a,= (x+4)\(^3\)

b,= (x-2)\(^3\)

c,= x\(^3\)+8

d,=x\(^3\)-27

a,= x\(^3\)+3.4x\(^2\)+3.4\(^2\).x+4\(^3\)=(x+4)\(^3\)

-3.2.x\(^2\)+3.2\(^2\).x-2\(^3\)= (x-2)\(^3\)

còn c,d áp dụng HĐT là ra! ( đc chx bà nội)

A

A. (x-2)³ = x³ - 6x² +12x - 8 (hằng đẳng thức)

A

\(A=6x^2+23x+21-\left(6x^2+23x-55\right)=76\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ C=x^4+x^3-3x^2-2x-\left(x^4+x^3-x^2-2x^2-2x+2\right)\\ =-2\)

ảnh lỗi r ạ

Lỗi rùi

lưu ảnh rồi gửi vào đây nhé

a)
=(x-2)³
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
 

a) A = (x - 3)(x² + 3x + 9) - (x³ + 3)

= x³ - 3³ - x³ - 3

= (x³ - x³) + (-27 - 3)

= -30

b) B = (2x + 1)(4x² - 2x + 1) - 8(x + 1/2)(x² - 1/2 x + 1/4)

= (2x)³ + 1³ - 8[x³ + (1/2)³]

= 8x³ + 1 - 8(x³ + 1/8)

= 8x³ + 1 - 8x³ - 1

= (8x³ - 8x³) + (1 - 1)

= 0

Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)

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

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

Lời giải:

a. $=(x-y)(x+y)=[(-1)-(-3)][(-1)+(-3)]=2(-4)=-8$
b. $=3x^4-2xy^3+x^3y^2+3x^2y+12xy+15y-12xy-12$

$=3x^4-2xy^3+x^3y^2+3x^2y+15y-12$
=3-2.1(-2)^3+1^3.(-2)^2+3.1^2(-2)+15(-2)-12$
$=-25$
c.

$=2x^4+3x^3y-4x^3y-12xy+12xy=2x^4-x^3y$

$=x^3(2x-y)=(-1)^3[2(-1)-2]=-1.(-4)=4$

d. 

$=2x^2y+4x^2-5xy^2-10x+3xy^2-3x^2y$

$=(2x^2y-3x^2y)+4x^2+(-5xy^2+3xy^2)-10x$

$=-x^2y+4x^2-2xy^2-10x$

$=-3^2.(-2)+4.3^2-2.3(-2)^2-10.3=0$

Câu 1:

b: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

\(\dfrac{1}{x-3}-\dfrac{1}{x+3}+\dfrac{2x}{9-x^2}\)

\(=\dfrac{1}{x-3}-\dfrac{1}{x+3}-\dfrac{2x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x+3-x+3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}\)

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

c: ĐKXĐ: \(x\notin\left\{2;0\right\}\)

Sửa đề: \(\dfrac{x+1}{x-2}+\dfrac{4-5x}{x^3+4x}:\dfrac{x-2}{x^2+4}\)

\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x^2+4\right)}\cdot\dfrac{x^2+4}{x-2}\)

\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x-2\right)}\)

\(=\dfrac{x\left(x+1\right)+4-5x}{x\left(x-2\right)}=\dfrac{x^2+x-5x+4}{x\left(x-2\right)}\)

\(=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)