b: \(BC=\sqrt{89}\left(cm\right)\)

\(\sin\widehat{B}=\dfrac{5\sqrt{89}}{89}\)

\(\Leftrightarrow\widehat{B}\simeq32^0\)

\(\widehat{C}=58^0\)

Bài 10:

a: \(\overrightarrow{AB}+\overrightarrow{BO}+\overrightarrow{OA}\)

\(=\overrightarrow{AO}+\overrightarrow{OA}=\overrightarrow{0}\)

b: \(\overrightarrow{OA}+\overrightarrow{BC}+\overrightarrow{DO}+\overrightarrow{CD}\)

\(=\overrightarrow{OA}+\overrightarrow{DO}+\overrightarrow{BD}\)

\(=\overrightarrow{OA}+\overrightarrow{BO}=\overrightarrow{BA}\)

Bài 1: 

a: \(36a^4-y^2=\left(6a^2-y\right)\left(6a^2+y\right)\)

n: \(6x^2+x-2\)

\(=6x^2+4x-3x-2\)

\(=\left(3x+2\right)\left(2x-1\right)\)

4:

1: S=1,2^2*3,14=4,5216m³

2: 

a: góc ABC+góc ABF=180 độ

=>B,C,F thẳng hàng

góc CDF=góc CEF=90 độ

=>CDEF nội tiếp

mn ơi  giúp em

Bài 3:

\(a,=3x\left(y-4x+6y^2\right)\\ b,=5xy\left(x^2-6x+9\right)=5xy\left(x-3\right)^2\\ d,=\left(x+y\right)\left(x-12\right)\\ f,=2x\left(x-y\right)\left(5x-4y\right)\\ g,=\left(x-2\right)\left(x-2+3x\right)=\left(x-2\right)\left(4x-2\right)=2\left(x-2\right)\left(2x-1\right)\\ h,=x^2\left(1-5x\right)+3xy\left(5x-1\right)=x\left(1-5x\right)\left(x-3y\right)\\ i,=x\left(x-2\right)+4\left(x-2\right)=\left(x+4\right)\left(x-2\right)\\ j,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ k,=4x^2-12x+3x-9=\left(x-3\right)\left(4x+3\right)\\ l,=\left(x+5\right)^2-y^2=\left(x-y+5\right)\left(x+y+5\right)\\ m,=x^2-\left(2y-6\right)^2=\left(x-2y+6\right)\left(x+2y-6\right)\\ n,=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\\ =\left(x^2+5x+5\right)^2-1-24\\ =\left(x^2+5x+5\right)^2-25\\ =\left(x^2+5x\right)\left(x^2+5x+10\right)\\ =x\left(x+5\right)\left(x^2+5x+10\right)\)

Bài 3:

a)3xy-12x²+24xy²=3x(y-4x+8y²)

b)5x³y-30x²y+45xy=5xy(x²-6x+9)=5xy(x-6x+3²)=5xy(x-3)²

d)x(x+y)-12x-12y=x(x+y)-(12x+12y)=x(x+y)-12(x+y)=(x+y)(x-12)

f)10x²(x-y)-8xy(x-y)=(x-y)(10x²-8xy)

g)(x-2)³+3x²-6x=(x-2)³+(3x²-6x)=(x-2)³+3x(x-2)=(x-2)[(x-2)+3x]=(x-2)(x-2+3x)=(x-2)(4x-2)

h)x²+15x²y-3xy-5x³=(x²-3xy)+(15x²y-5x³)=x(x-3y)+5x²(3y-x)=x(x-3y)-5x²(x-3y)=(x-3y)(x-5x²)

i)x²-2x-8+4x=(x²-2x)+(4x-8)=x(x-2)+4(x-2)=(x-2)(x+4)

k)4x²-9x-9=4x²+3x-12x-9=(4x²+3x)-(12x+9)=x(4x+3)-3(4x+3)=(4x+3)(x-3)

l)x²-y²+10x+25=(x²+10x+25)-y²=(x²+10x+5²)-y²=(x+5)²-y²=[(x+5)+y][(x+5)-y]=(x+5+y)(x+5-y)

m)x²-4y²+24y-36=x²-(4y²-24y+36)=x²-[(2y)²-24y+6²]=x²-(2y-6)²=[x+(2y-6)][x-(2y-6)]=(x+2y-6)(x-2y+6)

n)(không biết làm)

1 Where did you go?

2 Who did you go with?

3 How did you get there?

4 What did you do during the day?

5 Did you have a good time?

1. Where did you go?

Where was you going?

2. Who did you go with?

Who was you going with?

3.  How did you get there?

How was you getting there?

5. What did you during the day?

What was you doing during the day?

6. Did you have a good time?

Was you having a good time?

Giúp e bài hình với ạ.

Bài 2: 

Xé ΔADH vuông tại H và ΔCBK vuông tại K có 

AD=BC

\(\widehat{ADH}=\widehat{CBK}\)

Do đó: ΔADH=ΔCBK

Suy ra: AH=CK

Xét tứ giác AHCK có 

AH//CK

AH=CK

Do đó: AHCK là hình bình hành

Bài 6:

Ta có:

\(sin^2x+cos^2x=1\)

\(\Leftrightarrow cos^2x=1-sin^2x\)

\(\Leftrightarrow cos^2x=1-\left(\dfrac{1}{3}\right)^2=\dfrac{8}{9}\)

\(\Leftrightarrow cosx=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt{2}}{3}\) 

Mà: \(tanx=\dfrac{sinx}{cosx}\)

\(\Leftrightarrow tanx=\dfrac{1}{3}:\dfrac{2\sqrt{2}}{3}=\dfrac{1}{2\sqrt{2}}=\dfrac{\sqrt{2}}{4}\) 

\(\Leftrightarrow\text{c}otx=\dfrac{1}{tanx}=1:\dfrac{\sqrt{2}}{4}=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)