gọi chiều dài quãng đường là x km

khi đó thời gian đi từ A đến B là :\(\frac{x}{60}\text{ giờ}\)

thời gian trở về A là :\(\frac{x-10}{50}\text{ giờ}\)

ta có phương trình : \(\frac{x-10}{50}-\frac{x}{60}=1\Leftrightarrow60x-600-50x=3000\Leftrightarrow x=360\left(km\right)\)

ta có 

\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

\(=\left[\left(x^2+10x+12\right)-12\right]\left[\left(x^2+10x+12\right)+12\right]+128\)

\(=\left(x^2+10x+12\right)^2-12^2+128=\left(x^2+10x+12\right)^2-16\)

\(=\left(x^2+10x+12-4\right)\left(x^2+10x+12+4\right)=\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)

\(=\left(x+2\right)\left(x+8\right)\left(x^2+10x+8\right)\)

\(1,x^3+2x^2y+xy^2-4x\)

\(x\left(x^2+2xy+y^2-4\right)\)

\(x\left[\left(x+y\right)^2-2^2\right]\)

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

\(2,5x-5y-x^2+2xy-y^2\)

\(5\left(x-y\right)-\left(x^2-2xy+y^2\right)\)

\(5\left(x-y\right)-\left(x-y\right)^2\)

\(\left(x-y\right)\left(5-x+y\right)\)

\(3,x^4-3x^2\)

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

\(\left(a-b\right)^4\)

(a-b)⁴

k cho tớ nhé!

cảm ơn trước1

4x^2-81=0

4x^2=81

x^2=81/4

x=\(\mp\frac{9}{2}\)

Vậy............

Trả lời:

\(4x^2-81=0\)

\(\Leftrightarrow\left(2x\right)^2-9^2=0\)

\(\Leftrightarrow\left(2x-9\right)\left(2x+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-9=0\\2x+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=-\frac{9}{2}\end{cases}}}\)

Vậy x = 9/2; x = - 9/2 là nghiệm của pt.

\(a)\)

\(\left(x^2+4x\right)^2+9x^2-6x\left(x^2+4x\right)\)

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

\(=\left(x^2+4x\right)\left(x^2-2x\right)+9x^2\)

\(=x\left(x+4\right)x\left(x-2\right)+9x^2\)

\(=x^2\left(x^2+4x-2x-8\right)+9x^2\)

\(=x^2\left(x^2+2x-8\right)+9x^2\)

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

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

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

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

\(b)\)

\(\left(-6x^3+7x^2-4x+1\right):\left(-2+1\right)\)

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

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

\(c)\)

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

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

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

\(=3x^3-13x^2+18x-8\)