1)

\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}\)

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

MTC: \(x\left(x-3\right)\left(x+3\right)\)

\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}=\dfrac{\left(x-3\right)\left(7x-12\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-12x-21x+36}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-33x+36}{x\left(x-3\right)\left(x+3\right)}\)

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

2)

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

\(\dfrac{x+1}{2-4x+2x^2}=\dfrac{x+1}{2\left(1-2x+x^2\right)}=\dfrac{x+1}{2\left(1-x\right)^2}\)

MTC: \(2x\left(1-x\right)^2\)

\(\dfrac{2x-1}{x-x^2}=\dfrac{2x-1}{x\left(1-x\right)}=\dfrac{2\left(1-x\right)\left(2x-1\right)}{2x\left(1-x\right)^2}=\dfrac{\left(2-2x\right)\left(2x-1\right)}{2x\left(1-x\right)^2}=\dfrac{4x-2-4x^2+2x}{2x\left(1-x\right)^2}=\dfrac{6x-2-4x^2}{2x\left(1-x\right)^2}\)

\(\dfrac{x+1}{2-4x+2x^2}=\dfrac{x+1}{2\left(1-2x+x^2\right)}=\dfrac{x+1}{2\left(1-x\right)^2}=\dfrac{ x\left(x+1\right)}{2x\left(1-x\right)^2}=\dfrac{x^2+x}{2x\left(1-x\right)^2}\)

/x-y/>0

/y-50/>0

=>/x-y/+/y-50/>0

Mà theo đề:/x-y/+/y-50/<0

=>/x-y/=/y-50/=0

=>x-y=0=>x=y

Y-50=0=>y=50

Mà x=y=>x=50

Vậy x+y=100

 Tick nhé

Tất cả các đơn thức bậc 6 cua hai biến x,y có hệ số bằng 1 là:

\(xy^4\) ; \(x^2y^3\) ; \(x^3y^2\) ; \(x^4y\) .

5/4x6/5x7/6x...x2016/2015

=5x6x7x...x2016/4x5x6x...x2015=2016/4=504

Tick nhé

\(A+B+C=xyz\)

\(VT=A+B+C\)

\(\Leftrightarrow VT=x^2yz+xy^2z+xyz^2\)

\(\Leftrightarrow VT=xyz\left(x+y+z\right)\)

\(\Leftrightarrow VT=xyz\)

\(\Rightarrow VT=VP\)

\(\Rightarrow A+B+C=xyz\left(dpcm\right)\)

a) \(30+\left(-30\right)+7\)

\(=30-30+7\)

\(=0+7\)

\(=7\)

b) \(12+\left(-31\right)\)

\(=12-31\)

\(=-31+12\)

\(=-\left(31-12\right)\)

\(=-19\)

c) \(\left(-40\right)+\left(-20\right)+\left(-40\right)\)

\(=-40-20-40\)

\(=-\left(40+20\right)-40\)

\(=-60-40\)

\(=-\left(60+40\right)\)

\(=-100\)

d) \(13+\left|-13\right|\)

\(=13+13\)

\(=26\)

e) \(\left|-13\right|+\left|-12\right|+\left|-17\right|\)

\(=13+12+17\)

\(=25+17\)

\(=42\)

f) \(-17+\left|-17\right|+12\)

\(=-17+17+12\)

\(=-\left(17-17\right)+12\)

\(=-0+12\)

\(=-\left(0-12\right)\)

\(=-\left(-12\right)\)

\(=12\)

\(\dfrac{2x+8}{4x-5}=\dfrac{-2}{9}\)

\(\Leftrightarrow9\left(2x+8\right)=-2\left(4x-5\right)\)

\(\Leftrightarrow9\left(2x+8\right)+2\left(4x-5\right)=0\)

\(\Leftrightarrow18x+72+8x-10=0\)

\(\Leftrightarrow26x+62=0\)

\(\Leftrightarrow26x=0-62\)

\(\Leftrightarrow26x=-62\)

\(\Leftrightarrow x=-62:26\)

\(\Leftrightarrow x=\dfrac{-31}{13}\)

Vậy \(x=\dfrac{-31}{13}\)

Tính : \( A=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)..........\left(1-\dfrac{1}{1+2+3+.......+1986}\right)\)

Đó làm đi.