3x(8^2-2(2^5-1))=2022

=>3x(64-2*31)=2022

=>3x=1011

=>x=337

3x[8² - 2(2⁵ - 1)] = 2022

3x[64 - 2(32 - 1)] = 2022

3x(64 - 2.31) = 2022

3x(64 - 62) = 2022

3x.2 = 2022

6x = 2022

x = 2022 : 6

x = 337

\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)

=13/54

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

18. iron-doing
19. wakes
còn lời giải chi tiết thì... mk ko biết .-.

18. help somebody do sth: giúp ai làm gì

hate doing sth: ghét làm gì

19. wakes - Đặt điều kiện trong tương lai thì mệnh đề điều kiện dùng thì hiện tại: KHi Harry thức giấc, chúng tôi sẽ nói cho anh ấy biết tin tức tốt lành này.

\(P=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)

Vì \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\)

=>P_min=(x-1)²+4=4

<=>(x-1)²=0

<=>x-1=0

<=>x=1

Vậy P_min=4 khi x=1

----------------------------------------------------------

\(Q=2x^2-6x=2\left(x^2-3x\right)=2\left[x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2\right]-\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow2\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)

=>Q_min=\(2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}=-\frac{9}{2}\)

<=>\(2\left(x-\frac{3}{2}\right)^2=0\)

<=>\(\left(x-\frac{3}{2}\right)^2=0\)

<=>\(x-\frac{3}{2}=0\)

<=>\(x=\frac{3}{2}\)

Vậy Q_min=\(-\frac{9}{2}\) khi \(x=\frac{3}{2}\)

Cảm ơn bạn nha

\(A=\dfrac{6x^2+21x+22}{x^2+4x+4}\)

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

\(=6+\dfrac{-3x-2}{\left(x+2\right)^2}\)

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

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

-Đặt \(a=\dfrac{1}{x+2}\) thì:

\(A=6-3a+4a^2=\left(2a\right)^2-2.2a.\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{87}{16}=\left(2a-\dfrac{3}{4}\right)^2+\dfrac{87}{16}\ge\dfrac{87}{16}\)

\(A_{min}=\dfrac{87}{16}\)\(\Leftrightarrow\left(2a-\dfrac{3}{4}\right)^2=0\Leftrightarrow2a-\dfrac{3}{4}=0\Leftrightarrow2a=\dfrac{3}{4}\)

\(\Leftrightarrow2.\dfrac{1}{x+2}=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{x+2}=\dfrac{3}{8}\Leftrightarrow x+2=\dfrac{8}{3}\Leftrightarrow x=\dfrac{2}{3}\)

\(a,=30x^3-6x^2-12x\\ b,=-6x^5-2x^3+x^2\)

a: \(=30x^3-6x^2-12x\)

b: \(=-6x^5-2x^3+x^2\)

Câu 5: B

Câu 6: C

Câu 7: A

Câu 8: A