?

\(A=\dfrac{x^3}{9y^2}-\dfrac{1}{8}x^2y+\dfrac{2}{15}xy^2\\ B=\dfrac{2a-b}{a+1}-\dfrac{\left(a-1\right)^2}{b-2}\cdot\dfrac{\left(b-2\right)\left(b+2\right)}{\left(a-1\right)\left(a+1\right)}\\ B=\dfrac{2a-b}{a+1}-\dfrac{\left(a-1\right)\left(b+2\right)}{a+1}\\ B=\dfrac{2a-b-\left(a-1\right)\left(b+2\right)}{a+1}\\ B=\dfrac{2a-b-ab-2a+b+2}{a+1}=\dfrac{2-ab}{a+1}\)

 câu b bài 2 thiếu nha e

B1

a, 638 +780 x 5 - 369 : 9 = 4497 

b, ( 273 + 485 ) x16 - 483 :3 x4 = 11484 

B2

a, 325 x 6 + 6 x 560 + 115 = 5425

B3

a, x = 1020

b, x = 36

B4

26 nha

ah có thể giải chi tiết b4 đc ko ạ

1. a) \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=14x^5+21x^7\)

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

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

2: \(x^2-8x+7=0\)

=>\(x^2-x-7x+7=0\)

=>\(x\left(x-1\right)-7\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x-7\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

1:

a: \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=21x^7+14x^5\)

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

\(=x^2+1\)

a: \(=x^2-2x-35\)

b: \(=x^2y^2+4xy-5\)

dễ mà nhân lại là được

a: =5x^3-5x^2y+5x-2x^2y+2xy^2-2y

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

b: =(x^2-1)(x+2)

=x^3+2x^2-x-2

c: =1/2x^2y^2(4x^2-y^2)

=2x^4y^2-1/2x^2y^4

d: =(x^2-1/4)(4x-1)

=4x^3-x^2-x+1/4

e: =x^2-2x-35+(2x+1)(x-3)

=x^2-2x-35+2x^2-6x+x-3

=3x^2-7x-38

a: \(=\dfrac{x+1+3}{x-3}=\dfrac{x+4}{x-3}\)

\(a,=4x^2-4x+1-4x^2+4-x^2-x+6=-x^2-5x+11\\ b,=8x^3+27-8x^3+72x=72x+27\)

a) \(=4x^2-4x+1-4\left(x^2-1\right)-\left(x^2-2x+3x-6\right)=4x^2-4x+1-4x^2+4-x^2-x+6=-x^2-5x+11\)

b) \(=8x^3+27-8x\left(x^2-9\right)=8x^3+27-8x^3+72x=72x+27\)

c: \(=4x^2-12x+9-4x^2+x=-11x+9\)

\(a,=2x^2-10x+x^2+x-6=3x^2-9x-6\\ b,=x^2+4x+4-x^2+8x-15=12x-11\\ c,=4x^2-12x+9-4x^2+x=-11x+9\)

Bài 2

A =  (x + 2)^ 2 - x - 3 x (x + 1) = x² + 4x + 4 - x² + 2x + 3 = 6x + 7

B = x^3 - 2x² + 5x - 10 = x² x (x - 2) + 5 x (x - 2) = (x - 2) x (x² + 5)

Vậy x^3 - 2x² + 5x - 10 : (x - 2) = x² + 5

a: Ta có: \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)

\(=\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{x^2-4y^2}\)

\(=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(\)\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x+2y}\)

b: \(\frac{4x+7}{2x+2}-\frac{3x+6}{2x+2}\)

\(=\frac{4x+7-3x-6}{2x+2}\)

\(=\frac{x+1}{2\left(x+1\right)}=\frac12\)

c: \(\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}\)

\(=\frac{x+9}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x\cdot\left(x+3\right)}\)

\(=\frac{x\left(x+9\right)-3\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2+6x+9}{x\left(x+3\right)\left(x-3\right)}\)

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

d: \(\frac{1}{x^2+3x+2}-\frac{1}{x^2-4}\)

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

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