nhanh lên các bạn nhé mai mình đi học rồi

a

\(xy+3x-7y-21\\ =\left(xy+3x\right)-\left(7y+21\right)\\ =x\left(y+3\right)-7\left(y+3\right)\\ =\left(y+3\right)\left(x-7\right)\)

b

\(2xy-15-6x+5y\\ =\left(2xy-6x\right)-\left(15-5y\right)\\ =2x\left(y-3\right)-5\left(3-y\right)\\ =2x\left(y-3\right)+5\left(y-3\right)\\ =\left(y-3\right)\left(2x+5\right)\)

c Đề phải là \(\left(2x^2y+2xy^2-x-y\right)\) mới phân tích được: )

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

d

\(7x^3y-3xyz-21x^2+9z\\ =\left(7x^3y-21x^2\right)-\left(3xyz-9z\right)\\ =7x^2\left(xy-3\right)-3z\left(xy-3\right)\\ =\left(xy-3\right)\left(7x^2-3z\right)\)

e

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

f

\(9x^2-25y^2-6x+10y\\ =\left(3x\right)^2-\left(5y\right)^2-\left(6x-10y\right)\\ =\left(3x-5y\right)\left(3x+5y\right)-2\left(3x-5y\right)\\ =\left(3x-5y\right)\left(3x+5y-2\right)\)

a: =x(y+3)-7(y+3)

=(y+3)(x-7)

b: \(=2xy-6x+5y-15\)

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

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

c: \(=2xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(2xy-1\right)\)

d: \(=xy\left(7x^2-3z\right)-3\left(7x^2-3z\right)\)

=(7x^2-3z)(xy-3)

e: =4x^2-y^2-2x-y

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

=(2x+y)(2x-y-1)

f: =(3x-5y)(3x+5y)-2(3x-5y)

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

P(x)+Q(x)

=3x^2y-2x+5xy^2-7y^2+3xy^2-7y^2-9x^2y-x-5

=8xy^2-14y^2-6x^2y-3x-5

=>Chọn A

chiu

j

x² + 2y² + 2xy - 6x - 2y + 13 = 0

<=> ( x² + 2xy + y² - 6x - 6y + 9 ) + ( y² + 4y + 4 ) = 0

<=> [ ( x² + 2xy + y² ) - ( 6x + 6y ) + 9 ] + ( y + 2 )² = 0

<=> [ ( x + y )² - 2( x + y ).3 + 3² ] + ( y + 2 )² = 0

<=> ( x + y - 3 )² + ( y + 2 )² = 0

Ta có : \(\hept{\begin{cases}\left(x+y-3\right)^2\\\left(y+2\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(x+y-3\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

Dấu "=" xảy ra <=> x = 5 ; y = -2

Thế x = 5 ; y = -2 vào A ta được :

\(A=\frac{5^2-7\cdot5\cdot\left(-2\right)+52}{5-\left(-2\right)}=\frac{25+70+52}{7}=\frac{147}{7}=21\)

Bài 1:
a) (2x - y) + (2x - y) + (2x - y) + 3y
= 3(2x - y) + 3y
= 3(2x - y + 3y)
= 3(2x + 2y)
= 3.2(x + y)
= 6(x + y)

b) (x + 2y) + (x - 2y) + (8x - 3y)
= x + 2y + x - 2y + 8x - 3y
= 9x - 3y
= 3(3x - y)

c) (x + 2y) - 2(x - 2y) - (2x - 3y)
= x + 2y - 2x + 4y - 2x + 3y
= 9y - 3x
= 3(3y - x)

Bài 2:
M + 2(x² - 4y²) + Q = 6x² - 4xy + 5y² + P
M + 2x² - 8y²   -3x² + 7xy - 2y² = 6x² - 4xy + 5y² + 9x² - 6xy + 3y²
M + 2x² - 3x² - 6x² - 9x² - 8y² - 2y² - 5y² - 3y² + 7xy + 4xy + 6xy = 0
M - 16x² - 18y² + 17xy = 0
M = 16x² + 18y² - 17xy

Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3

Bài 2:

a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)

= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y²

= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y²

= 6\(x^3\) - 5\(x^2\)y + 2y²

a) 6x² - 12x

= 6x(x - 2)

b) x² + 2x + 1 - y²

= (x² + 2x + 1) - y²

= (x + 1)² - y²

= (x + 1 - y)(x + 1 + y)

c) x + y + z + x² + xy + xz

= (x + x²) + (y + xy) + (z + xz)

= x(1 + x) + y(1 + x) + z(1 + x)

= (x + y + z)(x + 1)

d) xy + xz + y² + yz

= (xy + xz) + (y² + yz)

= x(y + z) + y(y + z)

= (x + y)(x + z)

e) x³ + x² + x + 1

= (x³ + x²) + (x + 1)

= x²(x + 1) + (x + 1)

= (x² + 1)(x + 1)

f) xy + y - 2x - 2

= (xy + y) - (2x + 2)

= y(x + 1) - 2(x + 1)

= (y - 2)(x + 1)

g) x³ + 3x - 3x² - 9

= (x³ - 3x²) + (3x - 9)

= x²(x - 3) + 3(x - 3)

= (x² + 3)(x - 3)

h) x² - y² - 2x - 2y

= (x² - y²) - (2x + 2y)

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

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

i) 7x² - 7xy - 5x = 5y

mk thấy con này sai sai ý

i) 7x² - 7xy - 5x + 5y

= (7x² - 7xy) - (5x - 5y)

= 7x(x - y) - 5(x - y)

= (7x - 5)(x - y)

a) \(xy+3x-7y-21\)
\(\Leftrightarrow\left(xy+3x\right)-\left(7y+21\right)\)
\(\Leftrightarrow x\left(y+3\right)-7\left(y+3\right)\)
\(\Leftrightarrow\left(x-7\right)\left(y+3\right)\)

b) \(2xy-15-6x+5y\)
\(\Leftrightarrow\left(2xy-6x\right)-\left(15-5y\right)\)
\(\Leftrightarrow x\left(2y-6\right)-5\left(3-y\right)\)
\(\Leftrightarrow2x\left(y-3\right)+5\left(y-3\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(y-3\right)\)