Question

Tìm x : 

a) x(x - 4 ) - ( x^2 - 8 ) = 0

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

c) x( x + 5 )( x - 5 ) - ( x + 2 )( x^2 - 2x + 4 ) = 17

Answer 1

a) \(x\left(x-4\right)-\left(x^2-8\right)=0\)

\(\Leftrightarrow x^2-4x-x^2+8=0\)

\(\Leftrightarrow-4\left(x-2\right)=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

b) \(\left(3x+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)

\(\Leftrightarrow3x^2-3x+2x-2-3\left(x^2-2x+x-2\right)=0\)

\(\Leftrightarrow3x^2-3x+2x-2-3x^2+6x-3x+6=0\)

\(\Leftrightarrow2x=-4\)

\(\Leftrightarrow x=-2\)

c) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\)

\(\Leftrightarrow x^3-25x-x^3-8=17\)

\(\Leftrightarrow-25x=25\)

\(\Leftrightarrow x=-1\)

Answer 2

a) x(x - 4 ) - ( x^2 - 8 ) = 0

=>x²-4x-x²+8=0

=>8-4x=0

=>4x=8

=>x=2

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

=>3x²-x-2-3x²+3x+6=4

=>2x+4=4

=>2x=0

=>x=0

c) x( x + 5 )( x - 5 ) - ( x + 2 )( x^2 - 2x + 4 ) = 17

=>x(x²-25)-(x³+8)=17

=>x³-25x-x³-8=17

=>-25x-8=17

=>-25x=25

=>x=-1