a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)

\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)

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

=-a-1

b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)

\(=\left|3a-5\right|-2a+4\)

\(=5-3a-2a+4\)

=9-5a

c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)

\(=4a-3-\left|2a-1\right|\)

\(=4a-3-2a+1\)

\(=2a-2\)

d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)

\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)

\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)

\(=-a^2\)

a) \(\sqrt{4\left(a-3\right)^2}=2\left(a-3\right)=2a-6\)

b) \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)

c) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)

a: \(\sqrt{4\left(a-3\right)^2}=2\cdot\left(a-3\right)=2a-6\)

b: \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)

c: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)

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

\(=\left(a-b\right)\left(a^2+ab+b^2-a-b\right)\) 

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

\(y^3+8+y^2-4=\left(y+2\right)\left(y^2-2y+4\right)+\left(y-2\right)\left(y+2\right)\) 

\(=\left(y+2\right)\left(y^2-y+2\right)\)

\(64a^3+125b^3+5b\left(16a^2-25b^2\right)=\left(4a+5b\right)\left(16a^2-20ab+25b^2\right)+5b\left(....\right)\) 

\(=\left(4a+5b\right)16a^2\) 

\(\left(x-a\right)^4-\left(x+a\right)^4=\left(x^2-2ax+a^2\right)^2-\left(x^2+2ax+a^2\right)^2\) 

\(=2\left(a^2+x^2\right)-4ax\)

.(-4ax) nha

a) \(\left(2x-1\right)^4+\left(2x-3\right)^4=0\)

\(\Leftrightarrow\left(2x-1\right)^4+\left(2x-3\right)^4=0^4\)

\(\Leftrightarrow\left(2x-1\right)+\left(2x-3\right)=0\)

\(\Rightarrow\hept{\begin{cases}2x-1=0\\2x-3=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\left(0+1\right):2\\x=\left(0+3\right):2\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)

b) \(a^4-4a^3+12^2-16a+8=0\)

a)\(\left(2x-1\right)^4+\left(2x-3\right)^4=0\\ \)

do \(\left(2x-1\right)^4\ge0\)và \(\left(2x-3\right)^4\ge0\)

=> \(\hept{\begin{cases}\left(2x-1\right)^4=0\\\left(2x-3\right)^4=0\end{cases}\Rightarrow\hept{\begin{cases}2x-1=0\\2x-3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}\end{cases}}}\).

Vậy x vừa = 1/2 vừa =3/2 => vô lý => không tồn tại nghiệm

b) \(\dfrac{\left(n+1\right)!-n!}{\left(n+1\right)!+n!}=\dfrac{n!.\left(n+1\right)-n!}{n!\left(n+1\right)+n!}=\dfrac{n!\left(n+1-1\right)}{n!\left(n+1+1\right)}=\dfrac{n}{n+2}\)

a) \(\dfrac{8a^{n+2}+a^{n-1}}{16a^{n+4}+4a^{n+2}+a^n}=\dfrac{8a^{n-1+3}+a^{n-1}}{16a^{n-1+5}+4a^{n-1+3}+a^{n-1+1}}\)

\(=\dfrac{8a^{n-1}.a^3+a^{n-1}}{16a^{n-1}a^5+4a^{n-1}a^3+a^{n-1}a}=\dfrac{a^{n-1}\left(8a^3+1\right)}{a^{n-1}\left(16a^5+4a^3+a\right)}\)

\(=\dfrac{8a^3+1}{16a^5+4a^3+a}\)