrightarrow\left\{{}\begin{matrix}m\in[\dfrac{-1}{2};\dfrac{3}{2})U\left(\dfrac{3}{2};\infty\right)\\\left(2m-3\right)x^2+4x-1=0\end{matrix}\right.\)
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)
coi như đoạn trên bạn đúng nhé (làm tiếp)
\(S=\dfrac{m^2+2m}{m^2+2m+2017}=\dfrac{m^2+2m+2017-2017}{m^2+2m+2017}=1-\dfrac{2017}{\left(m+1\right)^2+2016}\)
có \(s_1=\left(m+1\right)^2+2016\ge2016\Rightarrow\dfrac{1}{\left(m+1\right)^2+2016}\le\dfrac{1}{2016}\)\(\Rightarrow-\dfrac{1}{\left(m+1\right)^2+2016}\ge\dfrac{1}{2016}\)
\(\Rightarrow\Rightarrow-\dfrac{2017}{\left(m+1\right)^2+2016}\ge\dfrac{-2017}{2016}\)
\(\Rightarrow\Rightarrow\Rightarrow1-\dfrac{2017}{\left(m+1\right)^2+2016}\ge1-\dfrac{2017}{2016}=\dfrac{-1}{2016}\)
\(S\ge-\dfrac{1}{2016}\)
đẳng thức khi m =-1
\(\Delta=\left(2m-3\right)^2-4\left(2m-4\right)=\left(2m-5\right)^2\ge0;\forall m\)
Pt luôn có 2 nghiệm với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m-3\\x_1x_2=2m-4\end{matrix}\right.\)
\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2m-3}{2m-4}=\dfrac{1}{2}\)
\(\Rightarrow4m-6=2m-4\)
\(\Leftrightarrow2m=2\)
\(\Leftrightarrow m=1\) (thỏa mãn)