a) \(H=\left(\dfrac{a-3\sqrt{a}}{a-2\sqrt{a}-3}-\dfrac{2a}{a-1}\right):\dfrac{1-\sqrt{a}}{a-2\sqrt{a}+1}\)

\(H=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)}-\dfrac{2a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]:\dfrac{1-\sqrt{a}}{\left(\sqrt{a}-1\right)^2}\)

\(H=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{2a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]:\dfrac{-\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}\)

\(H=\dfrac{a-\sqrt{a}-2a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{-1}{\sqrt{a}-1}\)

\(H=\dfrac{-a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot-\left(\sqrt{a}-1\right)\)

\(H=\dfrac{-\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot-\left(\sqrt{a}-1\right)\)

\(H=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\)

\(H=\sqrt{a}\)

b) Thay x = 2023 vào ta có: 

\(H=\sqrt{2023}\)

\(H=\dfrac{a^2\left(a^{-2}b^3\right)^2\cdot b^{-1}}{\left(a^{-1}\cdot b\right)\cdot a^{-5}\cdot b^{-2}}\)

\(=\dfrac{a^2\cdot a^{-4}\cdot b^6\cdot b^{-1}}{a^{-1-5}\cdot b^{1-2}}\)

\(=\dfrac{a^{-2}\cdot b^5}{a^{-4}\cdot b^{-1}}=a^{-2+4}\cdot b^{5+1}=a^2b^6\)

\(H=\dfrac{a^2.a^{-4}.b^6.b^{-1}}{a^{-1}.b.a^{-5}.b^{-2}}=\dfrac{a^{2-4}.b^{6-1}}{a^{-1-5}.b^{1-2}}=\dfrac{a^{-2}.b^5}{a^{-6}.b^{-1}}=a^{-2-\left(-6\right)}.b^{5-\left(-1\right)}=a^4b^6\)

1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5

This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured

Bài 1: 

a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

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

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

Điều kiện: x ≠ 0; x ≠ 1 và x ≠ 3

\(a,ĐK:x\ne\pm1;x\ne0\\ M=\dfrac{1-x+2x}{\left(1+x\right)\left(1-x\right)}:\dfrac{1-x}{x}\\ M=\dfrac{x+1}{\left(x+1\right)\left(1-x\right)}\cdot\dfrac{x}{1-x}=\dfrac{x}{\left(1-x\right)^2}\\ b,ĐK:x\ge0;x\ne4\\ N=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ N=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

Tất cả đều phải tìm điều kiện

Sửa đề: a+3căn a

\(\dfrac{a+3\sqrt{a}}{\sqrt{a}+3}=\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\sqrt{a}+3}=\sqrt{a}\)

Rút gọn biểu thức: (a+b)\(^3\) – (a–b)\(^3\)

Ta có: \(\left(a+b\right)^3-\left(a-b\right)^3\)

= \(a^3+3a^2b+3ab^2+b^3-a^3-3a^2b+3ab^2-b^3\)

= \(6ab^2\)

(a+b)³-(a-b)³ = a³+3a²b+3ab²+b³-a³+3a²b-3ab²+b³

=6a²b+2b³ =2b(3a² +b²)