(x-1)^2+(x-2)^2=2(x+4)^2-(22x+27)
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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}
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\(a.2\left(x-1\right)^2+\left(x+3\right)^2=3\left(x-2\right)\left(x+1\right)\)
\(\Leftrightarrow2x^2-4x+2+x^2+6x+9=3x^2-3x-6\)
\(\Leftrightarrow2x^2+x^2-3x^2-4x+6x+3x+2+9+6=0\)
\(\Leftrightarrow5x+17=0\)
\(\Leftrightarrow x=-\dfrac{17}{5}\)
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..................
\(c.TươngTự\)
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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
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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)
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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)
\(x^3+3x^2+3x+1-x^3-3x^2=2\)
\(3x+1=2\)
\(3x=1\)
\(x=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(\left(x-1\right)^2+\left(x-2\right)^2=2\left(x+4\right)^2-\left(22x+27\right)\\ \Rightarrow\left(x^2-2x+1\right)+\left(x^2-4x+4\right)=2\left(x^2+8x+16\right)-\left(22x+27\right)\\ \Rightarrow x^2-2x+1+x^2-4x+4=2x^2+16x+32-22x-27\\ \Rightarrow2x^2-6x+5=2x^2-6x+5\left(luôn.đúng\right)\)
Vậy pt có vô số nghiệm