\(\dfrac{4x^3+4x^2}{x^2-1}=\dfrac{4x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x^2}{x-1}\)

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

d) Để phân thức \(\dfrac{4x^3+4x^2}{x^2-1}\) có nghĩa thì: \(x^2-1\ne0\Leftrightarrow x\ne\pm1\)

Khi đó: \(\dfrac{4x^3+4x^2}{x^2-1}=\dfrac{4x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x^2}{x-1}\)

e) Để phân thức \(\dfrac{b^2+b}{a+ab}\) có nghĩa thì: \(a+ab\ne0\Leftrightarrow a\ne-ab\)

Khi đó: \(\dfrac{b^2+b}{a+ab}=\dfrac{b\left(b+1\right)}{a\left(1+b\right)}=\dfrac{b}{a}\)

\(=\dfrac{-8xy\left(1-3x\right)^3:4x\left(1-3x\right)}{12x^3\left(1-3x\right):4x\left(1-3x\right)}=\dfrac{-2y\left(1-3x\right)^2}{3x^2}\)

\(\dfrac{8xy\left(3x-1\right)^3}{12x^3\left(1-3x\right)}=\dfrac{-2y\left(1-3x\right)^3}{3x^2\left(1-3x\right)}=\dfrac{-2y\left(1-3x\right)^2}{3x^2}\)

\(\dfrac{8x^3y^4\left(x-y\right)^2}{12x^2y^5\left(y-x\right)}=\dfrac{2x\left(y-x\right)^2}{3y\left(y-x\right)}=\dfrac{2x\left(y-x\right)}{3y}\)

ĐKXĐ: \(x\ge0;x\ne4\)

\(A=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

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

\(C=\left(\dfrac{3}{x-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{3+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{1}\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)

\(a,=\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\\ b,=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)

a: \(=-\sqrt{a}\)

b: \(=\sqrt{p}\)

2P = 24.(5^2 + 1 )( 5^4 + 1 ) (5^8 +  1 )(5^16 + 1 )

2P = ( 5^2 - 1 )((5^2  + 1 )( 5^4 + 1 ) (5^ 8 + 1 )( 5^ 16 + 1)

2P = ( 5 ^ 4 - 1 )( 5 ^ 4 + 1 ) (5^8 + 1 )(5^16 + 1 )

2P = ( 5 ^8 - 1 )( 5^8 + 1 )( 5^16 + 1)

2P = ( 5 ^16 - 1 )( 5^16 + 1 )

2P = 5^32 - 1 

=> P = \(\frac{5^{32}-1}{2}\)

a) Ta có: \(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)

\(=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)

\(=4\sqrt{x+1}\)

b) Để B=16 thì \(4\sqrt{x+1}=16\)

\(\Leftrightarrow x+1=16\)

hay x=15

\(a.B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\\ =2\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\\ =2\sqrt{x+1}\)

b. để B=16⇒\(2\sqrt{x+1}=16\)

\(2\sqrt{x+1}=16\\ \Leftrightarrow\sqrt{x+1}=8\\ \Rightarrow x+1=64\\ \Leftrightarrow x=63\)

vậy với x=63 thì B có giá trị là 16