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\(x^2-2\left(m-2\right)x+m^2+2m-3=0\left(1\right)\)
Để phương trình có hai nghiệm phân biệt thì Δ' > 0
\(\Rightarrow\left(m-2\right)^2-m^2-2m+3>0\Leftrightarrow m^2-4m+4-m^2-2m+3>0\Leftrightarrow-6m+7>0\Leftrightarrow m< \dfrac{7}{6}\)\)
Theo viét : \(\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-2\right)\\x_1x_2=m^2+2m-3\end{matrix}\right.\)\)
Lại có :\( \dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{5}\)
\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{x_1+x_2}{5}\)
\(\Rightarrow\left(x_1+x_2\right)\left(x_1x_2\right)=5\left(x_1+x_2\right)\)
\(\Leftrightarrow\left(2m-4\right)\left(m^2+2m-3\right)=5\left(2m-4\right)\)
\(\Leftrightarrow2m^3+4m^2-6m-4m^2-8m+12=10m-20\)
\(\Leftrightarrow2m^3-24m+32=0\) \(\Leftrightarrow\left[{}\begin{matrix}m=-4\left(n\right)\\m=2\left(l\right)\end{matrix}\right.\)
Vậy \(m=-4\) thì thỏa điều kiện
\(A=\dfrac{3}{2}-tana\cdot cos^2a\)
\(=\dfrac{3}{2}-\dfrac{sina}{cosa}\cdot cos^2a\)
\(=\dfrac{3}{2}-sina\cdot cosa\)
\(=\dfrac{3}{2}-\dfrac{1}{2}sin2a\)
\(0^0< a< 90^0\)
=>\(0< =2a< =180^0\)
=>\(sin2a\in\left[-1;1\right]\)
\(-1< =sin2a< =1\)
=>\(\dfrac{1}{2}>=-\dfrac{1}{2}sin2a>=-\dfrac{1}{2}\)
=>\(\dfrac{7}{2}>=-\dfrac{1}{2}sin2a+3>=\dfrac{5}{2}\)
=>\(\dfrac{5}{2}< =y< =\dfrac{7}{2}\)
\(y_{min}=\dfrac{5}{2}\) khi sin2a=1
=>\(2a=\dfrac{\Omega}{2}+k2\Omega\)
=>\(a=\dfrac{\Omega}{4}+k\Omega\)
mà 0<a<90
nên a=45
a, \(\sqrt{x^2-6x+9}=2\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\)
\(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
a: Ta có: \(\sqrt{x^2-6x+9}=2\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
\(e,=\dfrac{\left(3+\sqrt{2}\right)\left(2\sqrt{2}+1\right)}{7}-\sqrt{\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}\\ =\dfrac{7\sqrt{2}+7}{7}-\dfrac{\sqrt{2}+1}{1}=\sqrt{2}+1-\sqrt{2}-1=0\)
\(f,=\sqrt{\dfrac{\left(2\sqrt{3}-3\right)^2}{\left(2\sqrt{3}-3\right)\left(2\sqrt{3}+3\right)}}\left(2+\sqrt{3}\right)\\ =\dfrac{\left(2\sqrt{3}-3\right)\left(2+\sqrt{3}\right)}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}=1\)
\(h,=\sqrt{\dfrac{\left(3\sqrt{5}-1\right)\left(2\sqrt{5}-3\right)}{20-9}}\left(\sqrt{2}+\sqrt{10}\right)\\ =\sqrt{\dfrac{2\left(33-11\sqrt{5}\right)}{11}}\left(\sqrt{5}+1\right)\\ =\sqrt{\dfrac{22\left(3-\sqrt{5}\right)}{11}}\left(\sqrt{5}+1\right)\\ =\sqrt{6-2\sqrt{5}}\left(\sqrt{5}+1\right)=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4\)
\(b,\text{PT hoành độ giao điểm: }-x-7=-3x+1\Leftrightarrow x=4\Leftrightarrow y=-11\Leftrightarrow B\left(4;-11\right)\\ c,\text{Gọi }\left(D_3\right):y=ax+b\left(a\ne0\right)\\ \left(D_3\right)\text{//}\left(D_1\right)\Leftrightarrow a=-1;b\ne-7\Leftrightarrow\left(D_3\right):y=-x+b\\ \left(D_3\right)\cap\left(D_2\right)\text{tại điểm có tung độ }4\Leftrightarrow\left\{{}\begin{matrix}4=-3x+1\\4=-x+b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\b=3\end{matrix}\right.\\ \text{Vậy }\left(D_3\right):y=-x+3\)
\(d,\Leftrightarrow B\left(4;-11\right)\in\left(D_4\right)\\ \Leftrightarrow8m-4-5m-2=-11\\ \Leftrightarrow3m=-5\Leftrightarrow m=-\dfrac{5}{3}\\ e,\Leftrightarrow\left\{{}\begin{matrix}-x-7=0\\\left(2m-1\right)x-5m-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\7-14m-5m-2=0\end{matrix}\right.\\ \Leftrightarrow m=\dfrac{5}{19}\)
1. \(\hept{\begin{cases}x-2=a+1\\3x+y=7a+3\end{cases}}\) mà a = 2y + 1
\(\Leftrightarrow\hept{\begin{cases}x-2=2y+1+1\\3x+y=7\left(2y+1\right)+3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2=2y+2\\3x+y=14y+10\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x-6=6y+6\\3x+y=14y+10\end{cases}}\)
\(\Rightarrow y+6=8y+4\)
\(\Leftrightarrow7y=2\Leftrightarrow y=\frac{2}{7}\)
2. \(\hept{\begin{cases}x-2y=a+1\\3x+y=7a+3\end{cases}}\) trong đó a = 1; a' = 3; b = -2; b' = 1
\(\Rightarrow\hept{\begin{cases}\frac{a}{a'}=\frac{1}{3}\\\frac{b}{b'}=-2\end{cases}}\Rightarrow\frac{a}{a'}\ne\frac{b}{b'}\) nên hệ pt có nghiệm duy nhất với mọi a
a, chưa nghĩ ra