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Ta có: AB//CD

=>\(\widehat{BAD}+\widehat{ADC}=180^0\)

=>\(2\cdot\left(\widehat{KAD}+\widehat{KDA}\right)=180^0\)

=>\(\widehat{KAD}+\widehat{KDA}=90^0\)

=>ΔKAD vuông tại K

=>\(\widehat{AKD}=90^0\)

Gọi vận tốc riêng của ca là x ( x > 0 ) 

vận tốc ca nô xuôi dòng : x + 4 km/h 

vận tốc ca nô ngược dòng : x - 4 km/h 

Thời gian đi xuôi A đến B : \(\dfrac{30}{x+4}\)giờ

Thời gian đi từ B ngược về một điểm cách B 20 km nên ta có : \(\dfrac{20}{x-4}\)giờ

Tổng thời gian đi hết 2h30p = 5/2 h

Ta có pt \(\dfrac{30}{x+4}+\dfrac{20}{x-4}=\dfrac{5}{2}\Leftrightarrow x=20\)km/h 

1: \(\left(2x+1\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3\)

\(=8x^3+12x^2+6x+1\)

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

\(=x^3-2x^2+\dfrac{4}{3}x-\dfrac{8}{27}\)

3: \(\left(3x-1\right)^3=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3\)

\(=27x^3-27x^2+9x-1\)

5: \(\left(2-3y\right)^3=2^3-3\cdot2^2\cdot3y+3\cdot2\cdot\left(3y\right)^2-\left(3y\right)^3\)

\(=8-36y+54y^2-27y^3\)

6: \(\left(3x-2y\right)^3=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2y+3\cdot3x\cdot\left(2y\right)^2-\left(2y\right)^3\)

\(=27x^3-54x^2y+36xy^2-8y^3\)

7: \(\left(4x+\dfrac{2}{3}y\right)^3=\left(4x\right)^3+3\cdot\left(4x\right)^2\cdot\dfrac{2}{3}y+3\cdot4x\cdot\left(\dfrac{2}{3}y\right)^2+\left(\dfrac{2}{3}y\right)^3\)

\(=64x^3+32x^2y+\dfrac{16}{3}xy^2+\dfrac{8}{27}y^3\)

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

\(=x^6-9x^4+27x^2-27\)

9: \(\left(2x^2-3\right)^3=\left(2x^2\right)^3-3\cdot\left(2x^2\right)^2\cdot3+3\cdot2x^2\cdot3^2-3^3\)

\(=8x^6-36x^4+54x^2-27\)

10: \(\left(\dfrac{1}{2}x+y^2\right)^3\)

\(=\left(\dfrac{1}{2}x\right)^3+3\cdot\left(\dfrac{1}{2}x\right)^2\cdot y^2+3\cdot\dfrac{1}{2}x\cdot\left(y^2\right)^2+\left(y^2\right)^3\)

\(=\dfrac{1}{8}x^3+\dfrac{3}{4}x^2y^2+\dfrac{3}{2}xy^4+y^6\)

11: \(\left(2x-\dfrac{1}{2}y\right)^3=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot\dfrac{1}{2}y+3\cdot2x\cdot\left(\dfrac{1}{2}y\right)^2-\left(\dfrac{1}{2}y\right)^3\)

\(=8x^3-6x^2y+\dfrac{3}{2}xy^2-\dfrac{1}{8}y^3\)

12: \(\left(x-y^2\right)^2=x^2-2\cdot x\cdot y^2+\left(y^2\right)^2=x^2-2xy^2+y^4\)

1: Đặt A=\(5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x+2\right)\)

\(=20x^3-10x^2+5x-20x^3+10x^2-4x=x\)

Thay x=15 vào A, ta được:

A=x=15

2: Đặt \(B=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)

\(=6x^2y^2-6xy^3-8x^3+8x^2y^2+5x^2y^2-5xy^3\)

\(=19x^2y^2-11xy^3-8x^3\)

Thay x=0,5 và y=2 vào B, ta được:

\(B=19\cdot0,5^2\cdot2^2-11\cdot0,5\cdot2^3-8\cdot2^3\)

=19-44-64

=-89

3: x=4 nên x+1=5

\(x^5-5x^4+5x^3-5x^2+5x-1\)

\(=x^5-x^4\left(x+1\right)+x^3\left(x+1\right)-x^2\left(x+1\right)+x\left(x+1\right)-1\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-1\)

=x-1=4-1=3

4: x=7 nên x+1=8

\(x^{15}-8x^{14}+8x^{13}-8x^{12}+...-8x^2+8x-5\)

\(=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-5\)

\(=x^{15}-x^{15}-x^{14}+x^{14}+...+x^2+x-5\)

=x-5=7-5=2

5: \(M=\left(2x-1\right)^2+2\left(2x-1\right)\left(3x+1\right)+\left(3x+1\right)^2\)

\(=\left(2x-1+3x+1\right)^2=\left(5x\right)^2=25x^2\)

6: \(N=\left(3x-1\right)^2-2\left(9x^2-1\right)+\left(3x+1\right)^2\)

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

\(=\left(3x-1-3x-1\right)^2=\left(-2\right)^2=4\)

a:

ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

 \(Q=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\)

\(=\dfrac{3\left(x-3\right)+x+3+18}{x^2-9}\)

\(=\dfrac{3x-9+x+21}{\left(x-3\right)\left(x+3\right)}=\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\)

b: \(R=Q\cdot x=\dfrac{4x}{x-3}=\dfrac{4x-12+12}{x-3}=4+\dfrac{12}{x-3}\)

Để R nguyên thì \(12⋮x-3\)

=>\(x-3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(x\in\left\{4;2;5;1;6;0;7;-1;9;-3;15;-9\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{4;2;5;1;6;0;7;-1;9;15;-9\right\}\)

DE đâu ra?

a, Cho O là giao điểm AC;BD 

Theo bđt tam giác 

OA + OB > AB ; OC + OD > DC

Cộng vế vs vế : OA + OB + OC + OD = AC + BD > AB + CD 

\(C=x^7+x^2+1\)

\(=x^7+x^6+x^5-x^6-x^5-x^4+x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\)

\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)

\(D=x^8+x+1\)

\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)