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\(\left\{{}\begin{matrix}x_1+x_2=-2019\\x_1x_2=2\end{matrix}\right.\) \(\left\{{}\begin{matrix}x_3+x_4=-2020\\x_3x_4=2\end{matrix}\right.\)
\(Q=\left(x_1+x_3\right)\left(x_1+x_4\right)\left(x_2-x_3\right)\left(x_2-x_4\right)\)
\(Q=\left(x_1^2+x_1x_4+x_1x_3+x_3x_4\right)\left(x_2^2-x_2x_4-x_2x_3+x_3x_4\right)\)
\(Q=\left(x_1^2+x_1\left(x_3+x_4\right)+x_3x_4\right)\left(x_2^2-x_2\left(x_3+x_4\right)+x_3x_4\right)\)
\(Q=\left(x_1^2-2020x_1+2\right)\left(x_2^2+2020x_2+2\right)\)
Mặt khác do \(x_1\); \(x_2\) là nghiệm của \(x^2+2019x+2=0\) nên:
\(\left\{{}\begin{matrix}x_1^2+2019x_1+2=0\\x_2^2+2019x_2+2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1^2+2=-2019x_1\\x_2^2+2=-2019x_2\end{matrix}\right.\)
\(\Rightarrow Q=\left(-2019x_1-2020x_1\right)\left(-2019x_2+2020x_2\right)\)
\(Q=-4039x_1.x_2=-4039.2=-8078\)
Mình nghĩ thế này bạn à:
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
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)
\(a>0\)
\(\Rightarrow x_3=-1+2a;x_4=-1-2a\)
\(\Rightarrow S=\left(1+2a\right)^2+\left(1-2a\right)^2+\left(-1+2a\right)^2+\left(-1-2a\right)^2\)
\(=2\left(1+2a\right)^2+2\left(1-2a\right)^2\)
\(\Rightarrow S< +\infty\)
Đặt \(x^2=t\ge0\Rightarrow t^2-2mt+2m+6=0\) (1)
Để pt đã cho có 4 nghiệm phân biệt \(\Leftrightarrow\left(1\right)\) có 2 nghiệm dương phân biệt
\(\Rightarrow\left\{{}\begin{matrix}\Delta'=m^2-2m-6>0\\t_1+t_2=2m>0\\t_1t_2=2m+6>0\end{matrix}\right.\) \(\Rightarrow m>\sqrt{7}+1\)
Giả sử (1) có 2 nghiệm dương \(0< t_1< t_2\) và \(\Rightarrow\left[{}\begin{matrix}x^2=t_1\\x^2=t_2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=-\sqrt{t_2}\\x_2=-\sqrt{t_1}\\x_3=\sqrt{t_1}\\x_4=\sqrt{t_2}\end{matrix}\right.\) \(\Rightarrow\sqrt{t_2}-2\sqrt{t_1}-2\sqrt{t_1}+\sqrt{t_2}=0\)
\(\Leftrightarrow\sqrt{t_2}=2\sqrt{t_1}\Rightarrow t_2=4t_1\)
Kết hợp Viet: \(\left\{{}\begin{matrix}t_1+t_2=2m\\t_2=4t_1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}t_1=\frac{2m}{5}\\t_2=\frac{8m}{5}\end{matrix}\right.\)
\(t_1t_2=2m+6\Rightarrow\frac{16m^2}{25}=2m+6\)
\(\Rightarrow16m^2-50m-150=0\Rightarrow\left[{}\begin{matrix}m=5\\m=-\frac{15}{8}\left(l\right)\end{matrix}\right.\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2024\\x_1x_2=\dfrac{c}{a}=2\end{matrix}\right.\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_3+x_4=-\dfrac{b}{a}=-2025\\x_3x_4=\dfrac{c}{a}=2\end{matrix}\right.\)
\(A=\left(x_1+x_3\right)\left(x_2-x_3\right)\left(x_1+x_4\right)\left(x_2-x_4\right)\)
\(=\left(x_1+x_3\right)\left(x_2-x_4\right)\left(x_2-x_3\right)\left(x_1+x_4\right)\)
\(=\left(x_1x_2-x_1x_4+x_2x_3-x_3x_4\right)\left(x_1x_2+x_2x_4-x_3x_1-x_3x_4\right)\)
\(=\left(2-x_1x_4+x_2x_3-2\right)\left(2+x_2x_4-x_3x_1-2\right)\)
\(=\left(-x_1x_4+x_2x_3\right)\left(x_2x_4-x_3x_1\right)\)
\(=-x_1\cdot x_2\cdot x_4^2+x_1^2\cdot x_3\cdot x_4+x_2^2\cdot x_3\cdot x_4-x_2\cdot x_1\cdot x_3^2\)
\(=-2\cdot x_4^2+2x_1^2+2x_2^2-2x_3^2\)
\(=2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-2\left[\left(x_3+x_4\right)^2-2x_3x_4\right]\)
\(=2\left[\left(-2024\right)^2-2\cdot2\right]-2\left[\left(-2025\right)^2-2\cdot2\right]\)
\(=2\cdot2024^2-8-2\cdot2025^2+8=2\left(2024^2-2025^2\right)\)
\(=2\left(2024-2025\right)\left(2024+2025\right)=-8098\)
Em không biết giải phương trình
A=32788802
A=-8098
A=-8089
Theo x1x2= 2,x3 + x4=-2025 ,x3x4 = 2
Biến đổi
A=(x1 + x3)(x1 + x4)(x2 - x3)(x2 -x4)
Rút gọn
A=x1x2
Mà x1x2=2
A=2
A=0
Ta có:
x₁,x₂ là n° của:
x²+2014x+2=0
x₃,x₄ là n° của:
x₂+2025x+2=0
Theo vi ét:
x₁+x₂=-2024; x₁x₂ =2
x₃+x₄=-2025; x₃x₄=2
A = (x₁ + x₃)(x₂ − x₃)(x₁ + x₄)(x₂ − x₄)
A = [(x₁ + x₃)(x₁ + x₄)] · [(x₂ − x₃)(x₂ − x₄)] Xét từng nhóm: Ta có: (x₁ + x₃)(x₁ + x₄) = x₁² + x₁(x₃ + x₄) + x₃x₄ = x₁² + x₁(-2025) + 2 = x₁² - 2025x₁ + 2
Vì x₁là n°của:
x²+2024x+2=0
=> x₁²=-2024x₁-2
Thay vào :
= (-2024x₁ - 2) - 2025x₁ + 2 = -4049x₁
Tương tự: (x₂ − x₃)(x₂ − x₄) = x₂² - x₂(x₃ + x₄) + x₃x₄ = x₂² + 2025x₂ + 2
Mà: x²₂=-2024x₂-2
=>=(-2024x₂-2)+2025x₂+2
=x₂
Suy ra: A = (-4049x₁)·(x₂) = -4049(x₁x₂) = -4049·2 = -8098
-8098