bó tay! =_=

a) x⁴ - x⁵ = x⁴( x - 1 )

b) -8x²y² - 12xy³ - 4xy²

= -4xy( 2xy + 3y² + y )

c) ( x - y )³ - x³ + y³

= x³ - 3x²y + 3xy² - y³ - x³ + y³

= 3xy² - 3x²y

= 3xy( y - x )

a) 6x - x² - 5

= -x² + 6x - 9 + 4

= -( x² - 6x + 9 ) + 4

= -( x - 3 )² + 4 ≤ 4 ∀ x ( chưa kl luôn âm được :)) )

https://www.youtube.com/watch?v=ADwFwrLD-_I

\(A=\frac{7}{x+4}+\frac{8}{x-4}+\frac{14x}{x^2-16}=\frac{7}{x+4}+\frac{8}{x-4}+\frac{14x}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{7\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}+\frac{8\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{14x}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{7x-28+8x+32+14x}{\left(x-4\right)\left(x+4\right)}=\frac{29x+4}{\left(x-4\right)\left(x+4\right)}\)

\(B=\frac{x^2-2x+1}{x-1}+\frac{x^2-9}{x+3}=\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x-3\right)\left(x+3\right)}{x+3}\)

\(=x-1+x-3=2x-4\)

a) A=x³+3x²+3x

A=x³+3x².1+3x.1²+1³

A=(x+1)³

b)A=x³-3x²+3x-1

A=x³-3x².1+3x.1²-1³

A=(x-1)³

c)A=x³+6x²+12x

A=x³+3.2x²+3.2²x+1³

A=(x+1)³

A = x³ + 3x² + 3x = (x³ + 3x² + 3x + 1) - 1 =  (x + 1)³ - 1³ = (x + 1 - 1)[(x + 1)² + (x + 1) + 1] = x(x² + 3x + 3)

A = x³ - 3x² + 3x - 1 = (x - 1)³

A = x³ + 6x² + 12x = (x³ + 6x² + 12x + 8) - 8 = (x + 2)³ - 2³ = (x + 2 - 2)[(x  + 2)² + 2(x + 2) + 4) = x(x² + 6x + 12)

1, \(A=\frac{9}{x+1}-\frac{8}{1-x}-\frac{16}{x^2-1}\)

\(=\frac{9}{x+1}-\frac{8}{1-x}-\frac{16}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{9\left(1-x\right)\left(x-1\right)}{\left(x+1\right)\left(1-x\right)\left(x-1\right)}-\frac{8\left(x+1\right)\left(x-1\right)}{\left(1-x\right)\left(x+1\right)\left(x-1\right)}-\frac{16\left(1-x\right)}{\left(1-x\right)\left(x+1\right)\left(x-1\right)}\)

\(=\frac{9\left(1-x\right)\left(x-1\right)-8\left(x+1\right)\left(x-1\right)-16\left(1-x\right)}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}\)

\(=\frac{18x-9-9x^2-8x^2+8-16+16x}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}=\frac{-17x^2+34x-17}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}\)

\(=\frac{-17\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}=\frac{-17\left(x-1\right)}{\left(x+1\right)\left(1-x\right)}\)

2, \(B=\frac{x^2+10x+25}{x+5}-\frac{x^2-6x+9}{x-3}\)

\(=\frac{\left(x+5\right)^2}{x+5}-\frac{\left(x-3\right)^2}{x-3}=x+5-x+3=8\)

Ta có x = 100

=> x + 1 = 101

Khi đó A = x¹⁵ - 101x¹⁴ + 101x¹³ - 101x¹² + ... + 101x³ - 101x² + 101x + 2020

 = x¹⁵ - (x  + 1)x¹⁴ + (x + 1)x¹³ - (x + 1)x¹² + ... + (x + 1)x³ - (x + 1)x² + (x + 1)x + 2020

= x¹⁵ - x¹⁵ - x¹⁴ + x¹⁴ + x¹³ - x¹³ - x¹² + ... + x⁴ + x³ - x³ - x² + x² + x + 2020

= x + 2020

= 101 + 2020 (Vì  x = 100)

= 2121

Vậy A = 2121 khi x = 100

A = x¹⁵ - 101x¹⁴ + 101x¹³ - ... - 101x² + 101x + 2020 tại x = 100

x = 100 => 101 = x + 1

Thế vào A ta được

A = x¹⁵ - ( x + 1 )x¹⁴ + ( x + 1 )x¹³ - ... - ( x + 1 )x² + ( x + 1 )x + 2020

= x¹⁵ - ( x¹⁵ + x¹⁴ ) + ( x¹⁴ + x¹³ ) - ... - ( x³ + x² ) + ( x² + x ) + 2020

= x¹⁵ - x¹⁵ - x¹⁴ + x¹⁴ + x¹³ - ... - x³ - x² + x² + x + 2020

= x + 2020

= 100 + 2020 = 2120

Sửa: \(A=\frac{x^2+6x+9}{x+3}+\frac{x^2-16}{x-4}=\frac{\left(x+3\right)^2}{x+3}+\frac{\left(x-4\right)\left(x+4\right)}{x-4}=x+3+x+4=2x+7\) (đk: \(x\ne-3;x\ne4\))

\(B=\frac{5}{x+2}+\frac{6}{x-2}-\frac{10x}{x^2-4}\)(đk x\(\ne\)\(\pm\)2)

\(B=\frac{5\left(x-2\right)+6\left(x+2\right)-10x}{\left(x-2\right)\left(x+2\right)}\)

\(B=\frac{5x-10+6x+12-10x}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x-2}\)

Gọi 2 số tự nhiên cần tìm là a,b (a,b \(\in\)N)

Theo bài ra, ta có: \(\hept{\begin{cases}a+b=13\\ab=36\end{cases}}\)

<=> \(\hept{\begin{cases}a=13-b\\ab=36\end{cases}}\) <=> \(\hept{\begin{cases}a=13-b\\\left(13-b\right)b=36\end{cases}}\) <=> \(\hept{\begin{cases}a=13-b\\b^2-13b+36=0\end{cases}}\)

<=> \(\hept{\begin{cases}a=13-b\\b^2-4x-9x+36=0\end{cases}}\) <=> \(\hept{\begin{cases}a=13-b\\\left(b-4\right)\left(b-9\right)=0\end{cases}}\)

<=> a = 13 - b và b = 4 hoặc b = 9

Với b = 4 => a = 13 - 4 = 9

Với b = 9 => a = 13 - 9 = 4

a) vật thể nhân tạo

b) vật thể tự nhiên

c) vật thể nhân tạo

d) vật thể tự nhiên