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\(x^2-\left(m+1\right)x+m+4=0\left(1\right)\)
\(\Rightarrow\Delta>0\Leftrightarrow\left(m+1\right)^2-4\left(m+4\right)>0\Leftrightarrow\left[{}\begin{matrix}m< -3\\m>5\end{matrix}\right.\)\(\left(2\right)\)
\(ddkt-thỏa:\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)
\(x1=0\Rightarrow\left(1\right)\Leftrightarrow m=-4\Rightarrow\left(1\right)\Leftrightarrow x^2+3x=0\Leftrightarrow\left[{}\begin{matrix}x1=0\\x2=-3< 0\left(loại\right)\end{matrix}\right.\)
\(x1\ne0\) \(\Rightarrow0< x1< x2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+1>0\\m+4>0\end{matrix}\right.\)\(\Rightarrow m>-1\)\(\left(3\right)\)
\(\left(2\right)\left(3\right)\Rightarrow m>5\)
\(\Rightarrow\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)
\(\Leftrightarrow x1+x2+2\sqrt{x1x2}=12\Leftrightarrow m+1+2\sqrt{m+4}=12\)
\(\Leftrightarrow m+4+2\sqrt{m+4}-15=0\)
\(đặt:\sqrt{m+4}=t>5\Rightarrow t^2+2t-15=0\Leftrightarrow\left[{}\begin{matrix}t=-5\left(ktm\right)\\t=3\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow m\in\phi\)
Để pt có 2 nghiệm pb
\(\left(m+1\right)^2-4\left(m+4\right)=m^2+2m+1-4m-16\)
\(=m^2-2m-15>0\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=m+1\\x_1x_2=m+4\end{matrix}\right.\)
Ta có : \(\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=12\Leftrightarrow x_1+2\sqrt{x_1x_2}+x_2=12\)
Thay vào ta được \(m+1+2\sqrt{m+4}=12\Leftrightarrow2\sqrt{m+4}=11-m\)đk : m >= -4
\(\Leftrightarrow4\left(m+4\right)=121-22m+m^2\Leftrightarrow m^2-26m+105=0\)
\(\Leftrightarrow m=21\left(ktm\right);m=5\left(ktm\right)\)
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:
\(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{3}{2}\Leftrightarrow\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{2}{\sqrt{x_1x_2}}=\dfrac{9}{4}\\ \Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}+\dfrac{2}{\sqrt{x_1x_2}}=\dfrac{9}{4}\\ \Leftrightarrow\dfrac{5}{m-2}+\dfrac{2}{\sqrt{m-2}}=\dfrac{9}{4}\Leftrightarrow20+8\sqrt{m-2}=9\left(m-2\right)\\ \Leftrightarrow\left(\sqrt{m-2}-2\right)\left(9\sqrt{m-2}+10\right)=0\Leftrightarrow\sqrt{m-2}=2\Leftrightarrow m-2=4\Leftrightarrow m=6\left(t.m\right)\)
\(x^2-11x+m-2=0\left(1\right)\)
Để phương trình (1) có 2 nghiệm phân biệt thì:
\(\Delta>0\Rightarrow\left(-11\right)^2-4.1.\left(m-2\right)>0\)
\(\Leftrightarrow121-4m+8>0\)
\(\Leftrightarrow m< \dfrac{129}{4}\)
Theo hệ thức Vi-et ta có:
\(\left\{{}\begin{matrix}x_1+x_2=11\left(1'\right)\\x_1x_2=m-2\end{matrix}\right.\).
Ta có: \(\sqrt{x^2_1-10x_1+m-1}=5-\sqrt{x_2+1}\left(2\right)\)
Đk: \(\left\{{}\begin{matrix}x_1^2-10x_1+m-1\ge0\\-1\le x_2\le24\end{matrix}\right.\)
\(\left(2\right)\Rightarrow x^2_1-10x_1+m-1=25-10\sqrt{x_2+1}+x_2+1\)
\(\Leftrightarrow x_1^2-10x_1+\left(m-2\right)-25+10\sqrt{11-x_1+1}-x_2=0\)
\(\Rightarrow x_1^2-\left(x_1+x_2\right)-9x_1+x_1x_2-25+10\sqrt{12-x_1}=0\)
\(\Rightarrow x_1\left(x_1+x_2\right)-11-9x_1-25+10\sqrt{12-x_1}=0\)
\(\Rightarrow11x_1-9x_1-36+10\sqrt{12-x_1}=0\)
\(\Leftrightarrow2x_1+10\sqrt{12-x_1}-36=0\)
\(\Leftrightarrow x_1+5\sqrt{12-x_1}-18=0\)
\(\Leftrightarrow18-x_1=5\sqrt{12-x_1}\left(x_1\le12\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}18-x_1\ge0\\\left(18-x_1\right)^2=25\left(12-x_1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}18-x_1\ge0\\324-36x_1+x_1^2=300-25x_1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1\le18\\x_1^2-11x_1+24=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1\le18\\\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\left(nhận\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=3\\x_1=8\end{matrix}\right.\left(nhận\right)\)
Thay \(x_1=3\) vào (1') ta được:
\(3+x_2=11\Rightarrow x_2=8\left(nhận\right)\)
\(\Rightarrow m=x_1x_2+2=3.8+2=26\left(thỏa\Delta>0\right)\)
Thay \(x_1=8\) vào (1') ta được:'
\(8+x_2=11\Rightarrow x_2=3\left(nhận\right)\)
\(\Rightarrow m=x_1x_2+2=8.3+2=26\left(thỏa\Delta>0\right)\)
Vậy giá trị m cần tìm là 26.
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
pt. 2 mghiemej pb
`<=>Delta>0`
`<=>(m+2)^2-4(3m-6)>0`
`<=>m^2+4m+4-12m+24>0`
`<=>m^2-8m+28>0`
`<=>(m-4)^2+8>0` luôn đúng
Áp dụng vi-ét ta có:`x_1+x_2=m+2,x_1.x_2=-3m-6`
`đk:x_1,x_2>=0=>x_1+x_2,x_1.x_2>=0`
`=>m+2>=0,3m-6>=0`
`<=>m>=2`
`pt<=>x_1+x_2+2sqrt(x_1.x_2)=4`
`<=>m+2+2sqrt{3m-6}=4`
`<=>3m+6+6sqrt(3m-6)=12`
`<=>3m-6+6sqrt(3m-6)=0`
`<=>3m-6=0`
`<=>m=2(tmđk)`
Vậy m=2
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)
\(x^2-2\left(m+1\right)x+3m-3=0\left(1\right)\)
\(\Delta'>0\Leftrightarrow\left(m+1\right)^2-\left(3m-3\right)=m^2-m+4>0\left(đúng\forall m\right)\)
\(đk\) \(tồn\) \(tại:\sqrt{x1-1}+\sqrt{x2-1}\)
\(\Leftrightarrow1\le x1< x2\Leftrightarrow\left\{{}\begin{matrix}\left(x1-1\right)\left(x2-1\right)\ge0\\x1+x2-2>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x1x2-\left(x1+x2\right)+1\ge0\\2\left(m+1\right)-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3m-2-2\left(m+1\right)+1\ge0\\m>0\end{matrix}\right.\)
\(\Leftrightarrow m\ge4\)
\(\Rightarrow\sqrt{x1-1}+\sqrt{x2-1}=4\Leftrightarrow x1+x2-2+2\sqrt{\left(x1-1\right)\left(x2-1\right)}=16\)
\(\Leftrightarrow2\left(m+1\right)+2\sqrt{x1.x2-\left(x1+x2\right)+1}=18\)
\(\Leftrightarrow\left(m+1\right)+\sqrt{3m-3-2\left(m+1\right)+1}=9\)
\(\Leftrightarrow m-4+\sqrt{m-4}=4\)
\(đặt:\sqrt{m-4}=t\ge0\Rightarrow t^2+t=4\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-1+\sqrt{17}}{21}\left(tm\right)\\t=\dfrac{-1-\sqrt{17}}{21}\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{m-4}=\dfrac{-1+\sqrt{17}}{21}\Leftrightarrow m=....\)
\(\)
\(\Delta=9-4m>0\Rightarrow m< \dfrac{9}{4}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=m\end{matrix}\right.\)
\(\sqrt{x_1^2+1}+\sqrt{x_2^2+1}=3\sqrt{3}\)
\(\Leftrightarrow x_1^2+x_2^2+2+2\sqrt{\left(x_1^2+1\right)\left(x_2^2+1\right)}=27\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\sqrt{\left(x_1x_2\right)^2+\left(x_1+x_2\right)^2-2x_1x_2+1}=25\)
\(\Leftrightarrow9-2m+2\sqrt{m^2+9-2m+1}=25\)
\(\Leftrightarrow\sqrt{m^2-2m+10}=m+8\left(m\ge-8\right)\)
\(\Leftrightarrow m^2-2m+10=m^2+16m+64\)
\(\Rightarrow m=-3\) (thỏa mãn)
Pt trên có a=1, b=5, c=-3m+2
\(\Delta=b^2-4ac=25-4\cdot1\cdot\left(-3m+2\right)=17+12m\)
Để pt có hai nghiệm phân biệt thì \(\Delta>0\)<=> 17+12m >0 <=>m> 17/12
Theo hệ thức Viet, ta có:
\(\hept{\begin{cases}x_1+x_2=-5\\x_1\cdot x_2=-3m+2\end{cases}}\)
\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1\cdot x_2=25-4\left(-3m+2\right)=17+12m=10\)
=> 12m = -7 <=>m=-7/12 (thỏa đkxđ)
Vậy với m=-7/12 thì phương trình có hai nghiệm x1, x2 thỏa (x1 - x2)^2 =10