a) \(\frac{3x+6}{x^2-4}=\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3}{x-2}\)( ĐKXĐ : x ≠ ±2 )

\(\frac{2x+6}{x^3+3x^2-9x-27}=\frac{2\left(x+3\right)}{x^2\left(x+3\right)-9\left(x+3\right)}=\frac{2\left(x+3\right)}{\left(x+3\right)\left(x^2-9\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)( ĐKXĐ : x ≠ ±3 )

MTC : ( x - 2 )( x - 3 )( x + 3 )

=> \(\hept{\begin{cases}\frac{3}{x-2}=\frac{3\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3\left(x^2-9\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3x-27}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\\\frac{2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{4x-4}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\end{cases}}\)

b) \(\frac{x^2-4x+4}{2x^2-3x+1}=\frac{\left(x-2\right)^2}{2x^2-2x-x+1}=\frac{\left(x-2\right)^2}{2x\left(x-1\right)-\left(x-1\right)}=\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}\)( ĐKXĐ : \(\hept{\begin{cases}x\ne1\\x\ne\frac{1}{2}\end{cases}}\))

\(\frac{x+4}{2x-2}=\frac{x+4}{2\left(x-1\right)}\)( ĐKXĐ : x ≠ 1 )

MTC : \(2\left(x-1\right)\left(2x-1\right)\)

=> \(\hept{\begin{cases}\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}=\frac{2\left(x^2-4x+4\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2-8x+8}{2\left(x-1\right)\left(2x-1\right)}\\\frac{x+4}{2\left(x-1\right)}=\frac{\left(x+4\right)\left(2x-1\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2+7x-4}{2\left(x-1\right)\left(2x-1\right)}\end{cases}}\)

c) \(\frac{6a}{a-b}\)( ĐKXĐ : a ≠ b ) ; \(\frac{2b}{b-a}=\frac{-2b}{a-b}\)( ĐKXĐ : a ≠ b) ; \(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\)( ĐKXĐ : a ≠ ±b )

MTC : \(\left(a-b\right)\left(a+b\right)\)

=> \(\frac{6a}{a-b}=\frac{6a\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{6a^2+6ab}{\left(a-b\right)\left(a+b\right)}\)

\(\frac{-2b}{a-b}=\frac{-2b\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{-2ab-2b^2}{\left(a-b\right)\left(a+b\right)}\)

\(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\)

d) \(\frac{x}{x^2+11x+30}=\frac{x}{x^2+5x+6x+30}=\frac{x}{x\left(x+5\right)+6\left(x+5\right)}=\frac{x}{\left(x+5\right)\left(x+6\right)}\)( ĐKXĐ : x ≠ -5 ; x ≠ -6 )

\(\frac{5}{x^2+9x+20}=\frac{5}{x^2+4x+5x+20}=\frac{5}{x\left(x+4\right)+5\left(x+4\right)}=\frac{5}{\left(x+4\right)\left(x+5\right)}\)( ĐKXĐ : x ≠ -4 ; x ≠ -5 )

MTC : \(\left(x+4\right)\left(x+5\right)\left(x+6\right)\)

=> \(\hept{\begin{cases}\frac{x}{\left(x+5\right)\left(x+6\right)}=\frac{x\left(x+4\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{x^2+4x}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\\\frac{5}{\left(x+4\right)\left(x+5\right)}=\frac{5\left(x+6\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{5x+30}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\end{cases}}\)

Sai chỗ nào bạn bỏ qua nhé 

\(\left(8x^3-7x^2\right)\div x^2=3x+\sqrt{\frac{9}{25}}\)

\(\Leftrightarrow\left(8x^3\div x^2\right)-\left(7x^2\div x^2\right)=3x+\frac{3}{5}\)

\(\Leftrightarrow8x-7=3x+\frac{3}{5}\)

\(\Leftrightarrow8x-3x=\frac{3}{5}+7\)

\(\Leftrightarrow5x=\frac{38}{5}\)

\(\Leftrightarrow x=\frac{38}{25}\)

( x + 1 )( 2 - x ) - ( 5x + 5 )( x + 2 ) = -4x² + 2

⇔ -x² + x + 2 - ( 5x² + 15x + 10 ) = -4x² + 2

⇔ -x² + x + 2 - 5x² - 15x - 10 = -4x² + 2

⇔ -6x² - 14x - 8 + 4x² - 2 = 0

⇔ -2x² - 14x - 10 = 0

⇔ -2( x² + 7x + 5 ) = 0

⇔ x² + 7x + 5 = 0 (*)

Δ = b² - 4ac = 7² - 4.5.1 = 49 - 20 = 29

Δ > 0 nên (*) có hai nghiệm phân biệt :

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-7+\sqrt{29}}{2}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-7-\sqrt{29}}{2}\end{cases}}\)

Vậy ... ( sao nghiệm xấu thế nhỉ ? )

Ngu thế tự đi mà làm rảnh đâu mà chỉ tao còn ko biết làm còn đi tìm câu trả lời đây này nhá:v có câu trả lời thì nói chuyện nhá ko có cút đi đồ ngu

Ye ye let do it bạn đã làm những điều thật phi phàm mình ném nón vàng để giành quyền chọn bạn về đội let go tạo nên lịch sử rap việt nào let gooooo

\(a^3+b^3+3\left(a^2+b^2\right)+4\left(a+b\right)+4=0\)

<=> \(\left(a+1\right)^3+\left(b+1\right)^3+\left(a+1\right)+\left(b+1\right)=0\)

<=> \(\left(a+1+b+1\right)\left[\left(a+1\right)^2+\left(b+1\right)^2-\left(a+1\right)\left(b+1\right)+1\right]=0\)

<=> \(a+b+2=0\)

<=> a + b = - 2 

Khi đó: 2020 (a +b ) = 2020. ( -2) = -4040

x^3+ 64= 0

x^3 = -64

x^3 = (-4)^3

x=-4

x³ + 64 = 0

x³= 0+64

x³ = 64 

x³= 4³

=> x=4

2x³ - x² + 2x - 1 = 0

⇔ x²( 2x - 1 ) + ( 2x - 1 ) = 0

⇔ ( 2x - 1 )( x² + 1 ) = 0

⇔ 2x - 1 = 0 hoặc x² + 1 = 0

⇔ x = 1/2 ( x² + 1 ≥ 1 > 0 ∀ x )