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Theo bài ra, ta suy ra được:

32x^5 +1 -(32x^5 -1) =2

2 = 2

Vậy có vô số x thỏa mãn đề bài.

b: \(\Leftrightarrow32x^5+1-32x^5+1=2\)

=>0x=0(luôn đúng)

\(16x^2-9=\left(4x-3\right)\left(4x+3\right)\)

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

16x^2-9=(4x-3)(4x+3)

16x^2-8x+1=(4x-1)^2 

\(A=\dfrac{16x^2-1}{16x^2-8x+1}\)

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

a) ĐKXĐ:

\(\left(4x-1\right)^2\ne0\Leftrightarrow4x-1\ne0\Leftrightarrow x\ne\dfrac{1}{4}\)

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

a,đkxđ : \(16x^2\ne0\Leftrightarrow x\ne0\)

b, \(\dfrac{16x^2}{1}-\dfrac{1}{16x^2}-\dfrac{8x}{1}+1=\dfrac{256x^4}{16x^2}-\dfrac{1}{16x^2}-\dfrac{128x^3}{16x^2}+\dfrac{16x^2}{16x^2}\)

\(=\dfrac{256x^4-1-128x^3+16x^2}{16x^2}=\dfrac{256x^4-128x^3+16x^2-1}{16x^2}\)

\(=\dfrac{\left(256x^4-128x^3+16^2\right)-1}{16x^2}=\dfrac{16x^2\left(16x^2-8x+1\right)-1}{16x^2}\)

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

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

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

b: \(\Leftrightarrow32x^5+1-32x^5+1=2\)

=>2=2(luôn đúng)

a: \(\Leftrightarrow\left[\left(x-3\right)^2-\left(x+3\right)^2\right]\left[\left(x-3\right)^2+\left(x+3\right)^2\right]+24x^3=216\)

\(\Leftrightarrow-12x\left(2x^2+18\right)+24x^3=216\)

=>-216x=216

hay x=-1

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

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

Chúc bạn học tốt!!!

\(a.\)

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

\(b.\)

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

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: \(\dfrac{4x^2-4xy+y^2}{y^2-4x^2}\)

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

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

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