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PT1: \(x^2+2013x+2=0.\)Theo Hệ thức Vi-ét ta có: \(x_1+x_2=-2013\\ x_1.x_2=2\)
Tương tự với PT2 ta có:\(x_3+x_4=-2014\\ x_3.x_4=2\)
\(Q=\left[\left(x_1+x_3\right)\left(x_2-x_4\right)\right]\left[\left(x_2_{ }-x_3\right)\left(x_1+x_4\right)\right]\)
\(Q=\left(x_1.x_2+x_2.x_3-x_1.x_4-x_3.x_4\right)\left(x_1.x_2+x_2.x_4-x_1.x_3-x_3.x_4\right)\)
\(Q=\left(2+x_2.x_3-x_1.x_4-2\right)\left(2+x_2.x_4-x_1.x_3-2\right)\)
\(Q=\left(x_2.x_3-x_1.x_4\right)\left(x_2.x_4-x_1.x_3\right)\)
\(Q=x_2.x_3.x_4-x_3.x_1.x_2-x_4.x_1.x_2+x_1.x_3.x_4\)
\(Q=2x_2-2x_3-2x_4+2x_1\)
\(Q=2\left(x_1+x_2\right)-2\left(x_3+x_4\right)\)
\(Q=2.\left(-2013\right)-2.\left(-2014\right)\)
\(Q=2\)
Bài này hay quá. Chúc bạn học tốt nhé
\(\left\{{}\begin{matrix}x_1+x_2=-2019\\x_1x_2=2\end{matrix}\right.\) \(\left\{{}\begin{matrix}x_3+x_4=-2020\\x_3x_4=2\end{matrix}\right.\)
\(Q=\left(x_1+x_3\right)\left(x_1+x_4\right)\left(x_2-x_3\right)\left(x_2-x_4\right)\)
\(Q=\left(x_1^2+x_1x_4+x_1x_3+x_3x_4\right)\left(x_2^2-x_2x_4-x_2x_3+x_3x_4\right)\)
\(Q=\left(x_1^2+x_1\left(x_3+x_4\right)+x_3x_4\right)\left(x_2^2-x_2\left(x_3+x_4\right)+x_3x_4\right)\)
\(Q=\left(x_1^2-2020x_1+2\right)\left(x_2^2+2020x_2+2\right)\)
Mặt khác do \(x_1\); \(x_2\) là nghiệm của \(x^2+2019x+2=0\) nên:
\(\left\{{}\begin{matrix}x_1^2+2019x_1+2=0\\x_2^2+2019x_2+2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1^2+2=-2019x_1\\x_2^2+2=-2019x_2\end{matrix}\right.\)
\(\Rightarrow Q=\left(-2019x_1-2020x_1\right)\left(-2019x_2+2020x_2\right)\)
\(Q=-4039x_1.x_2=-4039.2=-8078\)
a.
Ta co:
\(\orbr{\begin{cases}x^2-2x-3=0\left(1\right)\left(x\ge0\right)\\x^2+2x-3=0\left(2\right)\left(x< 0\right)\end{cases}}\)
(1)\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\left(l\right)\\x=3\left(n\right)\end{cases}}\)
(2)\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(l\right)\\x=-3\left(n\right)\end{cases}}\)
b.
Ta lai co:
\(\orbr{\begin{cases}x^2-2x+1-4a^2=0\left(3\right)\left(x\ge0\right)\\x^2+2x+1-4a^2=0\left(4\right)\left(x< 0\right)\end{cases}}\)
Xet (3)
De phuong trinh dau co 4 nghiem thi PT(3) co nghiem
\(\Rightarrow\Delta^`>0\)
\(\Leftrightarrow4a^2>0\)
\(\Leftrightarrow a>0\)
\(\Rightarrow x_1=1+2a;x_2=1-2a\)
Tuong tu
(4)
\(a>0\)
\(\Rightarrow x_3=-1+2a;x_4=-1-2a\)
\(\Rightarrow S=\left(1+2a\right)^2+\left(1-2a\right)^2+\left(-1+2a\right)^2+\left(-1-2a\right)^2\)
\(=2\left(1+2a\right)^2+2\left(1-2a\right)^2\)
\(\Rightarrow S< +\infty\)
Vì P(x) là đa thức bậc 4 và có 4 nghiệm x1 , x2 , x3 , x4 nên P(x) có thể viết thành : \(P\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\)
Xét : \(Q\left(x\right)=x^2-4=\left(x-2\right)\left(x+2\right)=\left(2-x\right)\left(-2-x\right)\)
Ta có \(Q\left(x_1\right)=\left(2-x_1\right)\left(-2-x_1\right)\); \(Q\left(x_2\right)=\left(2-x_2\right)\left(-2-x_2\right)\);
\(Q\left(x_3\right)=\left(2-x_3\right)\left(-2-x_3\right)\) ; \(Q\left(x_4\right)=\left(2-x_4\right)\left(-2-x_4\right)\)
Suy ra : \(T=Q\left(x_1\right).Q\left(x_2\right).Q\left(x_3\right).Q\left(x_4\right)\)
\(=\left[\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\right].\left[\left(-2-x_1\right)\left(-2-x_2\right)\left(-2-x_3\right)\left(-2-x_4\right)\right]\)
\(=P\left(2\right).P\left(-2\right)=-5.3=-15\)
Vậy T = -15
Vì P(x) có các nghiệm là x1 , x2 , x3 , x4 nên P(x) có thể viết được dưới dạng : \(P\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\)
Ta có : \(Q\left(x\right)=x^2-4=\left(x-2\right)\left(x+2\right)=\left(2-x\right)\left(-2-x\right)\)
Xét : \(Q\left(x_1\right)=\left(2-x_1\right)\left(-2-x_1\right)\) ; \(Q\left(x_2\right)=\left(2-x_2\right)\left(-2-x_2\right)\)
\(Q\left(x_3\right)=\left(2-x_3\right)\left(-2-x_3\right)\) ; \(Q\left(x_4\right)=\left(2-x_4\right)\left(-2-x_4\right)\)
Suy ra :
\(T=Q\left(x_1\right).Q\left(x_2\right).Q\left(x_3\right).Q\left(x_4\right)=\left[\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\right]\left[\left(-2-x_1\right)\left(-2-x_2\right)\left(-2-x_3\right)\left(-2-x_4\right)\right]\)
\(=P\left(2\right).P\left(-2\right)\)
Bạn thay P(2) và P(-2) vào và tính nhé :)
Đa thức \(P\left(x\right)=x^4-5x^2-2x+3\)có bốn nghiệm là \(x_1;x_2;x_3;x_4\)nên P(x) có dạng \(\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\)(do P(x) là đa thức bậc bốn)
Ta có: \(Q\left(x\right)=x^2-3=\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\)
\(\Rightarrow T=Q\left(x_1\right).Q\left(x_2\right).Q\left(x_3\right).Q\left(x_4\right)\)
\(=\left[\left(x_1-\sqrt{3}\right)\left(x_2-\sqrt{3}\right)\left(x_3-\sqrt{3}\right)\left(x_4-\sqrt{3}\right)\right]\)
\(\left[\left(x_1+\sqrt{3}\right)\left(x_2+\sqrt{3}\right)\left(x_3+\sqrt{3}\right)\left(x_4+\sqrt{3}\right)\right]\)
\(=P\left(\sqrt{3}\right).P\left(-\sqrt{3}\right)=\left(-3-2\sqrt{3}\right)\left(-3+2\sqrt{3}\right)\)
\(=\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)=9-12=-3\)
Vậy \(T=Q\left(x_1\right).Q\left(x_2\right).Q\left(x_3\right).Q\left(x_4\right)=-3\)
what the đề yêu cầu ?