a) Ta có: \(5y^3-10xy^2+5yx^2-20y\)

\(=5y\left(y^2-2xy+x^2-4y\right)\)

b) Ta có: \(x^2+2xy+y^2-xz-yz\)

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

\(=\left(x+y\right)\left(x+y-z\right)\)

c) Ta có: \(9x^2+y^2+6xy\)

\(=\left(3x\right)^2+2\cdot3x\cdot y+y^2\)

\(=\left(3x+y\right)^2\)

d) Ta có: \(8-12x+6x^2-x^3\)

\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)

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

e) Ta có: \(125x^3-75x^2+15x-1\)

\(=\left(5x\right)^3-3\cdot\left(5x\right)^2\cdot1+3\cdot5x\cdot1^2-1^3\)

\(=\left(5x-1\right)^3\)

\(a.\left(8x^4-4x^3+x^2\right):2x^2=4x^2-2x+\frac{1}{2}\)

\(b.\left(2x^4-x^3+3x^2\right):\left(-\frac{1}{3x^2}\right)=-6x^6+3x^5-9x^4\)

\(c.\left(-18x^3y^5+12x^2y^2-6xy^3\right):6xy=-3x^2y^4+2xy-y^2\)

\(d.\left(\frac{3}{4x^3y^6}+\frac{6}{5x^4y^5}-\frac{9}{10x^5y}\right):-\frac{3}{5x^3y}=-\frac{5}{4y^5}-\frac{2}{xy^4}-\frac{3}{2x^2}\)

bài 2:

a)\(A=16x^2-8x+3\)

\(=\left[\left(4x\right)^2-2.4x.1+1^2\right]-1+3\)

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

ta thấy: \(\left(4x-1\right)^2\ge0\)

\(\left(4x-1\right)^2+2\ge2\)

vậy GTNN của A là 2 khi \(x=\dfrac{1}{4}\)

b) \(B=19-6x-9x^2\)

\(=-\left[\left(3x\right)^2+2.3x.1+1^2\right]+19\)

\(=-\left(3x-1\right)^2+19\)

ta thấy: \(-\left(3x-1\right)^2\le0\)

\(-\left(3x-1\right)^2+19\le19\)

vậy GTLN của B là 19 khi \(x=\dfrac{1}{3}\)

2.

A = xy + 2yz + 3xz = xy + xz + 2yz + 2xz = x(y + z) + 2z(y + z)

Áp dụng BĐT: (a+b)^2/4 ≥ ab dấu = khi a = b
Ta có:
(x + y + z)^2/4 ≥ x(y + z)
(x+ y +z)^2/4 ≥ z(y + z)
=> A ≤ 3(x + y + z)^2/4 = 3.36/4 = 27
=> A max = 27 xảy ra khi:
{x = y + z
{z = y + z
<=> y = 0 và x = z = 3

\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)

\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)

\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)

\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

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

a: \(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\left(3x-y\right)\cdot\left(3x+y\right)=9x^2-y^2\)

b: \(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\cdot\dfrac{1}{2}\)

\(=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\)

\(=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

c: \(\left(5y-3x\right)\cdot\dfrac{1}{4}\cdot\left(12x+20y\right)\)

\(=\left(5y-3x\right)\left(5y+3x\right)\)

\(=25y^2-9x^2\)

d: \(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(\dfrac{3}{2}y+x\right)\cdot2\)

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

\(=\dfrac{9}{4}y^2-x^2\)

e: \(\left(a+b+c\right)\left(a+b-c\right)\)

\(=\left(a+b\right)^2-c^2\)

\(=a^2+2ab+b^2-c^2\)

mọi người giúp với

1.

PT $\Leftrightarrow x^2+3xy+(3y^2-3y)=0$

Coi đây là pt bậc 2 ẩn $x$

PT có nghiệm $\Leftrightarrow \Delta=(3y)^2-4(3y^2-3y)\geq 0$

$\Leftrightarrow -3y^2+12y\geq 0$

$\Leftrightarrow -y^2+4y\geq 0$

$\Leftrightarrow 0\leq y\leq 4$

Vì $y$ nguyên nên $y\in \left\{0;1;2;3;4\right\}$

Để pt có nghiệm nguyên thì $\Delta$ là scp. Thử các giá trị $y$ trên vô $\Delta$ ta thấy $y=0; 2;4$

Thay vô pt ban đầu thì:

$y=0\Rightarrow x=0$ (thỏa)
$y=2\Rightarrow x=-3\pm \sqrt{3}$ (loại)

$y=4\Rightarrow x=-6$ (thỏa)

2.

PT $\Leftrightarrow x^2-2xy+(5y^2-y-1)=0$

Coi đây là pt bậc 2 ẩn $x$.

$\Delta'=y^2-(5y^2-y-1)=-4y^2+y+1$

Để pt có nghiệm thì $\Delta'\geq 0$

$\Leftrightarrow -4y^2+y+1\geq 0$

$\Leftrightarrow \frac{1-\sqrt{17}}{8}\leq y\leq \frac{1+\sqrt{17}}{8}$

Mà $y$ nguyên nên $y=0$

Thay vô pt ban đầu ta có $x^2=1\Rightarrow x=\pm 1$

Vậy $(x,y)=(\pm 1,0)$