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a) Ta có: \(\Delta\) = (-2m)2 - 4.1.(m-2) = 4m2 - 4m + 8 = (4m2 - 4m + 1) + 7 = (2m-1)2 + 7 \(\ge\) 7 > 0 x do đo (1) luôn có 2 nghiệm với mọi m.
\(\Delta'=4-m+1=5-m\ge0\Rightarrow m\le5\)
Theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m-1\end{matrix}\right.\)
a/ \(x_1^3+x_2^3=40\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-40=0\)
\(\Leftrightarrow4^3-12\left(m-1\right)-40=0\Rightarrow m=3\)
b/ \(P=\left(x_1x_2\right)^2+5\left(x_1+x_2\right)^2-10x_1x_2+4\)
\(=\left(m-1\right)^2+5.4^2-10\left(m-1\right)+4\)
\(=m^2-12m+95\)
\(=\left(7-m\right)\left(5-m\right)+60\)
Do \(m\le5\Rightarrow\left\{{}\begin{matrix}7-m>0\\5-m\ge0\end{matrix}\right.\) \(\Rightarrow\left(7-m\right)\left(5-m\right)\ge0\)
\(\Rightarrow P\ge60\Rightarrow P_{min}=60\) khi \(m=5\)
\(5\left(x^2_1+x_2^2\right)=5\left(x_1^2+x_2^2+2x_1x_2-2x_1x_2\right)=5\left(x_1+x_2\right)^2-10x_1x_2\)
Theo hệ thức vi-et ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-2m-6\\x_1x_2=m^2-3\end{matrix}\right.\)
\(P=5\left(x_1+x_2\right)-2x_1x_2=5\left(-2m-6\right)-2\left(m^2-3\right)\)
\(=-2m^2-10m-24\)
\(=-2\left[\left(m^2+5m+\frac{25}{4}\right)+\frac{23}{4}\right]\)
\(=-\frac{46}{4}-2\left(m+\frac{5}{2}\right)^2\le-\frac{46}{4}=-\frac{23}{2}\)
Vậy GTLN của P là \(-\frac{23}{2}\) khi \(m=-\frac{5}{2}\)
△= \(7^2+4.4.1=65\)
\(\Rightarrow x_1=\frac{7+\sqrt{65}}{8},x_2=\frac{7-\sqrt{65}}{8}\)
M = \(x_1^2+x_2^2=\left(\frac{7+\sqrt{65}}{8}\right)^2+\left(\frac{7-\sqrt{65}}{8}\right)^2=\frac{114+14\sqrt{65}+114-14\sqrt{65}}{64}=\frac{228}{64}=\frac{57}{16}\)
\(\Delta=49-4.\left(-1\right).4=65>0\) => pt có 2 n0 pb
\(Vi-et\Rightarrow\left\{{}\begin{matrix}x_1+x_2=\frac{7}{4}\\x_1x_2=-\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow M=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\left(\frac{7}{4}\right)^2-2.\left(-\frac{1}{4}\right)=\frac{57}{16}\)
áp dụng hệ thức vi ét ta có : \(\left\{{}\begin{matrix}x_1x_2=7\\x_1+x_2=3\end{matrix}\right.\)
ta có : \(\left(3x_1+x_2\right)\left(3x_2+x_1\right)=9x_1x_2+3x_1^2+3x_2^2+x_1x_2\)
\(=10x_1x_2+3\left(x_1^2+x_2^2\right)=10x_1x_2+3\left(\left(x_1+x_2\right)^2-2x_1x_2\right)\)
\(=10x_1x_2+3\left(x_1+x_2\right)^2-6x_1x_2=3\left(x_1+x_2\right)^2+4x_1x_2\)
\(=3.\left(3\right)^2+4\left(7\right)=55\)
Bài 2:
a: \(\text{Δ}=\left(4m+2\right)^2-4\left(4m+3\right)\)
\(=16m^2+16m+4-16m-12=16m^2-8\)
Để phương trình có hai nghiệm thì \(2m^2>=1\)
=>\(\left[{}\begin{matrix}m>=\dfrac{1}{\sqrt{2}}\\m< =-\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)
c: \(A=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)
\(=\left(4m+2\right)^3-3\cdot\left(4m+3\right)\left(4m+2\right)\)
\(=64m^3+96m^2+48m+8-3\left(16m^2+20m+6\right)\)
\(=64m^3+96m^2+48m+8-48m^2-60m-18\)
\(=64m^3+48m^2-12m-10\)
Áp dụng Viét: \(\left\{{}\begin{matrix}x_1+x_2=a\\x_1x_2=1\end{matrix}\right.\)
\(A=x_1^5+x_2^5=\left(x_1^2+x_2^2\right)\left(x_1^3+x_2^3\right)-x_1^2x_2^2\left(x_1+x_2\right)\\ \Leftrightarrow A=\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\left[\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\right]-a\\ \Leftrightarrow A=\left(a^2-2\right)\left(a^3-3a\right)-a\\ \Leftrightarrow A=a^5-5a^3+5a\)