\(A=\dfrac{6}{1\cdot4}+\dfrac{6}{4\cdot7}+...+\dfrac{6}{46\cdot49}\\ =2\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{46\cdot49}\right)\\=2\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{46}-\dfrac{1}{49}\right)\\ =2\cdot\left(1-\dfrac{1}{49}\right)\\ =\dfrac{96}{49}\)

Được 10 con bạn nhé

\(...\Rightarrow\left[{}\begin{matrix}\dfrac{2x-1}{3}=\dfrac{1}{2}\\\dfrac{2x-1}{3}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}+1\\2x=-\dfrac{3}{2}+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{5}{2}\\2x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

\(=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ =\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

\(\dfrac{\sqrt[]{10}-\sqrt[]{2}}{\sqrt[]{5}-1}+\dfrac{2-\sqrt[]{2}}{\sqrt[]{2}-1}\)

\(=\dfrac{\sqrt[]{2}\left(\sqrt[]{5}-1\right)}{\sqrt[]{5}-1}+\dfrac{\sqrt[]{2}\left(\sqrt[]{2}-1\right)}{\sqrt[]{2}-1}\)

\(=\sqrt[]{2}+\sqrt[]{2}=2\sqrt[]{2}\)

a) \(7^3.7^5=7^{3+5}=7^8\)

b)\(5^6.5^4=5^{6+4}=5^{10}\)

\(\left(x-36\right):\left(2.3^2\right)=2^2.3\)

\(\Rightarrow\left(x-36\right):18=12\)

\(\Rightarrow\left(x-36\right)=12.18\)

\(\Rightarrow x-36=216\)

\(\Rightarrow x=216+36\)

\(\Rightarrow x=252\)

(x-36):(2.3²)=2.2³

⇒ (x-36):(2.9)=2.8

⇒ (x-36):18=16

⇒ x-36=18.16=288

⇒ x=288+36=324

1574+(625-x)=2023

⇒ 625-x=2023-1574=449

⇒ x=625-449=176

Vậy x=176

1574 + (625 - x) = 2023

625 - x = 2023 - 1574

625 - x = 449

x = 625 - 449

x = 176

Gọi số đó là A ⇒ Số mới là: A5

Theo bài ra ta có:

A+3407= A5

⇒ A+3407= Ax10+5

⇒ 3407-5= 10xA-A

⇒ 3402= 9xA

⇒ A= 3402:9= 378

Vậy số cần tìm là: 378