import math

a = float(input("Nhập a: "))
b = float(input("Nhập b: "))
c = float(input("Nhập c: "))

d = b**2 - 4*a*c

if d > 0:
    x1 = (-b + math.sqrt(d)) / (2*a)
    x2 = (-b - math.sqrt(d)) / (2*a)
    print("Phương trình có hai nghiệm: x1 =", x1, "và x2 =", x2)
elif d == 0:
    x = -b / (2*a)
    print("Phương trình có nghiệm kép: x =", x)
else:
    print("Phương trình không có nghiệm thực."

\(a^3+b^3=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}\)

\(=\sqrt{6}-\sqrt{2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}=0\)

\(\Rightarrow a=-b\Rightarrow a^5+b^5=0\)

a: Xét (O) có

ΔANB nội tiếp

AB là đường kính

=>ΔANB vuông tại N

Xét tứ giác HMNB có

góc MHB+góc MNB=180 độ

=>HMNB là tư giác nội tiếp

b: Xét ΔMDN và ΔMAC có

góc MDN=góc MAC

góc DMN=góc AMC

=>ΔMDN đồng dạng với ΔMAC

=>MD/MA=MN/MC

=>MD*MC=MA*MN

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

\(x^2-x+1-m=0\left(1\right)\\ \text{PT có 2 nghiệm }x_1,x_2\\ \Leftrightarrow\Delta=1-4\left(1-m\right)\ge0\\ \Leftrightarrow4m-3\ge0\Leftrightarrow m\ge\dfrac{3}{4}\\ \text{Vi-ét: }\left\{{}\begin{matrix}x_1+x_2=1\\x_1x_2=1-m\end{matrix}\right.\\ \text{Ta có }5\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)-x_1x_2+4=0\\ \Leftrightarrow5\cdot\dfrac{x_1+x_2}{x_1x_2}-x_1x_2+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m-1+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m+3=0\\ \Leftrightarrow5+\left(1-m\right)\left(m+3\right)=0\\ \Leftrightarrow m^2+2m-8=0\\ \Leftrightarrow m^2-2m+4m-8=0\\ \Leftrightarrow\left(m-2\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=2\left(n\right)\\m=-4\left(l\right)\end{matrix}\right.\)

Vậy $m=2$

1 had gone away, I arrived at the party

2 reached the ground, the football match had ended

3 got to the airport, the plane had taken off

4 had done all the exercises, he went out with his friends

5 I arrived, Jack had left the office

6 he had done all his work, he went home

7 picked up their bags, they had eaten all the food

8 of 20, Madonna had become famous

9 I had heard the condition, I decided not to enter for the competition

10 Steven bought a new motorbike, he had saved enough money

1 A

2 A

3 A

4 A

ĐK: \(x\le3\)

Đặt \(a=\sqrt{3-x}\left(a\ge0\right)\) \(\Leftrightarrow3-a^2=x\)

Pttt: \(x^3+\left(3-a^2\right)\left(1+a\right)=4a\)

\(\Leftrightarrow x^3-a^3-a^2-a+3=0\)

\(\Leftrightarrow x^3-a^3+\left(3-a^2\right)-a=0\)

\(\Leftrightarrow\left(x-a\right)\left(x^2+ax+a^2\right)+\left(x-a\right)=0\)

\(\Leftrightarrow x-a=0\) \(\Leftrightarrow x=a\) \(\Leftrightarrow x=\sqrt{3-x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=3-x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x-3=0\end{matrix}\right.\)\(\Rightarrow x=\dfrac{-1+\sqrt{13}}{2}\)(thỏa)

Vậy...

Đkxđ:

y≥0

x-1≠0 => x≠1

Lời giải:
Để pt có 2 nghiê pb thì:

$\Delta'=1-(m-3)>0\Leftrightarrow m< 4$

Áp dụng định lý Viet: \(\left\{\begin{matrix} x_1+x_2=2\\ x_1x_2=m-3\end{matrix}\right.\)

Khi đó:
\(x_1^2-2x_2+x_1x_2=-12\)

\(\Leftrightarrow x_1^2-2(2-x_1)+x_1(2-x_1)=-12\)

\(\Leftrightarrow x_1=-2\Leftrightarrow x_2=2-x_1=4\)

$m-3=x_1x_2=(-2).4=-8$

$\Leftrightarrow m=-5$ (tm)