bài 2 :

a, X : 5 = 163 ( dư 4)

    X      = 163 + 4 x 5

   X       =  835

b, 506 : X = 8( dư 2)

             X = ( 506 - 2 ):8

             X = 63

Uk:)

\(A=4^1+4^2+4^3+...+4^{50}\)

\(4A=4^2+4^3+4^4+...+4^{51}\)

\(4A-A=\left(4^2+4^3+4^4+...+4^{51}\right)-\left(4^1+4^2+4^3+...+4^{50}\right)\)

\(3A=4^{51}-4\) 

\(A=\frac{4^{51}-4}{3}\)

Đề tự vt đi t giải luôn

\(4A=4^2+4^3+...+4^{51}\)

\(4A-A=\left(4^2+4^3+...+4^{51}\right)-\left(4+4^2+...+4^{50}\right)\)

\(3A=4^{51}-4\)

\(A=\frac{4^{51}-4}{3}\)

\(A=4^1+4^2+4^3+....+4^{50}\)

\(4A=4^2+4^3+4^4+...+4^{51}\)

\(4A-A=\left(4^2+4^3+4^4+...+4^{51}\right)-\)\(\left(4^1+4^2+4^3+...+4^{50}\right)\)

\(3A=4^{51}-4^1\)

\(A=\frac{4^{51}-4}{3}\)

1 My parents go shopping twice a week

2 Hoa's house has a balcony

3 My brother usually plays badminton with his friends

4 My favorite book is Tam and Cam. What is yours?

5 There are 10 pencil cases on the table

1, My parents go shopping twice a week.

2, Hoa's house has a balcony.

3, My brother usually plays badminton with his friends.

4, My favorite book is Tam and Cam. What's yours?

5, There are ten pencil cases on the table.

a) \(x^3+x^2y-x^2z-xyz\)

\(=x^2\left(x+y\right)-xz\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-xz\right)\)

\(=x\left(x+y\right)\left(x-z\right)\)

b) \(x^2-6x+9-9y^2\)

\(=\left(x^2-2\cdot x\cdot3+3^2\right)-\left(3y\right)^2\)

\(=\left(x-3\right)^2-\left(3y\right)^2\)

\(=\left(x-3-3y\right)\left(x-3+3y\right)\)

c) \(x^2+9x+20\)

\(=x^2+5x+4x+20\)

\(=x\left(x+5\right)+4\left(x+5\right)\)

\(=\left(x+5\right)\left(x+4\right)\)

d) \(x^4+4\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot2+4-2\cdot x^2\cdot2\)

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

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

a/\(x^3+x^2y-x^2z-xyz\)

\(=\left(x^3-x^2y\right)+\left(x^2y-xyz\right)\)

\(=x^2\left(x-z\right)+xy\left(x-z\right)\)

\(=\left(x-z\right)\left(x^2+xy\right)\)

b/\(x^2-6x+9-9y^2\)

\(=\left(x^2-6x+9\right)-9y^2\)

\(=\left(x-3\right)^2-\left(3y\right)^2\)

\(=\left(x-3+3y\right)\left(x-3-3y\right)\)

c/\(x^2+9x+20\)

\(=x^2+4x+5x+20\)

\(=\left(x^2+4x\right)+\left(5x+20\right)\)

\(=x\left(x+4\right)+5\left(x+4\right)\)

\(=\left(x+5\right)\left(x+4\right)\)

d/\(x^4+4\)

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

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

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

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

\(a,x^3+x^2y-x^2z-xyz=x^2\left(x+y\right)-xz\left(x+y\right).\)

\(=\left(x+y\right)\left(x^2-xz\right)\)

\(=x\left(x+y\right)\left(x-z\right)\)

\(b,x^2-6x+9-9y^2=\left(x-3\right)^2-9y^2\)

\(=\left(x-3+3y\right)\left(x-3-3y\right)\)

\(c,x^2+9x+20=\left(x^2+8x+16\right)+\left(x+4\right)\)

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

\(=\left(x+4\right)\left(x+4+1\right)=\left(x+4\right)\left(x+5\right)\)

\(d,x^4+4=\left(x^2\right)^2+2^2\)

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

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

5:

a: Xét ΔCBA vuông tại A và ΔCAH vuông tại H có

góc C chung

=>ΔCBA đồng dạg với ΔCAH

=>CB/CA=CA/CH

=>CA^2=CB*CH

b: HB=6^2/8=4,5cm

=>BC=8+4,5=12,5cm

Xét ΔHAC có HK là phân giác

nên AK/HA=CK/CH

=>AK/3=CK/4=(AK+CK)/(3+7)=10/7

=>AK=30/7cm; CK=40/7cm

Các bn lm câu 3 thui nha

Bài 4: 

a: Xét tứ giác OBAC có 

\(\widehat{OBA}+\widehat{OCA}=180^0\)

Do đó: OBAC là tứ giác nội tiếp

b: Xét (O) có

AB là tiếp tuyến

AC là tiếp tuyến

Do đó: AB=AC

hay A nằm trên đường trung trực của BC(1)

Ta có: OB=OC

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

Từ (1) và (2) suy ra OA là đường trung trực của BC

hay OA⊥BC

c: Xét ΔOBA vuông tại B có BA là đường cao

nên \(OH\cdot OA=OB^2=R^2\)