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a. thay m=-4 vào (1) ta có:
\(x^2-5x-6=0\)
Δ=b\(^2\)-4ac= (-5)\(^2\) - 4.1.(-6)= 25 + 24= 49 > 0
\(\sqrt{\Delta}=\sqrt{49}=7\)
x\(_1\)=\(\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+7}{2}\)=6
x\(_2\)=\(\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{5-7}{2}\)=-1
vậy khi x=-4 thì pt đã cho có 2 nghiệm x\(_1\)=6; x\(_2\)=-1
Lời giải:
Để pt có 2 nghiệm dương phân biệt thì:
\(\left\{\begin{matrix} \Delta=25-4(m-2)>0\\ S=5>0\\ P=m-2>0\end{matrix}\right.\Leftrightarrow 2< m< \frac{33}{4}\)
Khi đó:
\(2\left(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\right)=3\Leftrightarrow 4(\frac{1}{x_1}+\frac{1}{x_2}+\frac{2}{\sqrt{x_1x_2}})=9\)
\(\Leftrightarrow 4\left(\frac{5}{m-2}+\frac{2}{\sqrt{m-2}}\right)=9\)
\(\Leftrightarrow 4(5t^2+2t)=9\) với $t=\frac{1}{\sqrt{m-2}}$
$\Rightarrow t=\frac{1}{2}$
$\Leftrightarrow m=6$ (thỏa)
Để pt có 2 nghiệm dương phân biệt thì:
\(\left\{{}\begin{matrix}\Delta=25-4\left(m-2\right)>0\\P=5>0\\S=m-2>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< 8,25\\5>0\\m>2\end{matrix}\right.\)
\(\Leftrightarrow2< m< 8,25\)
Theo vi-et thì ta có: \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=m-2\end{matrix}\right.\)
Theo đề bài ta có:
\(2\left(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}\right)=3\)
\(\Leftrightarrow4\left(\dfrac{1}{x_1}+\dfrac{2}{\sqrt{x_1x_2}}+\dfrac{1}{x_2}\right)=9\)
\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}+\dfrac{2}{\sqrt{x_1x_2}}=\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{5}{m-2}+\dfrac{2}{\sqrt{m-2}}=\dfrac{9}{4}\)
Đặt \(\dfrac{1}{\sqrt{m-2}}=a>0\) thì ta có
\(5a^2+2a-2,25=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-0,9\left(l\right)\\a=0,5\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{\sqrt{m-2}}=0,5=\dfrac{1}{2}\)
\(\Leftrightarrow m-2=4\)
\(\Leftrightarrow m=6\)
\(x^2-2\left(m+1\right)x+3m-3=0\left(1\right)\)
\(\Delta'>0\Leftrightarrow\left(m+1\right)^2-\left(3m-3\right)=m^2-m+4>0\left(đúng\forall m\right)\)
\(đk\) \(tồn\) \(tại:\sqrt{x1-1}+\sqrt{x2-1}\)
\(\Leftrightarrow1\le x1< x2\Leftrightarrow\left\{{}\begin{matrix}\left(x1-1\right)\left(x2-1\right)\ge0\\x1+x2-2>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x1x2-\left(x1+x2\right)+1\ge0\\2\left(m+1\right)-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3m-2-2\left(m+1\right)+1\ge0\\m>0\end{matrix}\right.\)
\(\Leftrightarrow m\ge4\)
\(\Rightarrow\sqrt{x1-1}+\sqrt{x2-1}=4\Leftrightarrow x1+x2-2+2\sqrt{\left(x1-1\right)\left(x2-1\right)}=16\)
\(\Leftrightarrow2\left(m+1\right)+2\sqrt{x1.x2-\left(x1+x2\right)+1}=18\)
\(\Leftrightarrow\left(m+1\right)+\sqrt{3m-3-2\left(m+1\right)+1}=9\)
\(\Leftrightarrow m-4+\sqrt{m-4}=4\)
\(đặt:\sqrt{m-4}=t\ge0\Rightarrow t^2+t=4\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-1+\sqrt{17}}{21}\left(tm\right)\\t=\dfrac{-1-\sqrt{17}}{21}\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{m-4}=\dfrac{-1+\sqrt{17}}{21}\Leftrightarrow m=....\)
\(\)
\(\Delta=5^2-4\cdot1\cdot\left(m-2\right)\)
=25-4(m-2)
=25-4m+8=-4m+33
Để phương trình có hai nghiệm phân biệt thì -4m+33>0
=>-4m>-33
=>m<33/4
Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=-5;x_1x_2=\frac{c}{a}=m-2\)
\(\left(x_1-1\right)^2+\left(x_2-1\right)^2\)
\(=x_1^2+x_2^2-2\left(x_1+x_2\right)+2\)
\(=\left(x_1+x_2\right)^2-2x_1x_2-2\left(x_1+x_2\right)+2\)
\(=\left(-5\right)^2-2\left(m-2\right)-2\cdot\left(-5\right)+2=25-2\left(m-2\right)+10+2\)
=37-2(m-2)
=37-2m+4
=41-2m
\(\left(x_1-1\right)^2\cdot\left(x_2-1\right)^2=\left\lbrack\left(x_1-1\right)\left(x_2-1\right)\right\rbrack^2\)
\(=\left\lbrack x_1x_2-\left(x_1+x_2\right)+1\right\rbrack^2\)
=[m-2-(-5)+1]^2=(m-1+5)^2=(m+4)^2
\(\frac{1}{\left(x_1-1\right)^2}+\frac{1}{\left(x_2-1\right)^2}=1\)
=>\(\frac{\left(x_1-1\right)^2+\left(x_2-1\right)^2}{\left(x_1-1\right)^2\cdot\left(x_2-1\right)^2}=1\)
=>\(\frac{41-2m}{\left(m+4\right)^2}=1\)
=>\(m^2+8m+16=41-2m\)
=>\(m^2+10m-25=0\)
=>\(m^2+10m+25-50=0\)
=>\(\left(m+5\right)^2=50\)
=>\(\left[\begin{array}{l}m+5=5\sqrt2\\ m+5=-5\sqrt2\end{array}\right.\Rightarrow\left[\begin{array}{l}m=5\sqrt2-5\left(nhận\right)\\ m=-5\sqrt2-5\left(nhận\right)\end{array}\right.\)
a. Với m=6 thì phương trình (1) có dạng
x^2 - 5x +4= 0
<=> (x-1)(x-4)=0
<=> x=1 hoặc x=4
Vậy m=6 thì phương trình có nghiệm x=1 hoặc x=4
b. Xét \(\text{ Δ}=\left(-5\right)^2-4\cdot1\cdot\left(m-2\right)=33-4m\)
Để (1) có nghiệm phân biệt khi \(m< \dfrac{33}{4}\)
Theo Vi-et ta có: \(x_1x_2=m-2;x_1+x_2=5\)
Để 2 nghiệm phương trình (1) dương khi m>2
Ta có:
\(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{3}{2}\Leftrightarrow\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{2}{\sqrt{x_1x_2}}=\dfrac{9}{4}\\ \Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}+\dfrac{2}{\sqrt{x_1x_2}}=\dfrac{9}{4}\\ \Leftrightarrow\dfrac{5}{m-2}+\dfrac{2}{\sqrt{m-2}}=\dfrac{9}{4}\Leftrightarrow20+8\sqrt{m-2}=9\left(m-2\right)\\ \Leftrightarrow\left(\sqrt{m-2}-2\right)\left(9\sqrt{m-2}+10\right)=0\Leftrightarrow\sqrt{m-2}=2\Leftrightarrow m-2=4\Leftrightarrow m=6\left(t.m\right)\)