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

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

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

\(A=\frac{4x}{x+2\left(4-x\right)}.\frac{x+2}{4}\)

\(A=\frac{x}{4-x}\)

\(b,\frac{x}{4-x}>0\)

xét 2 trường hợp x>0 đồng thời 4-x>0 (điều kiện x\(\ne\)4) và x<0 ,4-x<0

\(TH1:0< x< \text{4}\)

\(TH2:\)ko có giá trị x

\(c,Ax=\frac{x}{4-x}x\)=\(\frac{x^2}{4-x}\)

\(\frac{x^2-16+16}{4-x}\)

\(\frac{\left(x-4\right)\left(x+4\right)+16}{4-x}\)

\(-\left(x+4\right)+\frac{16}{4-x}\)

để AX nguyên thì \(16⋮4-x\)

lập bảng ra tìm đc x = 0,2,-4,-12,5,6,8,12,20

a: Xét (O) có

CM là tiếp tuyến có M là tiếp điểm

CN là tiếp tuyến có N là tiếp điểm

Do đó: CM=CN

hay C nằm trên đường trung trực của MN(1)

Ta có: OM=ON

nên O nằm trên đường trung trực của MN(2)

Từ (1) và (2) suy ra OC là đường trung trực của MN

1. I was given a notebook as a reward by my teacher.

2. Grapes are picked and then turned into wine.

3. I wasn't allowed to stay up late by my father.

4. Pencils aren't used in your Maths exam.

5. I was allowed to go picnic with my friends by my parents yesterday.

6. I wasn't told about his story.

7. This cake is made by me.

8. When I was a child, my leg was bitten by a dog.

9. Many social skills are needed when you enter the workplace.

1, I was given a notebook by my teacher as a reward

  or A notebook  was given to me  by my teacher as a reward

2, Grapes are picked and wine is turned into

3, I am not allowed to stay up late by my father

4, pencils are not used  in your maths

5, i was not allowed to go picnic with my friends by my parents yesterday

6, i was not told about his story

7, this cake was made by me

8,  When I was a kid, my leg was bitten by a dog

9, many social kills are needed when the workplace is entered

b: Gọi giao của AH với BC là F

=>AH vuông góc BC tại F

góic CHI=góc AHD=90 độ-góc HAD=góc ABC=1/2*sđ cung AC

góc CIH=1/2*sđ cung CA

=>góc CHI=góc CIH

=>ΔCHI cân tại C

c:

góc BDC=góc BEC=90 độ

=>BDEC nội tiếp đường tròn đường kính BC

=>MD=ME

=>ΔMDE cân tại M

mà MN là trung tuyến

nên MN vuông góc DE

Kẻ tiếp tuyến Ax của (O)

=>góc xAC=góc ABC

=>góc xAC=góc AED

=>Ax//DE

=>DE vuông góc OA

=>MN//AO