\(a.\)

\(\dfrac{16x^2-1}{16x^2-8x+1}\\ =\dfrac{\left(4x\right)^2-1}{\left(4x-1\right)^2}\\ =\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\\ =\dfrac{4x+1}{4x-1}\)

\(b.\)

\(\dfrac{4x^2-4xy+y^2}{-\left(4x^2-y^2\right)}\\ =-\dfrac{\left(2x-y\right)^2}{\left(2x-y\right)\left(2x+y\right)}\\ =\dfrac{-\left(2x-y\right)}{2x+y}\\ =\dfrac{y-2x}{y+2x}\)

a) Ta có: \(\dfrac{16x^2-1}{16x^2-8x+1}\)

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

\(=\dfrac{4x+1}{4x-1}\)

b) Ta có: \(\dfrac{4x^2-4xy+y^2}{y^2-4x^2}\)

\(=\dfrac{\left(2x-y\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)

\(=\dfrac{\left(y-2x\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)

\(=\dfrac{y-2x}{y+2x}\)

a) x ≠ -5.

b) Ta có P = ( x + 5 ) 2 x + 5 = x + 5  

c) Ta có P = 1 Û x = -4 (TMĐK)

d) Ta có P = 0 Û x = -5 (loại). Do vậy x ∈ ∅ .

`đk:x-\sqrt{x^2-4x+4}>=0`

`<=>x>=\sqrt{x^2-4x+4}`

`<=>x^2>=x^2-4x+4(x>=0)`

`<=>4x-4>=0`

`<=>4x>=4<=>x>=1`

`b)A=sqrt{x-sqrt{(x-2)^2}}`

`=sqrt{x-|x-2|}`

`x>=2=>|x-2|=x-2`

`=>A=sqrt{x-x+2}=sqrt2`

`1<=x<=2=>|x-2|=2x-`

`=>A=\sqrt{x+x-2}=sqrt{2x-2}`

a) Rút gọn:

b) Để B = 16 thì:

⇔ x + 1 = 16 ⇔ x = 15 (thỏa mãn x ≥ -1)

a.

\(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\left(x\ge-1\right)\)

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

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

\(B=\left(4-3+2+1\right).\sqrt{x+1}\)

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

b.

\(B=16\\\)

\(\Rightarrow4\sqrt{x+1}=16\)

\(\Rightarrow\sqrt{x+1}=\dfrac{16}{4}=4\)

\(\Rightarrow x+1=4^2\)

\(\Rightarrow x+1=16\rightarrow x=16-1=15\) (thỏa mãn)

vậy x=15

\(a,\left(2x-1\right)\left(4x^2+2x-1\right)-8x^3+16x\)

\(\Leftrightarrow\)\(8x^3+4x^2-2x-4x^2-2x+1-8x^3+16x\)

\(\Leftrightarrow\)\(4x^2-2x-4x^2-2x+1+16x\)

\(\Leftrightarrow12x+1\)

\(b,\left(x+2\right)^3-x^2\left(x+6\right)-2x-5\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2-2x-5\)

\(\Leftrightarrow6x^2+12x+8-6x^2-2x-5\)

\(\Leftrightarrow10x+3\)

Mình cũng không biết đúng hay sai nữa :V