A = |\(x\) + 5| + 2023

|\(x\) + 5| ≥ 0 ⇒| \(x\) + 5| + 2023 ≥ 2023⇒ A(min) = 2023 xảy ra khi \(x\) = -5

B = (\(x+2\))² - 2023

(\(x\) + 2)2 ≥ 0 ⇒ (\(x\) + 2)2 ≥ - 2023 ⇒ A(min) = -2023  xảy ra khi \(x\) = -2

C = \(x^2\) - 6\(x\) + 20

C = (\(x^2\) - 3\(x\)) - ( 3\(x\) - 9) + 11

C = \(x\)(\(x-3\)) - 3(\(x\) -3) + 11

C = (\(x-3\))(\(x\)-3) + 11

C = (\(x-3\))² + 11

(\(x\) -3)² ≥ 0 ⇒ (\(x\) - 3)² + 11 ≥ 11 vậy C(min) = 11 xảy ra khi \(x=3\)

D = \(x^2\) + 10\(x\) - 25

D = \(x^2\) + 5\(x\) + 5\(x\) + 25 - 55

D = (\(x^2\) + 5\(x\)) + (5\(x\) + 25) - 50

D = \(x\)(\(x\) + 5) + 5(\(x\) + 5)  - 50

D = (\(x\) +5)(\(x\) + 5) - 50

D = ( \(x\) + 5)² - 50

(\(x+5\))² ≥ 0 ⇒ (\(x\) + 5)² - 50 ≥ -50 ⇒ D(min) = -50 xảy ra khi \(x\) = -5

-4\(x^3\) + 4\(x\) = 0

- 4\(x\) ( \(x^2\) - 1) = 0

\(\left[{}\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)

\(-4x^3+4x=0\)

Áp dụng công thức phương trình bậc 3, ta có:

\(a=-4,b=0,c=4,d=0\)

\(\Rightarrow\Delta=b^2-3ac=0^2-3\cdot-4\cdot4=0+48=48\)

\(\Rightarrow k=\dfrac{9abc-2b^3-27a^2d}{2\sqrt{\left|\Delta\right|^3}}\)

\(\Rightarrow k=\dfrac{9\cdot-4\cdot0\cdot4-2\cdot0^3-27\cdot\left(-4\right)^2\cdot0}{2\sqrt{\left|48\right|^3}}\)

\(\Rightarrow k=\dfrac{0}{2\sqrt{\left|48\right|^3}}=0\)

Vì Δ = 48 > 0 và k = 0 < 1

\(\Rightarrow x_1=\dfrac{2\sqrt{\Delta}cos\left(\dfrac{arccos\left(k\right)}{3}\right)-b}{3a}\)

\(x_1=\dfrac{2\sqrt{48}cos\left(\dfrac{arccos\left(0\right)}{3}\right)-0}{3\cdot-4}\)

\(x_1=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{\pi}{2}}{3}\right)}{-12}\)

\(x_1=\dfrac{8\sqrt{3}cos\left(\dfrac{\pi}{6}\right)}{-12}\)

\(x_1=\dfrac{8\sqrt{3}\cdot\dfrac{\sqrt{3}}{2}}{-12}\)

\(x_1=\dfrac{\dfrac{8\sqrt{3}\cdot\sqrt{3}}{2}}{-12}\)

\(x_1=\dfrac{4\cdot3}{-12}=\dfrac{12}{-12}=-1\)

\(\Rightarrow x_2=\dfrac{2\sqrt{\Delta}cos\left(\dfrac{arccos\left(k\right)}{3}-\dfrac{2\pi}{3}\right)-b}{3a}\)

\(x_2=\dfrac{2\sqrt{48}cos\left(\dfrac{arccos\left(0\right)-2\pi}{3}\right)-0}{3\cdot-4}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{arccos\left(0\right)-2\pi}{3}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{\pi}{2}-2\pi}{3}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{-3\pi}{2}}{3}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{-3\pi}{6}\right)}{-12}=\dfrac{8\sqrt{3}cos\left(\dfrac{-\pi}{2}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}\cdot0}{-12}=0\)

\(\Rightarrow x_3=\dfrac{2\sqrt{\Delta}cos\left(\dfrac{arccos\left(k\right)}{3}+\dfrac{2\pi}{3}\right)-b}{3a}\)

\(x_3=\dfrac{2\sqrt{48}cos\left(\dfrac{arccos\left(0\right)+2\pi}{3}\right)-0}{3\cdot-4}\)

\(x_3=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{\pi}{2}+2\pi}{3}\right)}{-12}=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{5\pi}{2}}{3}\right)}{-12}\)

\(x_3=\dfrac{8\sqrt{3}cos\left(\dfrac{5\pi}{6}\right)}{-12}=\dfrac{8\sqrt{3}\cdot\dfrac{-\sqrt{3}}{2}}{-12}\)

\(x_3=\dfrac{\dfrac{8\sqrt{3}\cdot-\sqrt{3}}{2}}{-12}\)

\(x_3=\dfrac{\dfrac{8\cdot-3}{2}}{-12}\)

\(x_3=\dfrac{\dfrac{-24}{2}}{-12}\)

\(x_3=\dfrac{-12}{-12}=1\)

Vậy: \(x_1=-1,x_2=0,x_3=1\)

Lời giải:
\((a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})=\frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}\)

$\Leftrightarrow 2018.\frac{1}{2018}=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$

$\Leftrightarrow 1=1+1+1+S$

$S=1-1-1-1=-2$

O x và y khác nhau ở điểm truc nên ta có phuong trình x +y bằng 65% tỉ lệ hành hóa

Lời giải:
$C(2)=a.2^2+b.2+c=4a+2b+c$
$C(-1)=a(-1)^2+b(-1)+c=a-b+c$

$\Rightarrow C(2)+C(-1)=4a+2b+c+(a-b+c)=5a+b+2c=0$

$\Rightarrow C(-1)=-C(2)$

$\Rightarrow C(2)C(-1)=-C(2)^2\leq 0$ 

Ta có đpcm.

\(x^2+4x-5=0\)  ( a = 1 ; b^, = 2 ; c = - 5 ) 

\(\Delta^,=\left(b^,\right)^2-a.c\)

      \(=2^2-1.\left(-5\right)\)

      \(=9>0\)

pt có 2 n_o phân biệt :

\(x_1=\dfrac{-b^,+\sqrt{\Delta}}{a}=1\)

\(x_2=\dfrac{-b^,-\sqrt{\Delta}}{a}=-5\)

Vậy pt có n_o : x = 1

                      x = - 5

x^2+4x-4-1=0

x-1)(x+1)+4(x-1)=0

(x-1)(x+5)=0

=>x=1 ;x=-5

Tham khảo :
 

3​.98​.1615​.....100009999​

=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{99.101}{100.100}=2.21.3​.3.32.4​.4.43.5​.....100.10099.101​

=\dfrac{\left(1.2.3.....99\right)}{\left(2.3.4.....100\right)}.\dfrac{\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)}=(2.3.4.....100)(1.2.3.....99)​.(2.3.4.....100)(3.4.5.....101)​

=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}=1001​.2101​=200101​