Bài 1 :

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

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

Bài 2:

\(b,2x^3+3x^2+2x+3=0\)

\(\Leftrightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)

\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3=0\left(x^2+1>0\right)\)

\(\Leftrightarrow x=-\dfrac{3}{2}\)

mấy bn ơi, giúp mk nhanh vs nha!!!!!!!!!!!

a/ A = 2x² + y² - 2xy - 2x + 3

= (x² - 2xy + y²) + (x² - 2x + 1) + 2

= (x - y)² + (x - 1)² + 2\(\ge2\)

Câu 4:

5x + 7y = 112

5(x+ y) = 112 - 2y

5(x + y) = 2(56 - y)

\(\begin{cases}x+y=2\\ 56-y=5\end{cases}\)

\(\begin{cases}x+y=2\\ y=56-5\end{cases}\)

\(\begin{cases}x=2-y\\ y=51\end{cases}\)

\(\begin{cases}x=2-51\\ y=51\end{cases}\)

\(\begin{cases}x=-49\\ y=51\end{cases}\)

Vậy (x ; y) = (-49; 51)

1)<=> \(\left(x^2+2x+1\right)-\left(y^2+4x+4\right)-7=0\)

=> \(\left(x+1\right)^2-\left(y+2\right)^2=7\)

\(\Rightarrow\left(x+1-y-2\right)\left(x+1+y+2\right)=7\)

\(\left(x-y-1\right)\left(x+y+3\right)=7\)

vì x;y là số nguyên dương=> \(x;y\ge1\Rightarrow x+y+3\ge1+1+3=5\)

=> \(x+y+3=7\Rightarrow x+y=4\)

\(x-y-1=1\Rightarrow x-y=2\)

cộng hai phương trình ta có:

=> \(\left(x+y\right)+\left(x-y\right)=4+2\)

=> \(2x=6\)

=> x=3

=> y=4-3=1

2)<=> \(\left(x^2+xy+2x\right)+\left(2xy+2y^2+4y\right)+\left(x+y+2\right)=17\)

=> \(x\left(x+y+2\right)+2y\left(x+y+2\right)+1\left(x+y+2\right)=17\)

=> \(\left(x+y+2\right)\left(x+2y+1\right)=17\)

a: \(=3y^2-5x^2y^3-2y^2+3x^2y^3=y^2-2x^2y^3\)

b: \(=6x-y+2x^2+3y^2-2x^2+x=7x-y+3y^2\)

c: \(=x-y+4y^2-6xy+\dfrac{10x^2}{y}\)

\(a.\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)y^2\)

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

\(=y^2-2x^2y^3\)

\(b.\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)

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

\(=2x^2+7x+2y-2\)

\(c.\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^3\right):\dfrac{3}{2}x^2y^3\)

\(=x-y+4y^2-6xy+10x^2\)