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Đây là câu bđt của chuyên Quảng Nam vừa thi mà:vvv
Ta có: \(xy+yz+zx=xyz\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\left(a,b,c>0\right)\)
Khi đó: \(H=\frac{a}{9b^2+1}+\frac{b}{9c^2+1}+\frac{c}{9a^2+1}\)
\(=\left(a+b+c\right)-\left(\frac{9ab^2}{9b^2+1}+\frac{9bc^2}{9c^2+1}+\frac{9ca^2}{9a^2+1}\right)\)
\(\ge1-\left(\frac{9ab^2}{6b}+\frac{9bc^2}{6c}+\frac{9ca^2}{6a}\right)\)
\(=1-\frac{3}{2}\left(ab+bc+ca\right)\ge1-\frac{3}{2}\cdot\frac{\left(a+b+c\right)^2}{3}=1-\frac{3}{2}\cdot\frac{1}{3}=\frac{1}{2}\)
Dấu "=" xảy ra khi: \(x=y=z=3\)
Vậy Min(H) = 1/2 khi x = y = z = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Delta=\left(-2m\right)^2-4\left(2m-10\right)\)
=4m^2-8m+40
=4m^2-8m+4+36=(2m-2)^2+36>0
=>(1) luôn có hai nghiệm phân biệt
x1+x2=2m và 2x1+x2=-4
=>-x1=2m+4 và x1+x2=2m
=>x1=-2m-4 và x2=2m+2m+4=4m+4
x1x2=2m-10
=>(-2m-4)(4m+4)=2m-10
=>-8(m-2)(m+1)=2(m-5)
=>-4(m-2)(m+1)=(m-5)
=>-4(m^2-m-2)=m-5
=>-4m^2+4m+8-m+5=0
=>-4m^2+3m+13=0
=>\(m=\dfrac{3\pm\sqrt{217}}{8}\)
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\(\Delta'=m^2-\left(m-1\right)=m^2-m+1=\left(m-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
Vậy pt luôn có 2 nghiệm pb
Ta có : \(x_1+x_2+2\sqrt{x_1x_2}=4\Leftrightarrow2m+2\sqrt{m-1}=4\)
\(\Leftrightarrow\sqrt{m-1}=2-m\)
đk : m =< 2
TH1 \(m-1=2-m\Leftrightarrow m=\dfrac{3}{2}\)(tm)
TH2 \(m-1=m-2\)( vô lí )
\(\Delta'=m^2-m+1=\left(m-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0;\forall m\Rightarrow\) pt luôn có 2 nghiệm pb
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m-1\end{matrix}\right.\)
Để biểu thức \(\sqrt{x_1}+\sqrt{x_2}=2\) xác định \(\Rightarrow x_1;x_2\ge0\Rightarrow\left\{{}\begin{matrix}x_1+x_2\ge0\\x_1x_2\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2m\ge0\\m-1\ge0\end{matrix}\right.\) \(\Rightarrow m\ge1\)
Khi đó:
\(\sqrt{x_1}+\sqrt{x_2}=2\Leftrightarrow x_1+x_2+2\sqrt{x_1x_2}=4\)
\(\Leftrightarrow2m+2\sqrt{m-1}=4\)
\(\Leftrightarrow m+\sqrt{m-1}=2\)
Đặt \(\sqrt{m-1}=t\ge0\Rightarrow m=t^2+1\)
\(\Rightarrow t^2+1+t=2\Rightarrow t^2+t-1=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+\sqrt{5}}{2}\\t=\dfrac{-1-\sqrt{5}}{2}< 0\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{m-1}=\dfrac{-1+\sqrt{5}}{2}\Rightarrow m-1=\dfrac{3-\sqrt{5}}{2}\)
\(\Rightarrow m=\dfrac{5-\sqrt{5}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Delta=b^2-4ac=\left(m+1\right)^2+1>0\forall m\)
\(\Leftrightarrow\) pt luôn có 2 nghiệm phân biệt với mọi m
Theo định lí Viet ta có :
\(\left\{{}\begin{matrix}x_1+x_2=m+1\\x_1.x_2=-1\end{matrix}\right.\)
\(2\left|x_1\right|+\left|x_2\right|=3\)
\(\Leftrightarrow\left(2\left|x_1\right|+\left|x_2\right|\right)^2=9\)
\(\Leftrightarrow4x_1^2+x_2^2+4x_1.x_2=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2 :
\(\Delta'=m^2-\left(2m-1\right)=\left(m-1\right)^2\ge0\)
Để pt có 2 nghiệm pb
\(m-1\ne0\Leftrightarrow m\ne1\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=2m\left(1\right)\\x_1x_2=2m-1\left(2\right)\end{matrix}\right.\)
Ta có : \(2x_1-3x_2=4\left(3\right)\)
Từ (1) ; (3) ta có hệ \(\left\{{}\begin{matrix}x_1+x_2=2m\\2x_1-3x_2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x_1+2x_2=4m\\2x_1-3x_2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x_2=4m-4\\x_1=2m-x_2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{4m-4}{5}\\x_1=2m-\dfrac{4m-4}{5}=\dfrac{6m+4}{5}\end{matrix}\right.\)
Thay vào (3) ta được \(\left(\dfrac{6m+4}{5}\right)\left(\dfrac{4m+4}{5}\right)=2m-1\)
\(\Rightarrow\left(6m+4\right)\left(4m+4\right)=50m-25\Leftrightarrow24m^2+40m+16=50m-25\)
\(\Leftrightarrow24m^2-10m+41=0\)
\(\Delta'=10-41.24< 0\)Vậy pt vô nghiệm hay ko có gtri m
5.
\(\Delta'=9-\left(2m+1\right)=8-2m>0\Rightarrow m< 4\)
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=6\\x_1x_2=2m+1\end{matrix}\right.\)
Kết hợp Viet và điều kiện đề bài:
\(\left\{{}\begin{matrix}x_1+x_2=6\\x_1^2=x_2-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_2=6-x_1\\x_1^2=6-x_1-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=6-x_1\\x_1^2+x_1-2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x_1=1;x_2=5\\x_1=-2;x_2=8\end{matrix}\right.\)
Thế vào \(x_1x_2=2m+1\Rightarrow\left[{}\begin{matrix}2m+1=5\\2m+1=-16\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m=2\\m=-\dfrac{17}{2}\end{matrix}\right.\) (thỏa mãn)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\Leftrightarrow\left\{{}\begin{matrix}8m+9=41\\m+0=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=4\\m=4\end{matrix}\right.\Leftrightarrow m=4\)
\(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{y}}=\dfrac{\sqrt{xy}+y}{y}\)