a) (x + 2)² - 2 (x - 3) = (x + 1)²

<=> x² + 4x + 4 - 2x + 6 = x² + 2x + 1

<=> x² + 2x + 10           = x² + 2x + 1

<=> x² + 2x - x² - 2x       = 1 - 10

<=> 0x                           = -9

Vậy PT vô nghiệm

b) (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

<=> 2x² + 6x + 5                        = 2x² - 6x + 5

<=> 2x² + 6x - 2x² + 6x               = 5 - 5

<=> 12x                                     = 0

<=>   x                                       = 0 : 12

<=>   x                                       = 0

Vậy PT có nghiệm x = {0}

\(a.2\left(x-1\right)^2+\left(x+3\right)^2=3\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow2x^2-4x+2+x^2+6x+9=3x^2-3x-6\)

\(\Leftrightarrow2x^2+x^2-3x^2-4x+6x+3x+2+9+6=0\)

\(\Leftrightarrow5x+17=0\)

\(\Leftrightarrow x=-\dfrac{17}{5}\)

KL.............

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

\(\Leftrightarrow x^2+4x+4-2x+6=x^2+2x+1\)

\(\Leftrightarrow x^2-x^2+4x-2x-2x+4+6-1=0\)

\(\Leftrightarrow9=0\left(vôly\right)\)

KL..................

\(c.TươngTự\)

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

⇔\(2x^2-4x+2+x^2+6x+9=3x^2+3x-6x-6\)

⇔\(2x^2+x^2-3x^2-4x+6x-3x+6x=-2-9-6\)

⇔\(5x=-17\)

⇔\(x=\frac{-17}{5}\)

b: \(\Leftrightarrow x^2+4x+4-2x+6-x^2-2x-1=0\)

=>9=0(vô lý)

c: \(\Leftrightarrow x^2-2x+1+x^2-4x+4=2x^2+16x+32-22x-27\)

=>\(2x^2-6x+5-2x^2+6x-5=0\)

=>0x=0(luôn đúng)

a) ( 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 có vô số nghiệm

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

<=> x² + 4x + 4 - 2x + 6 = x² + 2x + 1

<=> x² + 2x - x² - 2x = 1 - 4 - 6

<=> 0x = -9 ( vô lí )

Vậy phương trình vô nghiệm

c) ( x + 1 )³ - x²( x + 3 ) = 2

<=> x³ + 3x² + 3x + 1 - x³ - 3x² = 2

<=> 3x + 1 = 2

<=> 3x = 1

<=> x = 1/3

a) 

\(x^2-2x+1+x^2-4x+4=2\left(x^2+8x+16\right)-22x-27\) 

\(2x^2-6x+5=2x^2+16x+32-22x-27\) 

\(-6x+5=-6x+5\) 

\(0=0\left(llđ\forall x\right)\) 

Vậy  \(x=R\) 

b) 

\(x^2+4x+4-2x+6=x^2+2x+1\) 

\(x^2+2x+10=x^2+2x+1\) 

\(10=1\) 

\(0=-9\left(sai\right)\) 

Vậy phương trình vô nghiệm 

c) 

\(x^3+3x^2+3x+1-x^3-3x^2=2\) 

\(3x+1=2\) 

\(3x=1\) 

\(x=\frac{1}{3}\)

ko có yêu cầu ai làm đc

lam gium minh 

ay

D=5x²+5x-27x+27/(x)²2(x)(2) +2²

D=5x(x+1)-27(x+1)/(x-2)²

D=(5x-27)(x+1)/(x-20(x-2)

bai cua ban Hakawa Genzo sai roi -27x+27 e thanh -27(x-1) ma