`(-x+1)^3`

`= (1-x)^3`

`= 1^3 + 3 . 1^2 . (-x) + 3 . 1 . (-x)^2 + (-x)^3`

`= 1 - 3x + 3x^2 - x^3`

\(a.3x^2y-12xy^2\\ =3xy\cdot x-4y\cdot3xy\\ =3xy\left(x-4y\right)\\ b.x^2y-8xy\\ =xy\cdot x-8\cdot xy\\ =xy\left(x-8\right)\)

\(x^2-2014x+2013\\ =x^2-2013x-x+2013\\ =\left(x^2-2013x\right)-\left(x-2013\right)\\ =x\left(x-2013\right)-\left(x-2013\right)\\ =\left(x-2013\right)\left(x-1\right)\)

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

\(=9x^2-30x+25-8x^2+2x\)

\(=x^2-28x+25\)

\(=x^2-28x+196-171\)

\(=\left(x-14\right)^2-171=\left(x-14-\sqrt{171}\right)\left(x-14+\sqrt{171}\right)\)

\(\dfrac{1}{3}\left(6xy^2\right)^2\cdot\left(-\dfrac{5}{4}x^4y^3\right)\\ =\dfrac{1}{3}\cdot36x^2y^4\cdot\left(-\dfrac{5}{4}x^4y^3\right)\\ =\left(\dfrac{1}{3}\cdot36\cdot\dfrac{-5}{4}\right)\cdot\left(x^2\cdot x^4\right)\cdot\left(y^4\cdot y^3\right)\\ =-15x^6y^7\)

Bậc: `6+7=13` 

\(\dfrac{1}{3}\left(6xy^2\right)^2\cdot\left(-\dfrac{5}{4}x^4y^3\right)=\dfrac{1}{3}\cdot36x^2y^4\cdot\dfrac{-5}{4}x^4y^3\)

\(=12\cdot\dfrac{-5}{4}\cdot x^6y^7=-15x^6y^7\)

Bậc là 6+7=13

Áp dụng định lý Pythagore cho tam giác ABC vuông tại A ta có: 

\(AB^2+AC^2=BC^2\\ =>BC=\sqrt{AB^2+AC^2}=\sqrt{3^2+4^2}=5\left(cm\right)\)

Áp dụng hệ thức lượng ta có:

\(AB^2=BC\cdot BH=>BH=\dfrac{AB^2}{BC}=\dfrac{3^2}{5}=\dfrac{9}{5}\left(cm\right)\) 

\(AB\cdot AC=AH\cdot BC=>AH=\dfrac{AB\cdot AC}{BC}=\dfrac{3\cdot4}{5}=\dfrac{12}{5}\left(cm\right)\)

\(=>S_{HAB}=\dfrac{1}{2}\cdot AH\cdot BH=\dfrac{1}{2}\cdot\dfrac{12}{5}\cdot\dfrac{9}{5}=\dfrac{54}{25}\left(cm^2\right)\)

\(B=5-x^2-8x\\ =\left(-x^2-8x-16\right)+21\\ =-\left(x^2+8x+16\right)+21\\ =-\left(x^2+2\cdot x\cdot4+4^2\right)+21\\ =-\left(x+4\right)^2+21\)

Ta có: `-(x+4)^2<=0` với mọi x 

`=>B=-(x+4)^2+21<=21` với mọi x 

Dấu "=" xảy ra: `x+4=0<=>x=-4`