Đề bài yêu cầu gì?

=>10y=18-8x

=>y=(9-4x)/5

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

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

b)\(2x^2 - 18\)

\(=2(x-3)(x+3)\)

c)\(x^2 - 7xy + 10y^2\)

\(=(2y-x)(5y-x)\)

d)\(x^4 - 7x^2 + 1\)

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

e)\(x^2 + 8x + 7\)

\(=(x+1)(x+7)\)

a: \(=\dfrac{3x}{5\left(x+y\right)}-\dfrac{x}{10\left(x-y\right)}\)

\(=\dfrac{6x\left(x-y\right)-x\left(x+y\right)}{10\left(x-y\right)\cdot\left(x+y\right)}\)

\(=\dfrac{6x^2-6xy-x^2-xy}{10\left(x-y\right)\left(x+y\right)}=\dfrac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}\)

b: \(=\dfrac{7}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{1}{x\left(2x+3\right)}-\dfrac{1}{2\left(2x-3\right)}\)

\(=\dfrac{7x+2\left(2x-3\right)-x\left(2x+3\right)}{2x\left(2x+3\right)\left(2x-3\right)}\)

\(=\dfrac{7x+4x-6-2x^2-3x}{2x\left(2x+3\right)\left(2x-3\right)}\)

\(=\dfrac{-2x^2-6}{2x\left(2x+3\right)\left(2x-3\right)}=\dfrac{-x^2-3}{x\left(2x+3\right)\left(2x-3\right)}\)

c: \(=\dfrac{5}{x+1}+\dfrac{10}{x^2-x+1}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x}{x^2-x+1}\)

\(-8x=-10y\Leftrightarrow\frac{-8x}{40}=\frac{-10y}{40}\Leftrightarrow\frac{-x}{5}=\frac{-y}{4}\Leftrightarrow\frac{-5x}{25}=\frac{-2y}{8}\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\Rightarrow\frac{-5x}{25}=\frac{-2y}{8}=\frac{-5x-2y}{25+8}=\frac{396}{33}=12\)

\(\Leftrightarrow\frac{-x}{5}=12\Leftrightarrow-x=60\Leftrightarrow x=-60\)

\(\Leftrightarrow\frac{-y}{4}=12\Leftrightarrow-y=48\Leftrightarrow y=-48\)

Vậy \(x=-60;y=-48\)

P= - (x^2-8x+16+y^2-10y+25)-124

P=-[(x-4)^2+(y-5)^2]-124

-[(x-4)^2+(y-5)^2] nhỏ hơn hoặc bằng 0 => P nhỏ hơn hoặc bằng -124

=> GTLN của P=-124 khi x=4 và y=5

A=3x^2-5x-6y^2+1

A+M=6x^2-8x+10y^2

=>M=6x^2-8x+10y^2-3x^2+5x+6y^2-1

=>M=3x^2-3x+16y^2-1