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

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

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

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

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

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

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

Câu 3:

a: 79,3826≃79,383

79,3826≃79,38

79,3826≃79,4

Câu 2:

a: y=-2x

=>y tỉ lệ thuận với x theo hệ số tỉ lệ là k=-2

b: xy=-2

=>x và y tỉ lệ nghịch với nhau theo hệ số tỉ lệ là a=-2

Bài 1:

a: \(\left(\frac15\right)^4\cdot5^4=\frac{1}{5^4}\cdot5^4=\frac{5^4}{5^4}=1\)

b: \(\left(x^2\right)^4=x^{2\cdot4}=x^8\)

c: \(3^{25}:3^5=3^{25-5}=3^{20}\)

d: \(\left(\frac13\right)^5\cdot3^5=\frac{1}{3^5}\cdot3^5=\frac{3^5}{3^5}=1\)

e: \(10^8\cdot2^8=\left(10\cdot2\right)^8=20^8\)

f: \(25^4\cdot2^8=25^4\cdot4^4=\left(25\cdot4\right)^4=100^4\)

g: \(\left(-\frac34\right)^2=\frac{\left(-3\right)^2}{4^2}=\frac{9}{16}\)

n: \(\frac23+\frac{-1}{6}=\frac46-\frac16=\frac36=\frac12\)

j: \(\left(-0,25\right)^5:\left(-0,25\right)^3=\left(-0,25\right)^{5-3}=\left(-0,25\right)^2=\left(\frac14\right)^2=\frac{1}{16}\)

a: \(B=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{99\cdot100}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}\)

\(=1+\frac12+\cdots+\frac{1}{100}-2\left(\frac12+\frac14+\cdots+\frac{1}{100}\right)\)

\(=1+\frac12+\cdots+\frac{1}{100}-1-\frac12-\cdots-\frac{1}{50}\)

\(=\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\)

=A

=>\(\frac{A}{B}=1\)

b: \(A=\frac{34}{7\cdot13}+\frac{51}{13\cdot22}+\frac{85}{22\cdot37}+\frac{68}{37\cdot49}\)

\(=17\left(\frac{2}{7\cdot13}+\frac{3}{13\cdot22}+\frac{5}{22\cdot37}+\frac{4}{37\cdot49}\right)\)

\(=\frac{17}{3}\left(\frac{6}{7\cdot13}+\frac{9}{13\cdot22}+\frac{15}{22\cdot37}+\frac{12}{37\cdot49}\right)\)

\(=\frac{17}{3}\left(\frac17-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)

\(=\frac{17}{3}\left(\frac17-\frac{1}{49}\right)\)

Ta có: \(B=\frac{39}{7\cdot16}+\frac{65}{16\cdot31}+\frac{52}{31\cdot43}+\frac{26}{43\cdot49}\)

\(=13\left(\frac{3}{7\cdot16}+\frac{5}{16\cdot31}+\frac{4}{31\cdot43}+\frac{2}{43\cdot49}\right)\)

\(=\frac{13}{3}\left(\frac{9}{7\cdot16}+\frac{15}{16\cdot31}+\frac{12}{31\cdot43}+\frac{6}{43\cdot49}\right)\)

\(=\frac{13}{3}\left(\frac17-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)

\(=\frac{13}{3}\left(\frac17-\frac{1}{49}\right)\)

Do đó: \(\frac{A}{B}=\frac{17}{3}:\frac{13}{3}=\frac{17}{13}\)

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