\(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)

\(=20x^3-10x^2+5x-20x^3+10x^2+4x\)

\(=9x\)

Thay x=15 \(\Rightarrow A=9.15=135\)

\(B=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)

\(=6x^2y^2-6xy^3-8x^3+8x^2y^2+5x^2y^2-5xy^3\)

\(=19x^2y^2-11xy^3-8x^3\)

Thay x=1/2 ; y=2 vào B \(\Rightarrow19.\left(\frac{1}{2}\right)^2.2^2-11\cdot\frac{1}{2}\cdot2^3-8\cdot\left(\frac{1}{2}\right)^3\)

\(=19-44-1\)

\(=-26\)

1) (x-1)² + (x- 4y)² + (y + 2)² +10 -1-4

GTNN = 5

2) tuong tu 

1) 

a) \(2x^2-12x+18+2xy-6y\)

\(=2x^2-6x-6x+18+2xy-6y\)

\(=\left(2xy+2x^2-6x\right)-\left(6y+6x-18\right)\)

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

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

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

b) \(x^2+4x-4y^2+8y\)

\(=x^2+4x-4y^2+8y+2xy-2xy\)

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

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

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

2)  \(5x^3-3x^2+10x-6=0\)

\(\Leftrightarrow x^2\left(5x-3\right)+2\left(5x-3\right)=0\Leftrightarrow\left(x^2+2\right)\left(5x-3\right)=0\)

Mà \(x^2+2>0\Rightarrow5x-3=0\Rightarrow x=\frac{3}{5}\)

\(x^2+y^2-2x+4y+5=0\)

\(\Leftrightarrow x^2+y^2-2x+4y+4+1=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

3)\(P\left(x\right)=x^2+y^2-2x+6y+12\)

\(P\left(x\right)=x^2+y^2-2x+6y+1+9+2\)

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

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)

Vậy \(P\left(x\right)_{min}=2\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

Bài làm

a) 2x² - 12x + 18 + 2xy - 6y

= 2x² - 6x - 6x + 18 + 2xy - 6y 

= ( 2xy + 2x² - 6x ) - ( 6y + 6x - 18 )

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

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

# Học tốt #

1.

a. x² - 2x + 1 = 0

x² - 2x*1 + 1² = 0

(x-1)² = 0

............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)

1, Tìm x biết:

a, x - 2x +1 = 0

(x-1)² = 0

x-1 = 0

x = 1. Vậy ...

b, ( 5x + 1) - (5x - 3) ( 5x + 3) = 30

25x² +10x + 1 - (25x² -9) = 30

25x² +10x + 1 - 25x² +9 = 30

10x + 10 =30

10(x+1) = 30

x+1 =3

x = 2. vậy ...

c, ( x - 1) ( x + x + 1) - x ( x +2 ) ( x - 2) = 5

(x³ - 1) - x(x² -4) = 5

x³ - 1 - x³ + 4x = 5

4x - 1 = 5

4x = 6

x = \(\dfrac{3}{2}\) .vậy ...

d, ( x - 2) - ( x - 3) ( x + 3x + 9 ) + 6 ( x + 1) = 15

x³ - 6x² + 12x - 8 - (x³ - 27) + 6 (x² + 2x +1) =15

x³ - 6x² + 12x - 8 - x³ + 27 + 6x² + 12x +6 =15

24x + 25 = 15

24x = -10

x = \(\dfrac{-5}{12}\) vậy ...

a/

\(\Leftrightarrow\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(y^2-6y+9\right)-19=0\)

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

Do 19 không thể phân tích thành tổng của 2 số chính phương nên pt vô nghiệm

b/

\(\left(4x^2+4y^2+8xy\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

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

Do x; y nguyên dương nên \(\left(2x+2y\right)^2>0\Rightarrow VT>0\)

Pt vô nghiệm

c/

\(\Leftrightarrow\left(x^2+4y^2+25-4xy+10x-20y+25\right)+\left(y^2-2y+1\right)+\left|x+y+z\right|=0\)

\(\Leftrightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+\left|x+y+z\right|=0\)

Do x;y;z nguyên dương nên \(\left|x+y+z\right|>0\Rightarrow VT>0\)

Vậy pt vô nghiệm

d/

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)

Do x;y;z nguyên dương nên vế phái luôn dương

Pt vô nghiệm

a, \(2x^2+3\left(x+1\right)\left(x-1\right)-5x\left(x+1\right)\)

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

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

\(=\left(2x^2+3x^2-5x^2\right)-3-5x\)

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

b, \(\left(4x+3y\right)\left(2x-5y\right)-\left(2x+6y\right)\left(3x-5y\right)\)

\(=8x^2-20xy+6xy-\left(15y^2-6x^2-10xy-18xy-30y^2\right)\)

\(=8x^2-20xy+6xy-15y^2+6x^2+10xy+18xy+30y^2\)

\(=\left(8x^2+6x^2\right)+\left(-20xy+6xy+10xy+18xy\right)+\left(-15y^2+30y^2\right)\)

\(=14x^2+14xy+15y^2\)

\(=14x.\left(x+y\right)+15y^2\)

Chúc bạn học tốt!!!

a) \(2x^2+3.\left(x+1\right).\left(x-1\right)-5x\left(x+1\right)\)

= \(2x^2+3.\left(x^2-1\right)-5x.\left(x+1\right)\)

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

= \(-5x-3\)

\(a,x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)

\(b,x\left(y+1\right)+y+1=\left(x+1\right)\left(y+1\right)\)

\(c,\left(x+y\right)^2-9z^2=\left(x+y-3z\right)\left(x+y+3z\right)\)

\(d,5\left(x^2-2x+1-y^2\right)=5\left(\left(x+1\right)^2-y^2\right)=5\left(x+1-y\right)\left(x+1+y\right)\)