a, (x+2)^2-2(x-3)=(x+1)^2
b, (x-1)^2+(X-2)^2=2.(x+4)^2-(22X+27)
Những câu hỏi liên quan
a) 2(x-1)2+(x+3)2=3(x-2)(x+1)
b) (x-1)2+(x-2)2=2(x+4)2-(22x+27)
c) x3-5x2+12x-5=0
d) (2x-1)2+(x+3)2-5(x+7) (x-7)=0
(x-1)^2+(x-2)^2=2(x+4)^2-(22x+27)
\(\left(x-1\right)^2+\left(x-2\right)^2=2\left(x+4\right)^2-\left(22x+27\right)\\ \Rightarrow\left(x^2-2x+1\right)+\left(x^2-4x+4\right)=2\left(x^2+8x+16\right)-\left(22x+27\right)\\ \Rightarrow x^2-2x+1+x^2-4x+4=2x^2+16x+32-22x-27\\ \Rightarrow2x^2-6x+5=2x^2-6x+5\left(luôn.đúng\right)\)
Vậy pt có vô số nghiệm
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Tìm x, biết;
a, 2.[x-1]2 cộng [x cộng 3]2 = 3.[x-2].[x cộng 1].
b, [x cộng 2]2 - 2.[x-3] = [x cộng 1]2
c, [x-1]2 cộng [x-2]2 = 2. [x cộng 4]2 - [22x cộng 27].
\(a.2\left(x-1\right)^2+\left(x+3\right)^2=3\left(x-2\right)\left(x+1\right)\)
\(\Leftrightarrow2x^2-4x+2+x^2+6x+9=3x^2-3x-6\)
\(\Leftrightarrow2x^2+x^2-3x^2-4x+6x+3x+2+9+6=0\)
\(\Leftrightarrow5x+17=0\)
\(\Leftrightarrow x=-\dfrac{17}{5}\)
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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Bài 1: Tìm x, biết:
a, 2(x-1)^2+(x+3)^2=3(x-2)(x+1)
b, (x+2)^2-2(x-3)=(x+1)^2
c, (x-1)^2+(x-2)^2=2(x+4)^2-(22x+27)
Làm giúp mình với! Cảm ơn!
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)
Xem thêm câu trả lời
Bài 1: Tìm x
a) 2(x-1)2+(x+3)2=3(x-2)(x+1)
b) (x+2)2-2(x-3)=(x+1)2
c) (x-1)2+(x-2)2=2(x+4)2-(22x+27)
a) \(2(x-1)\)2 + \((x + 3)\)2 = \(3(x-2)(x+1)\)
⇔\(2x^2-4x+2+x^2+6x+9=3x^2+3x-6x-6\)
⇔\(2x^2+x^2-3x^2-4x+6x-3x+6x=-2-9-6\)
⇔\(5x=-17\)
⇔\(x=\frac{-17}{5}\)
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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) 2(x-1)2+(x+3)2=3(x-2)(x+1)
b) (x-1)2+(x-2)2=2(x+4)2-(22x+27)
c) x3-5x2+12x-5=0
d) (2x-1)2+(x+3)2-5(x+7) (x-7)=0
Tìm x, biết;
a, 2.[x-1]2 cộng [x cộng 3]2 = 3.[x-2].[x cộng 1].
b, [x cộng 2]2 - 2.[x-3] = [x cộng 1]2
c, [x-1]2 cộng [x-2]2 = 2. [x cộng 4]2 - [22x cộng 27].
Tìm x (sử dụng HĐT)
a) (x - 1)2 + (x - 2)2 = 2(x + 4)2 - (22x + 27)
b) (x +2)2 - 2(x - 3) = (x + 1)2
c) (x + 1)3 - x2(x + 3) = 2
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)
\(x^3+3x^2+3x+1-x^3-3x^2=2\)
\(3x+1=2\)
\(3x=1\)
\(x=\frac{1}{3}\)
Tìm x:
a) (2x - 1) (x^2 - x + 1) = 2x^3 - 3x^2 + 2
b) (x + 1) (x^2 + 2x + 4) - x^3 - 3x^2 + 16 = 0
c) (x + 1) (x + 2) (x + 5) - x^3 - 8x^2 = 27
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
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