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Mất dấu nên xét 2 th.
TH1
`x^2-4x+4=0`
`<=>x^2-2.x.2+2^2=0`
`<=>(x-2)^2=0`
`<=>x-2=0`
`<=>x=2`
`=>S={2}`
TH2
`x^2+4x+4=0`
`<=>x^2+2.x.2+2^2=0`
`<=>(x+2)^2=0`
`<=>x+2=0`
`<=>x=-2`
`=>S={-2}`
\(S=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)
\(=\left[\dfrac{n\left(n+1\right)}{2}\right]^2=\dfrac{n^2\cdot\left(n+1\right)^2}{4}\)
Ta có:
\(x^2+4x+4\)
\(=x^2+2.2x+2^2\)
\(=\left(x+2\right)^2\)
`4x^2+4x+1=0`
`<=>(2x)^2+2.2x+1=0`
`<=>(2x+1)^2=0`
`<=>2x+1=0`
`<=>2x=-1`
`<=>x=-1/2`
Vậy `x=-1/2.`
a) \(2x\left(2x+5\right)-4x\left(x-3\right)=7\)
\(4x^2+10x-4x^2+12x=7\)
\(22x=7\Rightarrow x=0,31\)
b) \(\left(x+2\right)\left(x-2\right)-\left(x+1\right)^2=2\)
\(\left(x^2-4\right)-\left(x^2+2x+1\right)=2\)
\(x^2-4-x^2-2x-1=2\)
\(-2x=7\Rightarrow x=-3,5\)
c) \(\left(x+2\right)\left(x-1\right)-\left(x+3\right)\left(x-2\right)=0\)
\(x^2-x+2x-2-x^2+2x+3x-6=0\)
\(6x=8\Rightarrow x=1,3\)
1)3.x^2 - 75 = 0
3.x^2 - 3.25 = 0
3.(x^2-25)=0
x^2-5^2=0
(x-5)(x+5)=0
=> x-5=0 hoặc x+5=0
=> x=5 hoặc x=-5
1) \(3x^2-75=0\)
\(\Leftrightarrow3\left(x^2-25\right)=0\)
\(\Leftrightarrow x^2-25=0\)
\(\Leftrightarrow x^2=25\)
\(\Leftrightarrow x=\pm\sqrt{25}=\pm5\)
2) \(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
3) \(x^3+3x^2+3x=0\)
\(\Leftrightarrow x^3+3x^2+3x+1=1\)
\(\Leftrightarrow\left(x+1\right)^3=1^3\)
\(\Leftrightarrow x+1=1\Leftrightarrow x=0\)
\(\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\)
\(\left(x-2y\right)^2-4\left(x-2y\right)+4=\left(x-2y-2\right)^2\)
\(\left(a^2+1\right)^2-6\left(a^2+1\right)+9=\left(a^2+1-3\right)^2=\left(a^2-2\right)^2\)
\(\left(x+y\right)^2+\left(x+y\right)x+\frac{1}{4}x^2=\left(x+y+\frac{1}{2}x\right)^2=\left(\frac{3}{2}x+y\right)^2\)
\(16x^4-9x^2=x^2\left(16x^2-9\right)=x^2\left(4x-4\right)\left(4x+3\right)\)
\(a^2-b^4=\left(a-b^2\right)\left(a+b^2\right)\)
(x + 2y)2 - 16
= (x + 2y)2 - 42
= (x + 2y - 4).(x + 2y + 4)
(x - 2y)2 - 4.(x - 2y) + 4
= (x - 2y)2 - 2.(x - 2y).2 + 22
= (x - 2y - 2)2
(a2 + 1)2 - 6.(a2 + 1) + 9
= (a2 + 1)2 - 2.(a2 + 1).3 + 32
= (a2 + 1 - 3)2
= (a2 - 2)2
(x + y)2 + (x + y).x + 1/4.x2
= (x + y)2 + 2.(x + y).1/2.x + (1/2.x)2
= (x + y + 1/2.x)2
= (3/2.x + y)2
16x4 - 9x2
= (4x2)2 - (3x)2
= (4x2 - 3x).(4x2 + 3x)
a2 - b4
= a2 - (b2)2
= (a - b2).(a + b2)
=x^2 +2.x.2 +2^2
=(x+2)^2
`x^2+4x+4=0`
`⇔x^2+2.x.2+2^2=0`
`⇔(x+2)^2=0`
`⇔x+2=0⇔x=−2`
Vậy `x=-2`.