Ta có: 

\(1+2+3+...+n\)

Số lượng số hạng là: `(n-1):1+1=n` (số hạng) 

Tổng của dãy số là: `(n+1)*n/2` 

Áp dụng ta có:

\(\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+....+\dfrac{1}{1+2+3+...+100}\\ =\dfrac{1}{\dfrac{3\cdot\left(3+1\right)}{2}}+\dfrac{1}{\dfrac{4\cdot\left(4+1\right)}{2}}+...+\dfrac{1}{\dfrac{100\cdot\left(100+1\right)}{2}}\\ =\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{100\cdot101}\\ =2\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{100\cdot101}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\\ =2\cdot\dfrac{98}{303}\\ =\dfrac{196}{303}\)

Từ 2 đến 201 số lượng số hạng là: (201 - 2) : 1 + 1 = 200 (số hạng) 

Số lượng cặp là: 200 : 2 = 100 (cặp)

1 - 2 + 3 - 4 + 5 - ... + 199 - 200 + 201 

= 1 + (-2 + 3) + (-4 + 5) + ... + (-198 + 199) + (-200 + 201) 

= 1 + 1 + 1 + ... + 1 + 1

= 1 + 100*1 

= 1 + 100

= 101 

\(D=4-4^2+4^3-4^4+...+4^{2024}\\ 4D=4^2-4^3+4^4-4^5+...+4^{2025}\\ 4D+D=\left(4^2-4^3+4^4-4^5+...+4^{2025}\right)+\left(4-4^2+4^3-4^4+...+4^{2024}\right)\\ 5D=4^{2025}+4\\ D=\dfrac{4^{2025}+4}{5}\)

\(g.x^3-3x^2+3x-1=0\\ \Leftrightarrow\left(x-1\right)^3=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\\ h.x\left(2x-7\right)-4x+14=0\\ \Leftrightarrow x\left(2x-7\right)-2\left(2x-7\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=7\\x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\\ k.\left(2x-5\right)^2\left(x+2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\\ l.x\left(2x-9\right)=3x\left(x-5\right)\\ \Leftrightarrow3x^2-15x-2x^2+9x=0\\ \Leftrightarrow x^2-6x=0\\ \Leftrightarrow x\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\\ m.\left(x^2-2x+1\right)-4=0\\ \Leftrightarrow\left(x-1\right)^2=2^2\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+1=3\\x=-2+1=-1\end{matrix}\right.\)

a: (3x-2)(4x+5)=0

=>\(\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

c: \(\left(4x+2\right)\left(x^2+1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên 4x+2=0

=>4x=-2

=>\(x=-\dfrac{1}{2}\)

d: (2x+7)(x-5)(5x+1)=0

=>\(\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)

f: \(\left(x^2-4\right)\left(x-2\right)\left(3-2x\right)=0\)

=>\(\left(x-2\right)^2\cdot\left(x+2\right)\left(3-2x\right)=0\)

=>\(\left[{}\begin{matrix}x-2=0\\x+2=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Đặt:

\(A=1+3+3^2+...+3^{2023}\\ 3A=3+3^2+3^3+...+3^{2024}\\ 3A-A=\left(3+3^2+3^3+..+3^{2024}\right)-\left(1+3+3^2+...+3^{2023}\right)\\ 2A=3^{2024}-1\\ A=\dfrac{3^{2024}-1}{2}\)

`A =` \(1+3+3^2+...+3^{2023}\)

\(3A=3+3^2+3^3+...+3^{2024}\)

`3A - A =` \(\left(3+3^2+3^3+...+3^{2024}\right)-\left(1+3+3^2+...+3^{2023}\right)\)

`2A =` \(3^{2024}-1\)

`A =` \(\dfrac{3^{2024}-1}{2}\)

\(\left(4x+2\right)\left(x^2+1\right)=0\)(1) 

Ta có: `x^2>=0` với mọi x

`=>x^2+1>=1>0` với mọi x

`=>x^2+1≠0`

\(\left(1\right)\Leftrightarrow4x+2=0\\ \Leftrightarrow4x=-2\\ \Leftrightarrow x=-\dfrac{2}{4}=-\dfrac{1}{2}\)

`(4x + 2)(x^2 + 1) = 0`

Trường hợp 1: 

`4x + 2 = 0`

`<=> 4x = -2`

`<=> x =` \(-\dfrac{1}{2}\)

Trường hợp 2: 

`x^2 + 1 = 0`

`<=> x^2 = -1` (Không tồn tại `x`)

Vậy `x =` \(-\dfrac{1}{2}\)

\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{14}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{14}+\dfrac{5}{7}+1\\ =\dfrac{5}{7}\cdot\left(\dfrac{-2}{11}+\dfrac{-9}{14}+1\right)+1\\ =\dfrac{5}{7}\cdot\dfrac{27}{154}+1\\ =\dfrac{135}{1078}+1\\ =\dfrac{1213}{1078}\)

Mẹ mua 1kg cam có giá là:

52000 : 2 = 26000 (đồng)

Ba mua 1kg cam có giá là:

87000 : 3 = 29000 (đồng)

Ta có: 29000 > 26000 

=> Mẹ mua cam có giá rẻ hơn

ĐS: ... 

Giá 1 kg cam mẹ mua được là: 

`52000 : 2 = 26000` (đồng)

Giá 1kg cam ba mua được là: 

`87000 : 3 = 29000` (đồng)

Ta có: `26000 < 29000` (đồng)

Nên mẹ mua cam giá rẻ hơn

Mỗi kh cam mẹ mua rẻ hơn mỗi kg cam của ba là: 

`29000 - 26000 = 3000` (đồng)

Vậy ...

Ta có: 

\(A=3+3^2+3^3+...+3^{2024}\\ =\left(3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{2022}+3^{2023}+3^{2024}\right)\\ =12+3^3\cdot\left(1+3+3^2\right)+3^{2022}\cdot\left(1+3+3^2\right)\\ =12+13\cdot\left(3^3+...+3^{2022}\right)\)

=> A chia 13 dư 12 

\(A=1+4+4^2+...+4^{2021}\\ =\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{2019}+4^{2020}+4^{2021}\right)\\ =21+4^3\cdot\left(1+4+4^2\right)+...+4^{2019}\cdot\left(1+4+4^2\right)\\ =21+4^3\cdot21+...+4^{2019}\cdot21\\ =21\cdot\left(1+4^3+...+4^{2019}\right)\)

`x (2x - 9) = 3x(x - 5) `

`<=> 2x^2 - 9x = 3x^2 - 15x`

`<=> 3x^2 - 2x^2 - 15x + 9x =0`

`<=> x^2 - 6x = 0`

`<=> x(x-6) = 0`

`<=> x = 0` hoặc `x - 6 = 0`

`<=> x = 0` hoặc `x = 6`

Vậy ....