Ta có \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2-1+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\).

Từ đó \(1+\dfrac{1}{1.3}=\dfrac{2^2}{1.3}\); \(1+\dfrac{1}{2.4}=\dfrac{3^2}{2.4}\); \(1+\dfrac{1}{3.5}=\dfrac{4^2}{3.5}\); \(1+\dfrac{1}{4.6}=\dfrac{5^2}{4.6}\);...; \(1+\dfrac{1}{2022.2024}=\dfrac{2023^2}{2022.2024}\).

Suy ra \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2022.2024}\right)\)

\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}...\dfrac{2023^2}{2022.2024}\)

\(=\dfrac{2.2023}{2024}\) \(=\dfrac{2023}{1012}\)

Lời giải:
$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n(n+1)}=\frac{2022}{2023}$

$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{n(n+1)}=\frac{2022}{2023}$

$2[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{n(n+1)}]=\frac{2022}{2023}$

$2[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n(n+1)}]=\frac{2022}{2023}$
$2(\frac{1}{2}-\frac{1}{n+1})=\frac{2022}{2023}$

$1-\frac{2}{n+1}=1-\frac{1}{2023}$

$\Rightarrow \frac{2}{n+1}=\frac{1}{2023}$

$\Rightarrow n+1=2.2023=4046$

$\Rightarrow n=4045$

\(=\left(\dfrac{5}{7}+\dfrac{9}{7}\right)\times\dfrac{3}{2}=2\times\dfrac{3}{2}=3\)

\(=\dfrac{4}{5}\times\left(6+1+3\right)=\dfrac{4}{5}\times10=\dfrac{40}{5}=8\)

a) 5/7 x 3/2 + 3/2 x 9/7 
= (5/7 + 9/7) x 3/2
= 2 x 3/2
= 6/2 = 3
b) 4/5 + 4/5 x 6 + 3 x 4/5
= 4/5 x 1 + 4/5 x 6 + 3 x 4/5
= 4/5 x (1 + 6 + 3)
= 4/5 x 10
= 40/5 = 8

=2 × 3/2

b. =4/5 × (6 + 1 + 3)

=40/5

\(x+2y=3z-14\\ =>x+2y-3z=-14\)

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}\\ =>\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{18}\)

áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{18}=\dfrac{x+2y-3z}{3+8-18}=\dfrac{-14}{-7}=2\)

\(=>x=2\cdot3=6\\ y=2\cdot4=8\\ z=2\cdot6=12\)

\(x+2y=3z-14\\ =>x+2y-3z=-14\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{x+2y-3z}{3+2.4-3.6}=\dfrac{-14}{-7}=2\\ =>\left\{{}\begin{matrix}x=2.3=6\\y=2.4=8\\z=2.6=12\end{matrix}\right.\)

LÀM ƠN GIÚP MIK ĐI MÀ, NĂN NỈ CÁC BẠN ĐÓ!!!!!!!!!!!!!!!!!!!!!!

`2/5 - 3/8 + 7/6 : 7/5`

`= 2/5 - 3/8 + 7/6 xx 5/7`

`= 2/5 - 3/8 +5/6`

`= 1/40+5/6`

`=103/120`

\(\dfrac{16}{27}xX=\dfrac{10}{27}-\dfrac{6}{27}\)

\(\dfrac{16}{27}xX=\dfrac{4}{27}\)

\(X=\dfrac{4}{27}:\dfrac{16}{27}\)

X=\(\dfrac{1}{4}\)

16/27x X+2/9=10/27

16/27x X=10/27-2/9

16/27x X=4/27

X=4/27:16/27

X=1/4

x = 1/4

Ngại làm lắm