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`a,-x^3/8 + 3/(4x^2) - 3/(2x) +1`
`=-(x^3/8 - 3/(4x^2) + 3/(2x) - 1)`
`=-(x/2 - 1)^3`
`b,x^6 - 3/(2x^{4} y) + 3/(4x^{2}y^{2}) - 1/(8y^{3})`
`=(x^3 - 1/(2y))^{3}`
Bài 1:
a) \(a^2-6a+9=\left(a-3\right)^2\)
b) \(\dfrac{1}{4}x^2+2xy^2+4y^4=\left(\dfrac{1}{2}x+2y^2\right)^2\)
Bài 2:
a) \(\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\)
\(\Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\)
b) \(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)
a) \(x^2+2x+1\)
\(=\left(x+1\right)^2\)
b) \(9-24x+16x^2\)
\(=\left(3-4x\right)^2\)
c) \(4x^2+\dfrac{1}{4}+2x\)
\(=4x^2+2x+\dfrac{1}{4}\)
\(=\left(2x+\dfrac{1}{2}\right)^2\)
Bài 3:
b: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)+10=0\)
\(\Leftrightarrow6x^2+12-6x^2+12x-6=0\)
hay \(x=-\dfrac{1}{2}\)
Bài 2:
a: \(x^3+3x^2+3x+1=\left(x+1\right)^3\)
b: \(m^3+9m^2n+27mn^2+27n^3=\left(m+3n\right)^3\)
a) \(x^2+4x+4\)
\(=x^2+2\cdot2\cdot x+2^2\)
\(=\left(x+2\right)^2\)
b) \(4x^2-4x+1\)
\(=\left(2x\right)^2-2\cdot2x\cdot1+1^2\)
\(=\left(2x-1\right)^2\)
c) \(x^2-x+\dfrac{1}{4}\)
\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
d) \(4\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=\left[2\left(x+y\right)\right]^2-2\cdot2\left(x+y\right)\cdot1+1^2\)
\(=\left[2\left(x+y\right)-1\right]^2\)
\(=\left(2x+2y-1\right)^2\)
a) \(x^4+4x^2+4=\left(x^2+2\right)^2\)
b) \(\left(2y-x\right)^2+2\left(2y-x\right)+1=\left(2y-x+1\right)^2\)
c) \(\left(2a-4b\right)^2+4a-8b+1=\left(2a-4b\right)^2+2\cdot\left(2a-4b\right)\cdot1+1^2=\left(2a-4b+1\right)^2\)
a. $x^2+4x+4$
$=x^2+2\cdot x\cdot2+2^2$
$=(x+2)^2$
b. $x^2-6xy+9y^2$
$=x^2-2\cdot x\cdot3y+(3y)^2$
$=(x-3y)^2$
c. $4x^2+12x+9$
$=(2x)^2+2\cdot2x\cdot3+3^2$
$=(2x+3)^2$
d. $x^2-x+\dfrac14$
$=x^2-2\cdot x\cdot \dfrac12+\Bigg(\dfrac12\Bigg)^2$
$=\Bigg(x-\dfrac12\Bigg)^2$
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
a) ( 2 x + 1 ) 2 . b) ( 3 x – 2 ) 2 .
c) 1 2 ab 2 + 1 2 . d) ( 4 uv 2 – 1 ) 2 .