Bài 6:

\(A=3+\left(3^2+3^3\right)+\left(3^4+3^5\right)+...+\left(3^{2020}+3^{2021}\right)\\ A=3+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^{2021}\left(1+3\right)\\ A=3+4\left(3^2+3^4+...+3^{2021}\right)⋮̸4\left(3⋮̸4\right)\)

Bài 1

a) A=3¹+3²+...+3²⁰²²

A=(3¹+3²)+(3³+3⁴)+...+(3²⁰²¹+3²⁰²²)

A=4.3¹+4.3³+...+4.3²⁰²¹

A=4.(3¹+3³+...+3²⁰²¹)

⇒A⋮4

b) làm tương tự

c) A=3¹+3²+...+3²⁰²²

3A=3²+3³+...+3²⁰²³

3A-A=(3²+3³+...+3²⁰²³)-(3¹+3²+...+3²⁰²²)

2A=3²⁰²³-3¹

A=\(\dfrac{3^{2023}-3^1}{2}\)

Vậy A ko thể là SCP

thay x đi

Đặt \(P\left(x\right)=0\)

\(\Rightarrow x^2-4x+3=0\)

\(\Rightarrow x^2-x-3x+3=0\)

\(\Rightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy \(x=1,x=3\) là ngiệm của đa thức

P(x) = x² - 4x + 3

Thay x = 1 => 1² - 4 . 1 + 3 = 0

Vậy x = 1 là nghiệm của đa thức P(x)

Thay x = 3 => 3² - 4 . 3 + 3 = 0

Vậy x = 3 là nghiệm của đa thức P(x)

\(a,=2\sqrt{2a}+3\sqrt{2a}-4\sqrt{2a}=\sqrt{2a}\\ b,=3\sqrt{5}-3+\sqrt{5}=4\sqrt{5}-3\\ c,=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\\ d,=-7+5-2\cdot\dfrac{2}{3}+\dfrac{1}{3}\cdot3=-2-\dfrac{4}{3}+1=-\dfrac{7}{3}\)

is spoken

studied
spent

was broken

are made

was written

will be opened

was driving-was stopped

1 C

2 A

3 B

1 B => will

2 C => will turn

3 D => arrive

4 C => is going to have

5 C => ask

6 B => is going to leave

7 C => is going to come

8 B => is going to leave