\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)

a) \(\Leftrightarrow x^3-8-x^3+9x=1\)

\(\Leftrightarrow9x=9\Leftrightarrow x=1\)

b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)

\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)

a) 2x-5=3+2x-7x

2x-2x+7x=3+5

7x=8

  x=8/7

vậy x=8/7

a) 2x - 5 = 3 + 2x - 7x

=> 2x - 2x + 7x = 3 +5 

=> 7x = 8

=> x = 8/7

b) \(\left(2x-1\right)^2=\left(2x-1\right)^5\)

=> \(\left(2x-1\right)^2-\left(2x-1\right)^5=0\)

=> \(\left(2x-1\right)^2\left[1-\left(2x-1\right)^3\right]=0\)

=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)^3=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^3=1\end{cases}}\)

=> \(\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)

(2x + 1) . (2x + 2) . (2x + 3) . (2x + 4) - 5y = 11879

[(2x + 1). (2x + 4)].[(2x + 2) . (2x + 3)] -5y = 11879

(4x²+10x+4).(4x²+10x+6) -5y = 11879

Đặt t= 4x²+10x+4

t(t+2) -5y = 11879

t²+2t-5y = 11879

(t+1)² = 11880+5y

(4x²+10x+5)² = 5(2376+y)

=> x = 0; y=-2371

thanks

giải chi tiết cho mình từ chỗ thứ 8 đi

a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 ) 

<=> 2x - 3 - x + 5 = x + 7 - x - 2

<=> x = 3

b)(7x-5)-(6x+4)=(2x+3)-(2x+1)

<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1

<=> x = 11

c)(9x-3)-(8x+5)=(3x+2)

<=> 9x - 3 - 8x - 5 = 3x + 2

<=> -2x = 10

<=> x = -5

d)(x+7)-(2x+3)=(3x+5)-(2x+4)

<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4

<=> -2x = -3

<=> x = 3/2

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