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mk chắc chắn 100% là 99m<9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
\(\Leftrightarrow2m.2^x+\left(2m+1\right)\left(3-\sqrt{5}\right)^x+\left(3+\sqrt{5}\right)^x=0\)
\(\Leftrightarrow\left(\frac{3+\sqrt{5}}{2}\right)^x+\left(2m+1\right)\left(\frac{3-\sqrt{5}}{2}\right)^x+2m< 0\)
Đặt \(t=\left(\frac{3+\sqrt{5}}{2}\right)^x,0< t\le1\Rightarrow\frac{1}{t}=\left(\frac{3-\sqrt{5}}{2}\right)^x\)
Phương trình trở thành :
\(t+\left(2m+1\right)\frac{1}{t}+2m=0\) (*)
a. Khi \(m=-\frac{1}{2}\) ta có \(t=1\) suy ra \(\left(\frac{3+\sqrt{5}}{2}\right)^x=1\Leftrightarrow x=0\)
Vậy phương trình có nghiệm là \(x=0\)
b. Phương trình (*) \(\Leftrightarrow t^2+1=-2m\left(t+1\right)\Leftrightarrow\frac{t^2+1}{t+1}=-2m\)
Xét hàm số \(f\left(t\right)=\frac{t^2+1}{t+1};t\in\)(0;1]
Ta có : \(f'\left(t\right)=\frac{t^2+2t+1}{\left(t+1\right)^2}\Rightarrow f'\left(t\right)=0\Leftrightarrow=-1+\sqrt{2}\)
t f'(t) f(t) 0 1 0 - + 1 1 -1 + căn 2 2 căn 2 - 2
Suy ra phương trình đã cho có nghiệm đúng
\(\Leftrightarrow2\sqrt{2}-2\le-2m\le1\Leftrightarrow\sqrt{2}-1\ge m\ge-\frac{1}{2}\)
Vậy \(m\in\left[-\frac{1}{2};\sqrt{2}-1\right]\) là giá trị cần tìm
a/ \(\Leftrightarrow m^2x-m^2-x-m+2=0\)
\(\Leftrightarrow\left(m^2-1\right)x=m^2+m-2\)
Xét khi \(m^2-1=0\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0x=1+1-2=0\\0x=1-1-2=-2\left(l\right)\end{matrix}\right.\)
Vậy vs m= 1 pt vô số nghiệm (x>0)
Xét khi \(m^2-1\ne0\Leftrightarrow\left\{{}\begin{matrix}m\ne1\\m\ne-1\end{matrix}\right.\)
\(\Rightarrow x=\frac{m^2+m-2}{m^2-1}\)
Có \(x>0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left(m-1\right)\left(m+2\right)>0\\\left(m-1\right)\left(m+1\right)>0\end{matrix}\right.\\\left\{{}\begin{matrix}\left(m-1\right)\left(m+2\right)< 0\\\left(m-1\right)\left(m+1\right)< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m>-1\\m< -2\end{matrix}\right.\)
b/ \(\Leftrightarrow mx-m-x+1+m-2=0\)
\(\Leftrightarrow\left(m-1\right)x=1\)
Vs \(m\ne1\)
\(\Rightarrow x=\frac{1}{m-1}\)
Có \(x\ge3\Rightarrow\frac{1}{m-1}\ge3\Leftrightarrow1\ge3m-3\Leftrightarrow m\le\frac{4}{3}\)
Xét \(m=1\Rightarrow0x=1\left(l\right)\)
Vậy vs \(m\le\frac{4}{3}\) thì pt có nghiệm vs x\(\ge3\)
c/ ĐKXĐ: \(9-x^2>0\Leftrightarrow\left(3-x\right)\left(3+x\right)>0\Leftrightarrow-3< x< 3\)
hmm, xem lại hộ cái đề boài nhoa, vế phải trên tử có dấu bằng là sao nhể? =))
ĐKXĐ: \(x\ne0\)
Đặt \(x+\frac{2}{x}=t\Rightarrow\left(x+\frac{2}{x}\right)^2=t^2\Rightarrow x^2+4+\frac{4}{x^2}=t^2\Rightarrow x^2+\frac{4}{x^2}=t^2-4\)
\(\Rightarrow t^2-4-4t+m+1=0\)
\(\Leftrightarrow t^2-4t+m-3=0\)
Để pt có đúng nghiệm<=> \(\left\{{}\begin{matrix}\Delta'=0\\\left(x-1\right)^2>0\left(lđ\right)\end{matrix}\right.\Leftrightarrow4-m+3=0\Leftrightarrow m=7\)
a/ \(\left\{{}\begin{matrix}m+1>0\\\Delta'=\left(m-1\right)^2-3\left(m-1\right)\left(m+1\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>-1\\-m^2-m+2\le0\end{matrix}\right.\) \(\Rightarrow m\ge1\)
b/ \(\left\{{}\begin{matrix}m^2+4m-5< 0\\\Delta'=\left(m-1\right)^2-2\left(m^2+4m-5\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2+4m-5< 0\\-m^2-10m+11\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-5< m< 1\\\left[{}\begin{matrix}m\le-11\\m\ge1\end{matrix}\right.\end{matrix}\right.\)
Không tồn tại m thỏa mãn
c/ Do \(x^2-8x+20=\left(x-4\right)^2+4>0\) \(\forall x\) nên BPT nghiệm đúng với mọi x khi mẫu số âm với mọi x
\(\Rightarrow\left\{{}\begin{matrix}m< 0\\\Delta'=\left(m+1\right)^2-m\left(9m+4\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< 0\\-8m^2-2m+1< 0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m< 0\\\left[{}\begin{matrix}m< -\frac{1}{2}\\m>\frac{1}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m< -\frac{1}{2}\)
d/ Do \(3x^2-5x+4>0\) \(\forall x\) nên BPT luôn đúng khi:
\(\left\{{}\begin{matrix}m-4>0\\\left(m+1\right)^2-4\left(2m-1\right)\left(m-4\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>4\\-7m^2+38m-15< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>4\\\left[{}\begin{matrix}m< \frac{3}{7}\\m>5\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m>5\)
\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2-2m\left(x+\frac{1}{x}\right)-1=0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow\left|t\right|\ge2\)
Phương trình trở thành: \(t^2-2mt-1=0\) (1)
Để pt đã cho có nghiệm \(\Leftrightarrow\left(1\right)\) có ít nhất 1 nghiệm thỏa mãn \(\left|t\right|\ge2\)
Do \(ac=-1< 0\Rightarrow\left(1\right)\) luôn luôn có 2 nghiệm phân biệt trái dấu
Đặt \(f\left(t\right)=t^2-2mt-1\)
Để (1) có 2 nghiệm thỏa mãn \(\left|t\right|\ge2\)
\(\Leftrightarrow f\left(2\right).f\left(-2\right)\le0\)
\(\Leftrightarrow\left(3+4m\right)\left(3-4m\right)\le0\Leftrightarrow\left[{}\begin{matrix}m\le-\frac{3}{4}\\m\ge\frac{3}{4}\end{matrix}\right.\)