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Theo vi ét: \(\left\{{}\begin{matrix}x_1+x_2=6\\x_1x_2=8\end{matrix}\right.\)
Theo đề:
\(B=\dfrac{x_1\sqrt{x_1}-x_2\sqrt{x_2}}{x_1-x_2}=\dfrac{\left(\sqrt{x_1}-\sqrt{x_2}\right)\left(x_1+\sqrt{x_1x_2}+x_2\right)}{\left(\sqrt{x_1}-\sqrt{x_2}\right)\left(\sqrt{x_1}+\sqrt{x_2}\right)}\left(x_1,x_2\ge0\right)\)
\(=\dfrac{6+\sqrt{8}}{\sqrt{x_1}+\sqrt{x_2}}\)
Tính: \(\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=x_1+x_2+2\sqrt{x_1x_2}=6+2\sqrt{8}=6+4\sqrt{2}=\left(\sqrt{4}+\sqrt{2}\right)^2\)
\(\Rightarrow\sqrt{x_1}+\sqrt{x_2}=\sqrt{4}+\sqrt{2}\) (thỏa mãn \(x_1,x_2\ge0\))
Khi đó: \(P=\dfrac{6+\sqrt{8}}{\sqrt{4}+\sqrt{2}}=4-\sqrt{2}\)
\(\Delta=9-4m>0\Rightarrow m< \dfrac{9}{4}\)
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
\(x^2-4x+3=0\)
Theo vi-et, ta có: \(x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-4\right)}{1}=4;x_1x_2=\dfrac{c}{a}=\dfrac{3}{1}=3\)
Đặt \(A=\sqrt{x_1}+\sqrt{x_2}\)
=>\(A^2=x_1+x_2+2\sqrt{x_1x_2}\)
=>\(A^2=4+2\cdot\sqrt{3}\)
=>\(A=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(x^2-\left(m+1\right)x+m+4=0\left(1\right)\)
\(\Rightarrow\Delta>0\Leftrightarrow\left(m+1\right)^2-4\left(m+4\right)>0\Leftrightarrow\left[{}\begin{matrix}m< -3\\m>5\end{matrix}\right.\)\(\left(2\right)\)
\(ddkt-thỏa:\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)
\(x1=0\Rightarrow\left(1\right)\Leftrightarrow m=-4\Rightarrow\left(1\right)\Leftrightarrow x^2+3x=0\Leftrightarrow\left[{}\begin{matrix}x1=0\\x2=-3< 0\left(loại\right)\end{matrix}\right.\)
\(x1\ne0\) \(\Rightarrow0< x1< x2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+1>0\\m+4>0\end{matrix}\right.\)\(\Rightarrow m>-1\)\(\left(3\right)\)
\(\left(2\right)\left(3\right)\Rightarrow m>5\)
\(\Rightarrow\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)
\(\Leftrightarrow x1+x2+2\sqrt{x1x2}=12\Leftrightarrow m+1+2\sqrt{m+4}=12\)
\(\Leftrightarrow m+4+2\sqrt{m+4}-15=0\)
\(đặt:\sqrt{m+4}=t>5\Rightarrow t^2+2t-15=0\Leftrightarrow\left[{}\begin{matrix}t=-5\left(ktm\right)\\t=3\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow m\in\phi\)
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
\(\Leftrightarrow4\left(m+4\right)=121-22m+m^2\Leftrightarrow m^2-26m+105=0\)
\(\Leftrightarrow m=21\left(ktm\right);m=5\left(ktm\right)\)
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
\(x^2-11x+m-2=0\left(1\right)\)
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.
x1;x2 là nghiệm của pt
=> \(x^2_1-3\sqrt{2}x_1-\sqrt{2}=0\Rightarrow x^2_1=3\sqrt{2}x_1+\sqrt{2}\)
\(x^2_2-3\sqrt{2}x_2-\sqrt{2}=0\Rightarrow x^2_2=3\sqrt{2}x_2+\sqrt{2}\)
=> \(A=\frac{2}{3\sqrt{2}x_1+3\sqrt{2}x_2+\sqrt{2}-3\sqrt{2}}+\frac{3\sqrt{2}x_2+3\sqrt{2}x_1+\sqrt{2}-3\sqrt{2}}{2}\)
\(A=\frac{2}{3\sqrt{2}\left(x_1+x_2\right)-2\sqrt{2}}+\frac{3\sqrt{2}\left(x_2+x_1\right)-2\sqrt{2}}{2}\)
Theo VI ét => \(x_1+x_2=3\sqrt{2}\). Thay vào A
=> quy đồng.....