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Phương trình có nghiệm \(\Leftrightarrow\Delta'\ge0\Leftrightarrow1-m\ge0\Leftrightarrow m\le1\)
Theo hệ thức Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m\end{matrix}\right.\) (1)
Ta có: \(\dfrac{1}{x^2}+\dfrac{1}{x^2}=1\Leftrightarrow\dfrac{x^2_1+x^2_2}{x^2_1x^2_2}=1\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{\left(x_1x_2\right)^2}=1\) (2)
Từ (1) và (2) \(\Rightarrow4-2m=m^2\Leftrightarrow m^2+2m-4=0\)
\(\Delta'=1+4=5\Rightarrow\sqrt{\Delta'}=\sqrt{5}\Rightarrow\left[{}\begin{matrix}m=-1+\sqrt{5}\left(\text{loại}\right)\\m=-1-\sqrt{5}\left(\text{nhận}\right)\end{matrix}\right.\)
Vậy \(m=-1-\sqrt{5}\)
a: Khi m = -4 thì:
\(x^2-5x+\left(-4\right)-2=0\)
\(\Leftrightarrow x^2-5x-6=0\)
\(\Delta=\left(-5\right)^2-5\cdot1\cdot\left(-6\right)=49\Rightarrow\sqrt{\Delta}=\sqrt{49}=7>0\)
Pt có 2 nghiệm phân biệt:
\(x_1=\dfrac{5+7}{2}=6;x_2=\dfrac{5-7}{2}=-1\)
\(ac=-3< 0\Rightarrow\) pt đã cho luôn có 2 nghiệm pb trái dấu với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-3\end{matrix}\right.\)
\(\dfrac{x_1}{x_2^2}+\dfrac{x_2}{x_1^2}=m-1\Leftrightarrow\dfrac{x_1^3+x_2^3}{\left(x_1x_2\right)^2}=m-1\)
\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)}{9}=m-1\)
\(\Leftrightarrow8\left(m-1\right)^3+18\left(m-1\right)=9\left(m-1\right)\)
\(\Leftrightarrow\left(m-1\right)\left[8\left(m-1\right)^2+9\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=1\\8\left(m-1\right)^2+9=0\left(vô-nghiệm\right)\end{matrix}\right.\)
PT có 2 nghiệm `<=> \Delta' >0 <=> 2^2-1.(m+1)>0<=> m<3`
Viet: `x_1+x_2=-4`
`x_1 x_2=m+1`
`(x_1)/(x_2)+(x_2)/(x_1)=10/3`
`<=> (x_1^2+x_2^2)/(x_1x_2)=10/3`
`<=> ((x_1+x_2)^2-2x_1x_2)/(x_1x_2)=10/3`
`<=> (4^2-2(m+1))/(m+1)=10/3`
`<=> m=2` (TM)
Vậy `m=2`.
Xét \(\Delta=4\left(m-1\right)^2-4.\left(-3\right)=4\left(m-1\right)^2+12>0\forall m\)
=>Pt luôn có hai nghiệm pb
Theo viet:\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1.x_2=-3\ne0\forall m\end{matrix}\right.\)
Có \(\dfrac{x_1}{x_2^2}+\dfrac{x_2}{x_1^2}=m-1\)
\(\Leftrightarrow x_1^3+x_2^3=\left(m-1\right)x_1^2.x_2^2\)
\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=\left(m-1\right).\left(-3\right)^2\)
\(\Leftrightarrow8\left(m-1\right)^3-3\left(-3\right).2\left(m-1\right)=9\left(m-1\right)\)
\(\Leftrightarrow8\left(m-1\right)^3+9\left(m-1\right)=0\)
\(\Leftrightarrow\left(m-1\right)\left[8\left(m-1\right)^2+9\right]=0\)
\(\Leftrightarrow m=1\)(do \(8\left(m-1\right)^2+9>0\) với mọi m)
Vậy m=1
Vì \(ac< 0\) \(\Rightarrow\) Phương trình luôn có 2 nghiệm phân biệt
Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1x_2=-3\end{matrix}\right.\)
Mặt khác: \(\dfrac{x_1}{x_2^2}+\dfrac{x_2}{x_1^2}=m-1\) \(\Rightarrow\dfrac{\left(x_1+x_2\right)\left(x_1^2+x_2^2-x_1x_2\right)}{x_1^2x_2^2}=m-1\)
\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-3x_1x_2\right]}{x_1^2x_2^2}=m-1\)
\(\Rightarrow\dfrac{\left(2m-2\right)\left(4m^2-8m+13\right)}{9}=m-1\)
\(\Leftrightarrow...\)
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: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-2\end{matrix}\right.\)
\(\dfrac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1-1\right)\left(x_2-1\right)}=4\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1^2+x_2^2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)
\(\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1+x_2\right)^2+4x_1x_2+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)
\(\Rightarrow\dfrac{\left(m-2\right)^2-2m^2+4\left(m-2\right)+4}{m-2-m+1}=4\)
\(\Rightarrow-m^2=-4\Rightarrow m=\pm2\)
Ta có: \(\Delta=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(m+1\right)\)
\(=\left(-2m+2\right)^2-4\left(m+1\right)\)
\(=4m^2-8m+4-4m-4\)
\(=4m^2-12m\)
Để phương trình có nghiệm thì \(\text{Δ}\ge0\)
\(\Leftrightarrow4m^2-12m\ge0\)
\(\Leftrightarrow4m\left(m-3\right)\ge0\)
\(\Leftrightarrow m\left(m-3\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}m\ge3\\m\le0\end{matrix}\right.\)
Khi \(\left[{}\begin{matrix}m\ge3\\m\le0\end{matrix}\right.\), Áp dụng hệ thức Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)=2m-2\\x_1\cdot x_2=m+1\end{matrix}\right.\)
Ta có: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=4\)
\(\Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1\cdot x_2}=4\)
\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=4\)
\(\Leftrightarrow\dfrac{\left(2m-2\right)^2-2\cdot\left(m+1\right)}{m+1}=4\)
\(\Leftrightarrow4m^2-8m+4-2m-2=4\left(m+1\right)\)
\(\Leftrightarrow4m^2-10m+2-4m-4=0\)
\(\Leftrightarrow4m^2-14m-2=0\)
Đến đây bạn tự làm nhé, chỉ cần tìm m và đối chiều với điều kiện thôi
Pt có 2 nghiệm
\(\to \Delta=[-2(m-1)]^2-4.1.(m+1)=4m^2-8m+4-4m-4=4m^2-12m\ge 0\)
\(\leftrightarrow m^2-3m\ge 0\)
\(\leftrightarrow m(m-3)\ge 0\)
\(\leftrightarrow \begin{cases}m\ge 0\\m-3\ge 0\end{cases}\quad or\quad \begin{cases}m\le 0\\m-3\le 0\end{cases}\)
\(\leftrightarrow m\ge 3\quad or\quad m\le 0\)
Theo Viét
\(\begin{cases}x_1+x_2=2(m-1)\\x_1x_2=m+1\end{cases}\)
\(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=4\)
\(\leftrightarrow \dfrac{x_1^2+x_2^2}{x_1x_2}=4\)
\(\leftrightarrow \dfrac{(x_1+x_2)^2-2x_1x_2}{x_1x_2}=4\)
\(\leftrightarrow \dfrac{[2(m-1)]^2-2.(m+1)}{m+1}=4\)
\(\leftrightarrow 4m^2-8m+4-2m-2=4(m+1)\)
\(\leftrightarrow 4m^2-10m+2-4m-4=0\)
\(\leftrightarrow 4m^2-14m-2=0\)
\(\leftrightarrow 2m^2-7m-1=0 (*)\)
\(\Delta_{*}=(-7)^2-4.2.(-1)=49+8=57>0\)
\(\to\) Pt (*) có 2 nghiệm phân biệt
\(m_1=\dfrac{7+\sqrt{57}}{2}(TM)\)
\(m_2=\dfrac{7-\sqrt{57}}{2}(TM)\)
Vậy \(m=\dfrac{7\pm \sqrt{57}}{2}\) thỏa mãn hệ thức
Có\(\Delta=4\left(m+1\right)^2-4\left(2m-3\right)=4m^2+16>0\forall m\)
=> pt luôn có hai nghiệm pb
Theo viet có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m-3\end{matrix}\right.\)
Có :\(P^2=\left(\dfrac{x_1+x_2}{x_1-x_2}\right)^2=\dfrac{4\left(m+1\right)^2}{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\dfrac{4\left(m+1\right)^2}{4\left(m+1\right)^2-4\left(2m-3\right)}=\dfrac{4\left(m+1\right)^2}{4m^2+16}\)\(\ge0\)
\(\Rightarrow P\ge0\)
Dấu = xảy ra khi m=-1
a) để phương trình có 2 nghiệm : \(\Leftrightarrow\left\{{}\begin{matrix}m-3\ne0\\\Delta'\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m-3\ne0\\\left(m+2\right)^2-\left(m-3\right)\left(m+1\right)\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m\ne3\\6m+7\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne3\\m\ge\dfrac{7}{6}\end{matrix}\right.\)
thay \(x_1=2\) vào phương trình ta có :
\(4\left(m-3\right)-4\left(m+2\right)+m+1=0\Leftrightarrow m=19\)
áp dụng hệ thức vi ét ta có : \(x_1+x_2=\dfrac{2\left(m+2\right)}{m-3}=\dfrac{2\left(21\right)}{16}=\dfrac{21}{8}\)
\(\Rightarrow x_2=\dfrac{21}{8}-x_1=\dfrac{21}{8}-2=\dfrac{5}{8}\)
vậy ....................................................................................................
b) áp dụng hệ thức vi ét ta có : \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m+2\right)}{m-3}\\x_1x_2=\dfrac{m+1}{m-3}\end{matrix}\right.\)
ta có : \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=10\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=10\Leftrightarrow\dfrac{2\left(m+2\right)}{m-3}:\dfrac{m+1}{m-3}=10\)
\(\Leftrightarrow\dfrac{2m+4}{m+1}=10\Leftrightarrow2m+4=10m+10\Leftrightarrow m=\dfrac{-3}{4}\left(L\right)\)
vậy không có m thỏa mãn điều kiện bài toán .
câu 2) a) để phương trình có 2 nghiệm cùng dấu \(\Leftrightarrow\left\{{}\begin{matrix}m-2\ne0\\\Delta'\ge0\\p>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m-2\ne0\\\left(m+1\right)^2-\left(m-2\right)\left(m-1\right)\ge0\\\dfrac{m-1}{m-2}>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\5m-1\ge0\\\left(m-1\right)\left(m-2\right)>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\m\ge\dfrac{1}{5}\\\left[{}\begin{matrix}m>2\\m< 1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m>2\) vậy \(m>2\)
b) áp dụng hệ thức vi ét ta có : \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-2\left(m+1\right)}{m-2}\\x_1x_2=\dfrac{m-1}{m-2}\end{matrix}\right.\)
ta có : \(x_1^3+x_2^3=64\Leftrightarrow\left(x_1+x_2\right)^3-3\left(x_1x_2\right)\left(x_1+x_2\right)=64\)
\(\left(\dfrac{2m+2}{2-m}\right)^3+6\left(\dfrac{m-1}{m-2}\right)\left(\dfrac{m+1}{m-2}\right)=64\)
\(\Leftrightarrow\dfrac{\left(-2m-2\right)^3}{\left(m-2\right)^3}+\dfrac{6\left(m-1\right)\left(m+1\right)\left(m-2\right)}{\left(m-2\right)^3}=64\)
\(\Leftrightarrow\dfrac{-8m^3-24m^2-24m-8+6m^2-12m^3-6m+12}{m^2-6m^2+12m-8}=64\)
\(\Leftrightarrow\dfrac{-20m^3-18m^2-30m+4}{m^3-6m^2+12m-8}=64\)
\(\Leftrightarrow84m^3-402m^2+798m-516=0\)
giải nốt nha .