b: \(=3\left[\left(x+y\right)^2-2xy\right]-2\left[\left(x-y\right)^3+3xy\left(x-y\right)\right]\)

\(=3\left(1-2xy\right)-2\left(1+3xy\right)\)

\(=3-6xy-2-6xy=-12xy+1\)

c: \(=\left(x+y\right)^3-3\left(x^2+y^2+2xy\right)+3\left(x+y\right)+2012\)

\(=101^2-3\cdot101^2+3\cdot101+2012\)

=1002013

Nhân ra thôi chứ sao?

thì bạn nhân đi !

a)\(x^4-6x^2+2x+28\)

\(=\left(x^4-x^3\right)+\left(x^3-x^2\right)-\left(5x^2-5x\right)-\left(3x-3\right)+25\)

\(=\left(x-1\right)\left(x^3+x^2-5x-3\right)+25\)

=> số dư là 25

b) Cách làm tương tự câu a nhé

\(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\le0\)

Nhận thấy:  \(\left|2x+1\right|\ge0\);     \(\left|x+y-\frac{1}{2}\right|\ge0\)

=>   \(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\ge0\)

Dấu "=" xảy ra  <=>  \(\hept{\begin{cases}2x+1=0\\x+y-\frac{1}{2}=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=-\frac{1}{2}\\y=1\end{cases}}\)

đến đây bạn thay x,y tìm đc vào A để tính nhé

1

2x . 3=3y .4

=> x=2y=>\(\frac{x}{2}=y\Rightarrow\frac{x}{4}=\frac{y}{2}\)

\(\frac{x}{4}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{2}=\frac{z}{5}=\frac{x-2y+3z}{4-4+15}=\frac{1}{15}=\)

x=1/15x4=4/15

y=1/15x2=2/15

z=1/15x6=1/10

\(\Rightarrow x-y-z=\frac{4}{15}-\frac{2}{15}-\frac{1}{10}=\frac{1}{30}\)

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

4\(x^2\)-12x+9-2(9\(x^2\)+6x+1)=2\(x^2\)-4x+\(x^2\)+2x-x-2

4\(x^2\)-12x+9-18\(x^2\)-12x-2=2\(x^2\)-4x+\(x^2\)+2x-x-2

(4\(x^2\)-18\(x^2\)-2\(x^2\)-\(x^2\)) +(-12x-12x+4x-2x+x)+(9-2+2)=0

-17\(x^2\)-21x+9=0

=1-1/2+1/2-1/3+....................+1/99-1/100

=1+(1/2-1/2+1/3-1/3+.............+1/99-1/99)-1/100

=1-1/100

=99/100

A = 1 - 1/2 + 1/2 - 1/3 +.............+ 1/99 - 1/100

   = 1 - 1/100 = 99/100

a) (2x - 3y)² = (2x)² + 2.2x.3y + (3y)² = 4x² + 12xy + 9y²

b) 99² = (100 - 1)² = 100² - 2.100.1 + 1² = 10000 - 200 + 1 = 9801

A) 2²-2.2.3+3²

 = 4-12+9

= 1

B)  =(100-1)²

     = 100²-2.100.1+1²

     = 10000-200+1 

   =9801