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Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{1}{1}=1\\x_1x_2=\dfrac{c}{a}=-\dfrac{3}{1}=-3\end{matrix}\right.\)
a
\(A=x_1^2+x_2^2=x_1^2+2x_1x_2+x_2^2-2x_1x_2\)
\(=\left(x_1+x_2\right)^2-2x_1x_2=1^2-2.\left(-3\right)=1+6=7\)
b
\(B=x_1^2x_2+x_1x_2^2=x_1x_2\left(x_1+x_2\right)=\left(-3\right).1=-3\)
c
\(C=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_2}{x_1x_2}+\dfrac{x_1}{x_1x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{-3}=-\dfrac{1}{3}\)
d
\(D=\dfrac{x_2}{x_1}+\dfrac{x_1}{x_2}=\dfrac{x_2^2}{x_1x_2}+\dfrac{x_1^2}{x_1x_2}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=\dfrac{1^2-2.\left(-3\right)}{-3}=\dfrac{1+6}{-3}=\dfrac{7}{-3}=-\dfrac{3}{7}\)
Δ=2^2-4(m-3)
=4-4m+12=16-4m
Để phương trình có hai nghiệm thì 16-4m>=0
=>m<=4
m(x1^3+x2^3)+(x1*x2)^2=9
=>m[(x1+x2)^3-3x1x2(x1+x2)]+(m-3)^2=9
=>m[(-2)^3-3(m-3)*(-2)]+(m-3)^2=9
=>m[-8+6(m-3)]+(m-3)^2=9
=>m^2-6m+9-9+m[-8+6m-18]=0
=>m^2-6m+m[6m-26]=0
=>m^2-6m+6m^2-26m=0
=>7m^2-32m=0
=>m=0(nhận) hoặc m=32/7(loại)
△ = 4-4m+12 = 16-4m
ptr có 2 ngh \(x_1;x_2\) ⇔△≥0 ⇔m≤4
Theo viet: \(x_1+x_2=-2;x_1x_2=m-3\)
Ta có\(m\left(x_1^3+x_2^3\right)+x_1^2x_2^2=9\\ \Leftrightarrow m\left(x_1+x_2\right)\left(x_1^2+x_2^2-x_1x_2\right)+x_1^2x_2^2=9\\ \Leftrightarrow m\left(-2\right)\left(x_1+x_2\right)^2-3x_1x_2m\left(-2\right)+\left(x_1x_2\right)^2=9\\ \Leftrightarrow-8m+6m\left(m-3\right)+\left(m-3\right)^2=9\\ \Leftrightarrow6m^2-18m-8m+m^2-6m+9=9\Leftrightarrow7m^2-32m=0\\ \)
⇔m=0(tm) hoặc m=32/7 (loại)
kl....
Δ=2^2-4(m-3)
=4-4m+12=16-4m
Để phương trình có hai nghiệm phân biệt thì 16-4m>0
=>m<4
m(x1^3+x2^3)+(x1*x2)^2=9
=>m[(x1+x2)^3-3x1x2(x1+x2)]+(m-3)^2=9
=>m[(-2)^3-3(m-3)*(-2)]+(m-3)^2=9
=>m[-8+6(m-3)]+(m-3)^2=9
=>m^2-6m+9-9+m[-8+6m-18]=0
=>m^2-6m+m[6m-26]=0
=>m^2-6m+6m^2-26m=0
=>7m^2-32m=0
=>m=0(nhận) hoặc m=32/7(loại)
vậy nếu cho x1x2 là hai nghiệm thì sao ạ ( không có phân biệt)
1, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=-5\\x_1x_2=-6\end{matrix}\right.\)
\(A=\left(x_1-2x_2\right)\left(2x_1-x_2\right)\\ =2x_1^2-4x_1x_2-x_1x_2+2x_1^2\\ =2\left(x_1^2+x_2^2\right)-5x_1x_2\\ =2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5x_1x_2\\ =2\left(-5\right)^2-4.\left(-6\right)-5.\left(-6\right)\\ =104\)
2, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=-3\end{matrix}\right.\)
\(B=x_1^3x_2+x_1x_2^3\\ =x_1x_2\left(x_1^2+x_2^2\right)\\ =\left(-3\right)\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =\left(-3\right)\left[5^2-2\left(-3\right)\right]\\ =-93\)
\(\text{Δ}=\left(-m\right)^2-4\left(m-5\right)\)
\(=m^2-4m+20\)
\(=m^2-4m+4+16=\left(m-2\right)^2+16>0\forall m\)
=>Phương trình luôn có 2 nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-m\right)}{1}=m\\x_1\cdot x_2=\dfrac{c}{a}=m-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1+2x_2=1\\x_1+x_2=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_2=1-m\\x_1=m-x_2=m-1+m=2m-1\end{matrix}\right.\)
\(x_1\cdot x_2=m-5\)
=>\(\left(1-m\right)\left(2m-1\right)=m-5\)
=>\(2m-1-2m^2+m-m+5=0\)
=>\(-2m^2+2m+4=0\)
=>\(m^2-m-2=0\)
=>(m-2)(m+1)=0
=>\(\left[{}\begin{matrix}m-2=0\\m+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=2\left(nhận\right)\\m=-1\left(nhận\right)\end{matrix}\right.\)
\(\Delta'=m-1\ge0\Rightarrow m\ge1\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m^2-m+1\end{matrix}\right.\)
\(A=x_1^3+x_2^3-2\left(x_1+x_2\right)\)
\(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-2\left(x_1+x_2\right)\)
\(=8m^3-3.2m\left(m^2-m+1\right)-4m\)
\(=2m^3+6m^2-10m\)
\(=2\left(m^3+3m^2-5m+1\right)-2\)
\(=2\left(m-1\right)\left[\left(m^2-1\right)+4m\right]-2\)
Do \(m\ge1\Rightarrow\left\{{}\begin{matrix}m-1\ge0\\\left(m^2-1\right)+4m>0\end{matrix}\right.\)
\(\Rightarrow2\left(m-1\right)\left[\left(m^2-1\right)+4m\right]\ge0\)
\(\Rightarrow A\ge-2\)
\(A_{min}=-2\) khi \(m=1\)
\(A=\dfrac{\left(x_1+x_2\right)^2+3x_1x_2}{4x_1x_2\left(x_1+x_2\right)}=\dfrac{9+3}{4\cdot1\left(-3\right)}=\dfrac{12}{-12}=-1\)
\(\text{Δ}=\left(-5\right)^2-4\cdot1\cdot\left(m+2\right)\)
\(=25-4m-8=-4m+17\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>-4m+17>0
=>-4m>-17
=>\(m< \dfrac{17}{4}\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-5\right)}{1}=5\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{m+2}{1}=m+2\end{matrix}\right.\)
\(P=x_1^2\cdot x_2+x_1\cdot x_2^2-x_1^2\cdot x_2^2-4\)
\(=x_1x_2\left(x_1+x_2\right)-\left(x_1x_2\right)^2-4\)
\(=5\left(m+2\right)-\left(m+2\right)^2-4\)
\(=5m+10-m^2-4m-4-4\)
\(=-m^2+m+2\)
\(=-\left(m^2-m-2\right)\)
\(=-\left(m^2-m+\dfrac{1}{4}-\dfrac{9}{4}\right)\)
\(=-\left(m-\dfrac{1}{2}\right)^2+\dfrac{9}{4}< =\dfrac{9}{4}\forall m\)
Dấu '=' xảy ra khi \(m=\dfrac{1}{2}\)
\(\Delta=25-4\left(m+2\right)=17-4m>0\Rightarrow m< \dfrac{17}{4}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=m+2\end{matrix}\right.\)
\(P=x_1x_2\left(x_1+x_2\right)-\left(x_1x_2\right)^2-4\)
\(=5\left(m+2\right)-\left(m+2\right)^2-4\)
\(=-\left[\left(m+2\right)-\dfrac{5}{2}\right]^2+\dfrac{9}{4}\le\dfrac{9}{4}\)
\(P_{max}=\dfrac{9}{4}\) khi \(m+2=\dfrac{5}{2}\Rightarrow m=\dfrac{1}{2}\)