(x^2+4)^2=x^4+8x^2+16

MS=(x^2+4)^2-4x(x^2+4)=(x^2+4)(x^2-4x+4)=(x^2+4)(x-2)^2

ĐK x khác 2

A=(x+2)/(x-2)=1+4/(x-2)

(x-2)= Uocs (4) 

hết

a/ \(A=\frac{2x^3-6x^2+x-8}{x-3}=2x^2+1-\frac{5}{x-3}\)

Từ đây ta thấy A nguyên khi x - 3 là ước nguyên của 5 hay

\(\left(x-3\right)=\left(-5,-1,1,5\right)\)

\(\Rightarrow x=\left(-2,2,4,8\right)\)

b/ \(B=\frac{x^4-16}{x^4-4x^3+8x^2-16x+16}=\frac{\left(x^2+4\right)\left(x-2\right)\left(x+2\right)}{\left(x^2+4\right)\left(x-2\right)^2}\)

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

Để B nguyên thì x - 2 phải là ước nguyên của 4 hay

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

\(\Rightarrow x=\left(-2,0,1,3,4,6\right)\)

Ta có : Để M=\(\left(\frac{4}{x-4}-\frac{4}{x+4}\right)\left(\frac{x^2+8x+16}{32}\right)=0\)

<=> M=\(\left(\frac{4\left(x+4\right)-4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)=0\)

<=>M=\(\left(\frac{4x+16-4x+16}{\left(x+4\right)\left(x-4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)\)

<=>M=\(\left(\frac{32}{\left(x-4\right)\left(x+4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)\)

<=>M=\(\frac{x+4}{x-4}\)

b) Thay x=\(\frac{-3}{8}\) vào M:

M=\(\frac{x+4}{x-4}=\frac{\frac{-3}{8}+4}{\frac{-3}{8}-4}=\frac{-29}{35}\)

c)Hình như sai!

d)

Điều kiện: \(x\ne2\)

Phân tích tử thức: \(x^4-16=\left(x^2\right)^2-4^2=\left(x^2-4\right)\left(x^2+4\right)=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)

Phân tích mẫu thức: \(x^4-4x^3+8x^2-16x+16=\left(x^4-4x^3+4x^2\right)+\left(4x^2-16x+16\right)\)

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

Ta có: \(P=\frac{\left(x-2\right)\left(x+2\right)\left(x^2+4\right)}{\left(x-2\right)^2\left(x^2+4\right)}=\frac{x+2}{x-2}=\frac{\left(x-2\right)+4}{x-2}=1+\frac{4}{x-2}\)

Để P là số nguyên thì \(x-2\inƯ\left(4\right)\)

\(\Rightarrow x-2\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow x\in\left\{-2;0;1;3;4;6\right\}\)

Điều kiện: x\ne2x̸​=2

Phân tích tử thức: x^4-16=\left(x^2\right)^2-4^2=\left(x^2-4\right)\left(x^2+4\right)=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)x4−16=(x2)2−42=(x2−4)(x2+4)=(x−2)(x+2)(x2+4)

Phân tích mẫu thức: x^4-4x^3+8x^2-16x+16=\left(x^4-4x^3+4x^2\right)+\left(4x^2-16x+16\right)x4−4x3+8x2−16x+16=(x4−4x3+4x2)+(4x2−16x+16)

=x^2\left(x^2-4x+4\right)+4\left(x^2-4x+4\right)=\left(x-2\right)^2\left(x^2+4\right)=x2(x2−4x+4)+4(x2−4x+4)=(x−2)2(x2+4)

Ta có: P=\frac{\left(x-2\right)\left(x+2\right)\left(x^2+4\right)}{\left(x-2\right)^2\left(x^2+4\right)}=\frac{x+2}{x-2}=\frac{\left(x-2\right)+4}{x-2}=1+\frac{4}{x-2}P=(x−2)2(x2+4)(x−2)(x+2)(x2+4)​=x−2x+2​=x−2(x−2)+4​=1+x−24​

Để P là số nguyên thì x-2\inƯ\left(4\right)x−2∈Ư(4)

\Rightarrow x-2\in\left\{-4;-2;-1;1;2;4\right\}⇒x−2∈{−4;−2;−1;1;2;4}

\Rightarrow x\in\left\{-2;0;1;3;4;6\right\}⇒x∈{−2;0;1;3;4;6}

oc cho

a: ĐKXĐ: x∉{2;-2;3}

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\right):\frac{16}{x^2-x-6}\)

\(=\left(\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\frac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right):\frac{16}{\left(x-3\right)\left(x+2\right)}\)

\(=\left(\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\frac{\left(x-3\right)\cdot\left(x+2\right)}{16}\)

\(=\frac{4x\left(x+2\right)-4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\cdot\frac{\left(x-3\right)\left(x+2\right)}{16}=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)}\cdot\frac{x-3}{16}\)

\(=\frac{-16\left(x-3\right)}{16\left(x+2\right)}=\frac{-x+3}{x+2}\)

b: A<0

=>\(\frac{-x+3}{x+2}<0\)

=>\(\frac{x-3}{x+2}>0\)

=>x>3 hoặc x<-2

c: A>=5

=>\(\frac{-x+3}{x+2}-5\ge0\)

=>\(\frac{-x+3-5x-10}{x+2}\ge0\)

=>\(\frac{-6x-7}{x+2}\ge0\)

=>\(\frac{6x+7}{x+2}\le0\)

=>-2<x<=-7/6