a) \(2\left(x-1\right)^2+\left(x+3\right)^2=3\left(x-2\right)\left(x+1\right)\)
\(\Leftrightarrow2x^2-4x+2+x^2+6x+9=3x^2-3x-6\)
\(\Leftrightarrow5x=-17\)
\(\Rightarrow x=-\frac{17}{5}\)
b) \(\left(x+2\right)^2-2\left(x-3\right)=\left(x+1\right)^2\)
\(\Leftrightarrow x^2+4x+4-2x+6=x^2+2x+1\)
\(\Leftrightarrow10=1\)
=> vô nghiệm
c) \(\left(x-1\right)^2+\left(x-2\right)^2=2\left(x+4\right)^2-\left(22x+27\right)\)
\(\Leftrightarrow x^2-2x+1+x^2-4x+4=2x^2+8x+8-22x-27\)
\(\Leftrightarrow8x=-24\)
\(\Rightarrow x=-3\)
a) 2( x - 1 )2 + ( x + 3 )2 = 3( x - 2 )( x + 1 )
<=> 2( x2 - 2x + 1 ) + x2 + 6x + 9 = 3( x2 - x - 2 )
<=> 2x2 - 4x + 2 + x2 + 6x + 9 = 3x2 - 3x - 6
<=> 2x2 - 4x + x2 + 6x - 3x2 + 3x = -6 - 2 - 9
<=> 5x = -17
<=> x = -17/5
b) ( x + 2 )2 - 2( x - 3 ) = ( x + 1 )2
<=> x2 + 4x + 4 - 2x + 6 = x2 + 2x + 1
<=> x2 + 4x - 2x - x2 - 2x = 1 - 4 - 6
<=> 0x = -9 ( vô lí )
Vậy phương trình vô nghiệm
c) ( x - 1 )2 + ( x - 2 )2 = 2( x + 4 )2 - ( 22x + 27 )
<=> x2 - 2x + 1 + x2 - 4x + 4 = 2( x2 + 8x + 16 ) - 22x - 27
<=> 2x2 - 6x + 5 = 2x2 + 16x + 32 - 22x - 27
<=> 2x2 - 6x - 2x2 - 16x + 22x = 32 - 27 - 5
<=> 0x = 0 ( đúng ∀ x ∈ R )
Vậy phương trình nghiệm đúng ∀ x ∈ R
a) 2(x - 1)2 + (x + 3)2 = 3(x - 2)(x + 1)
=> 2(x2 - 2x + 1) + x2 + 6x + 9 = 3(x2 - x - 2)
=> 2x2 - 4x + 2 + x2 + 6x + 9 = 3x2 - 3x - 6
=>3x2 + 2x + 11 = 3x2 - 3x - 6
=> 3x2 + 2x + 11 - 3x2 + 3x + 6 = 0
=> 5x + 17 = 0
=> 5x = -17
=> x = -17/5
b) (x + 2)2 - 2(x - 3) = (x + 1)2
=> x2 + 4x + 4 - 2x + 6 = x2 + 2x + 1
=> x2 + 4x + 4 - 2x + 6 - x2 - 2x - 1 = 0
=> (x2 - x2) + (4x - 2x - 2x) + (4 + 6 - 1) = 0
=> 9 = 0(vô lí)
c) (x - 1)2 + (x - 2)2 = 2(x + 4)2 - (22x + 27)
=> x2 - 2x + 1 + x2 - 4x + 4 = 2(x2 + 8x + 16) -22x - 27
=> x2 - 2x + 1 + x2 - 4x + 4 = 2x2 + 16x + 32 - 22x - 27
=> (x2 + x2) + (-2x - 4x) + (1 + 4) = 2x2 + (16x - 22x) + (32 - 27)
=> 2x2 - 6x + 5 = 2x2 - 6x + 5
=> 2x2 - 6x + 5 - 2x2 + 6x - 5 = 0
=> (2x2 - 2x2) + (-6x + 6x) + (5 - 5) = 0
=> 0 = 0(đúng)
a,\(2\left(x-1\right)^2+\left(x+3\right)^2=3\left(x-2\right)\left(x+1\right)\)
\(2\left(x^2-2x+1\right)+x^2+6x+9=3\left(x^2-x-2\right)\)
\(2x^2-4x+2+x^2+6x+9=3x^2-3x-6\)
\(2x^2-4x+2+x^2+6x+9-3x^2+3x+6=0\)
\(5x+17=0\)
\(5x=-17\)
\(x=\frac{-17}{5}\)
