1 ) a) \(4x^2-x^2+8x^2\)

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

\(=12.x^3\)

b) \(\frac{1}{2}.x^2.y^2-\frac{3}{4}.x^2.y^2+x^2.y^2\)

\(\left(\frac{1}{2}-\frac{3}{4}\right).x^2.x^2.x^2.+y^2+y^2+y^2\)

\(=-\frac{1}{4}.x^6+y^6\)

c) \(3y-7y+4y-6y\)

\(=\left(3-7+4-6\right).y.y.y.y\)

\(=-6.y^4\)

2) 

\(\left(-\frac{2}{3}.y^3\right)+3y^2-\frac{1}{2}.y^3-y^2\)

\(\left(-\frac{2}{3}+3-\frac{1}{2}\right).y^3.y^3-y\)

\(=\frac{25}{6}.y^5\)

b) \(5x^3-3x^2+x-x^3-4x^2-x\)

\(=\left(5-3-4\right).\left(x^3.x^2+x-x^3-x^2-x\right)\)

\(=-2.0=0\)

hông chắc

3)a)  \(5xy^2.\frac{1}{2}x^2y^2x\)

\(\left(5.\frac{1}{2}\right).x^2.x^2.x.y^2.y^2\)

\(=\frac{5}{2}.x^5.y^4\)

b) Tổng các bậc của đơn thức là

5+4 = 9

Hệ số của đơn thức là \(\frac{5}{2}\)

Phần biến là x;y

Thay x=1;y=-1 vào đơn thức

\(\frac{5}{2}.1^5.\left(-1\right)^4\)

\(\frac{5}{2}.1.\left(-1\right)\)

\(\frac{5}{2}.\left(-1\right)=-\frac{5}{2}\)

Vậy ....

chắc không đúng đâu uwu

đã tắt máy chưa để cho mình giải nha

Giúp mik nha mọi người :)

Chưa tắt máy, ai giúp mik giải với!!!

\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)

\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)

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

\(=6x^2y\)

\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)

\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)

\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)

1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy

2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3

=6x^2y

3: =(x+y-x+y)^2=(2y)^2=4y^2

4: =(2x+3-2x-5)^2=(-2)^2=4

5: =18^8-18^8+1=1

a) (x + 2)(2 - x)  - (2x - 1)(x + 3)

= 4 - x² - 2x² - 5x + 3

= -3x² - 5x + 7

b) (x + 1)² - 2(x² - 1) + (x - 1)² 

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

= (x + 1 - x + 1)² 

= 2² = 4

c) (x + y)³ - (x - y)² - 6x²y

= x³ + 3x²y + 3xy² + y³ - x³ + 3x²y - 3xy² + y³ - 6x²y

= 2y³ 

B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

a) (x - y)(x² + xy + y²)

Ta có :

(x - y)(x² + xy + y²) + 2y³

= x³ - y³ + 2y³

= x³ + y³

thay x = 2; y = (-3) vào biểu thức đại số, ta có:

2³ + (-3)³ = 8 + (-27)

= -19

b) (x + y)(x² - xy + y²) - 2y³

= x³ + y³ - 2y³

= x³ - y³

Thay x = 2 : y = -3 vào biểu thức đại số, ta có:

2³ - (-3)³ = 8 - (-27)

= 8 + 27

= 35

Mk không chắc lắm nhưng hình như con b) bn chép sai đầu bài nên mk đã sửa,

MONG BN THÔNG CẢM

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

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

\(\left(2y-3\right)^3=8y^3-36y^2+54y-27\)

a: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

A = (\(x-y\)).(\(x^2\) + \(xy\) + y²) + 2y³

A = \(x^3\) - y³ + 2y³

A = \(x^3\) + y³

Thay \(x=\dfrac{2}{3}\); y = \(\dfrac{1}{3}\) vào biểu thức

A = \(x\)³ +  y³ ta có:

A = (\(\dfrac{2}{3}\))³ + (\(\dfrac{1}{3}\))³

A = \(\dfrac{8}{27}\) + \(\dfrac{1}{27}\)

A = \(\dfrac{9}{27}\)

A = \(\dfrac{1}{3}\) 

\(B=\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}\)

\(=\frac{x^2y-x^2z+zy^2-xy^2+z^2x-z^2y}{x^2\left(y-z\right)-y^2\left(y-z\right)}\)

\(=\frac{\left(x^2y-z^2y\right)-\left(xy^2-zy^2\right)-\left(x^2z-z^2x\right)}{\left(x^2-y^2\right)\left(y-z\right)}\)

\(=\frac{\left[y\left(x+z\right)-y^2-xz\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)

\(=\frac{\left(xy+zy-y^2-xz\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)

\(=\frac{\left[\left(xy-y^2\right)-\left(xz-zy\right)\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)

\(=\frac{\left[y\left(x-y\right)-z\left(x-y\right)\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)

\(=\frac{\left(y-z\right)\left(x-y\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)

\(=\frac{x-z}{x+y}\)

\(A=\frac{\left(x^2-y\right)\left(y+1\right)+x^2y^2-1}{\left(x^2+y\right)\left(y+1\right)+x^2y^2+1}\)

\(=\frac{x^2y-y^2+x^2-y+x^2y^2-1}{x^2y+y^2+x^2+y+x^2y^2+1}\)

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

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

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

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

\(=\frac{x^2-1}{x^2+1}\)