2:

a: 5/x-y/3=1/6

=>\(\dfrac{15-xy}{3x}=\dfrac{1}{6}\)

=>\(\dfrac{30-2xy}{6x}=\dfrac{x}{6x}\)

=>30-2xy=x

=>x(2y+1)=30

=>(x;2y+1) thuộc {(30;1); (-30;-1); (10;3); (-10;-3); (6;5); (-6;-5)}

=>(x,y) thuộc {(30;0); (-30;-1); (10;1); (-10;-2); (6;2); (-6;-3)}

b: x/6-2/y=1/30

=>\(\dfrac{xy-12}{6y}=\dfrac{1}{30}\)

=>\(\dfrac{5xy-60}{30y}=\dfrac{y}{30y}\)

=>5xy-60=y

=>y(5x-1)=60

=>(5x-1;y) thuộc {(-1;-60); (4;15); (-6;-10)}(Vì x,y là số nguyên)

=>(x,y) thuộc {(0;-60); (1;15); (-1;-10)}

\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-5.\left(\dfrac{1}{2}\right)^3+3\left(\dfrac{1}{2}\right)^2+\dfrac{2}{2}+5-5\left(\dfrac{1}{2}\right)^3+6\left(\dfrac{1}{2}\right)^2+\dfrac{2}{2}+5\)

\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-\dfrac{5.1}{8}+\dfrac{3.1}{4}+6-\dfrac{5.1}{8}+\dfrac{6.1}{4}+6\)

\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-\dfrac{5}{8}+\dfrac{3}{4}+6-\dfrac{5}{8}+\dfrac{3}{2}+6\)

\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=13\)

\(Q\left(x\right)-P\left(x\right)=6\)

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

\(3x^2=6\)

\(x^2=2\)

\(=>x=\pm\sqrt{2}\)

\(\left(3x+1\right)^2=9\left(x-2\right)^2\)

\(\Leftrightarrow9x^2+6x+1=9\left(x^2-4x+4\right)\)

\(\Leftrightarrow9x^2+6x+1=9x^2-36x+36\)

\(\Leftrightarrow9x^2+6x+1-9x^2+36x-36=0\)

\(\Leftrightarrow42x-35=0\)

\(\Leftrightarrow42x=35\)

\(\Leftrightarrow x=\dfrac{35}{42}=\dfrac{5}{6}\)

Vậy: \(S=\left\{\dfrac{5}{6}\right\}\)

1) \(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=8\left(x+4\right)\left(x-1\right)\)

2) \(A=x^2+2y^2+2xy-2y+2021=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+2020=\left(x+y\right)^2+\left(y-1\right)^2+2020\ge2020\)

\(minA=2020\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

\(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=2.\left(x+4\right).4\left(x-1\right)=8\left(x-1\right)\left(x+4\right)\)

Bài 1: 

Ta có: \(\left(3x+2\right)^2-\left(x-6\right)^2\)

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

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

\(=8\left(x+4\right)\left(x-1\right)\)

\(\Leftrightarrow a\cdot\dfrac{13}{15}=\dfrac{28}{13}:2=\dfrac{14}{13}\)

=>\(a=\dfrac{14}{13}:\dfrac{13}{15}=\dfrac{210}{169}\)

Lời giải:
$117=(2y+1)^2-x^2=(2y+1-x)(2y+1+x)$

Vì $x,y$ nguyên nên $2y+1-x, 2y+1+x$ nguyên. Do đó ta có bảng sau:

Chia câu hỏi ra cho thành nhiều phần cho dễ trả lời á bạn

Giúp dới

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

=>x=-1

bn ghi bắng công thức đi