a) ABCD là hình thang

=> \(\widehat{A}+\widehat{D}=180^o=>\widehat{A}=180^o-\widehat{D}\) 

Mà: 

\(\widehat{A}-\widehat{D}=20^o=>180^o-\widehat{D}-\widehat{D}=20^o\\ =>2\widehat{D}=180^o-20^o=160^o\\ =>\widehat{D}=\dfrac{160^o}{2}=80^o\)

=> \(\widehat{A}=180^o-80^o=100^o\)

\(\widehat{B}+\widehat{C}=180^o\) (ABCD là hình thang) 

Mà: \(\widehat{B}=2\widehat{C}=>2\widehat{C}+\widehat{C}=180^o=>3\widehat{C}=180^o=>\widehat{C}=60^o\)

\(=>\widehat{B}=2\cdot60^o=120^o\) 

b) Xét ΔABC có: \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^o\)

\(=>\widehat{BAC}=180^o-120^o-50^o=10^o\)  

Mà: \(\widehat{DAC}+\widehat{BAC}=\widehat{DAB} =>\widehat{DAC}=\widehat{DAB}-\widehat{BAC}=100^o-10^o=90^o\)

`=> AD⊥AC`

a) Ta có:

\(x>1\\ \Leftrightarrow\left\{{}\begin{matrix}2x-1>0\\1-x< 0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left|2x-1\right|=2x-1\\\left|1-x\right|=x-1\end{matrix}\right.\)

Thay vào A được:

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

Vậy...

b) Ta có: 

\(x< \dfrac{5}{4}\\ \Leftrightarrow5-4x>0\\ \Leftrightarrow\left|5-4x\right|=5-4x\)

Thay vào B được:

\(B=5-4x+3\\ B=-4x+8\)

Vậy...

\(\left(-2x^2y\right)^2\cdot8x^3\cdot yz^3\\ =\left(-2\right)^2\cdot\left(x^2\right)^2\cdot y^2\cdot8x^3\cdot yz^3\\ =4x^4y^2\cdot8x^3\cdot yz^3\\ =4\cdot8\cdot\left(x^4\cdot x^3\right)\cdot\left(y^2\cdot y\right)\cdot z^3\\ =32x^7y^3z^3\)

`#3107.101107`

\((-2x^2y)^2\cdot8x^3\cdot yz^3 \\ = 4x^4y^2 \cdot8x^3 \cdot yz^3 \\ = (4 \cdot 8) \cdot (x^4y^2 \cdot x^3 \cdot yz^3) \\ = 32x^7y^3z^3\)

Gọi vận tốc lượt đi là x(km/h)

(Điều kiện: x>0)

vận tốc lượt về là \(x\left(1+20\%\right)=1,2x\left(\dfrac{km}{h}\right)\)

Thời gian đi là \(\dfrac{120}{x}\left(giờ\right)\)

Thời gian về là \(\dfrac{120}{1,2x}=\dfrac{100}{x}\left(giờ\right)\)

Tổng thời gian cả đi lẫn về là 4h24p=4,4 giờ nên ta có:

\(\dfrac{120}{x}+\dfrac{100}{x}=4,4\)

=>\(\dfrac{220}{x}=4,4\)

=>\(x=\dfrac{220}{4,4}=50\left(nhận\right)\)

Vậy: vận tốc lượt đi là 50km/h

\(2x^3-5x-6\\ =2x^3-4x^2+4x^2-8x+3x-6\\ =2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\\=\left(2x^2+4x+3\right)\left(x-2\right)\)

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

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

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

Em ghi là đường  cao H là sai, phải ghi là BH mới đúng vì vậy Olm bảo em làm  sai em hiểu  chưa nhỉ?

\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\\ =x^2y-x^2z+y^2z-xy^2+xz^2-yz^2\\ =\left(x^2y-x^2z\right)+\left(y^2z-yz^2\right)-\left(xy^2-xz^2\right)\\ =x^2\left(y-z\right)+yz\left(y-z\right)-x\left(y^2-z^2\right)\\ =x^2\left(y-z\right)+yz\left(y-z\right)-x\left(y+z\right)\left(y-z\right)\\ =\left(y-z\right)\left[x^2+yz-x\left(y+z\right)\right]\\ =\left(y-z\right)\left(x^2+yz-xy-xz\right)\\ =\left(y-z\right)\left[x\left(x-y\right)-z\left(x-y\right)\right]\\ =\left(y-z\right)\left(x-z\right)\left(x-y\right)\)

a: \(\left(x-2\right)\left(3x-1\right)\left(x^2-4x+1\right)\)

\(=\left(3x^2-x-6x+2\right)\left(x^2-4x+1\right)\)

\(=\left(3x^2-7x+2\right)\left(x^2-4x+1\right)\)

\(=3x^4-12x^3+3x^2-7x^3+28x^2-7x+2x^2-8x+2\)

\(=3x^4-19x^3+33x^2-15x+2\)

b: \(x\left(3-4x\right)\left(2x^2-3x\right)\)

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

\(=-8x^4+12x^3+6x^3-9x^2\)

\(=-8x^4+18x^3-9x^2\)

a) 

\(\left(x-2\right)\left(3x-1\right)\left(x^2-4x+1\right)\\ =\left(3x^2-6x-x+2\right)\left(x^2-4x+1\right)\\ =\left(3x^2-7x+2\right)\left(x^2-4x+1\right)\\ =3x^4-12x^3+3x^2-7x^3+28x^2-7x-8x+2\\ =3x^4-19x^3+31x^2-15x+2\) 

b) 

\(x\left(3-4x\right)\left(2x^2-3x\right)\\ =\left(3x-4x^2\right)\left(2x^2-3x\right)\\ =6x^3-9x^2-8x^4+12x^3\\ =-8x^4+18x^3-9x^2\)

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

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

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

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