Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Để x;y;z ra ngoài làm thừa số chung rồi quất hết phần còn lại vào ngoặc thì thành 2 nhân tử thôi bạn, kiểu như phân phối ý.
a) \(9x^2+6x+1=\left(3x+1\right)^2\)
b)\(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
c)\(x^2y^4-2xy^2+1=\left(xy^2-1\right)^2\)
d) \(x^2+\frac{2}{3}x+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2\)
a) 9x2 + 6x + 1 = ( 3x + 1 )2
b) x2 - x + 1/4 = ( x - 1/2)2
c) x2 . y4 - 2xy2 + 1 = ( xy2 - 1 ) 2
d) x2 + 2/3x + 1/9 = (x+1/3)2
a)
\(\begin{array}{l}P + \frac{1}{{x + 2}} = \frac{x}{{{x^2} - 2{\rm{x}} + 4}}\\P = \frac{x}{{{x^2} - 2{\rm{x}} + 4}} - \frac{1}{{x + 2}}\\P = \frac{{x\left( {x + 2} \right) - {x^2} + 2{\rm{x}} - 4}}{{\left( {{x^2} - 2{\rm{x}} + 4} \right)\left( {x + 2} \right)}}\\P = \frac{{{x^2} + 2{\rm{x}} - {x^2} + 2{\rm{x}} + 4}}{{{x^3} + 8}}\\P = \frac{{4{\rm{x}} - 4}}{{{x^3} + 8}}\end{array}\)
b)
\(\begin{array}{l}P - \frac{{4\left( {x - 2} \right)}}{{x + 2}} = \frac{{16}}{{x - 2}}\\P = \frac{{16}}{{x - 2}} + \frac{{4\left( {x - 2} \right)}}{{x + 2}}\\P = \frac{{16\left( {x + 2} \right) + 4\left( {x - 2} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\P = \frac{{16{\rm{x}} + 32 + 4{{\rm{x}}^2} - 16{\rm{x}} + 16}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\P = \frac{{4{{\rm{x}}^2} + 48}}{{{x^2} - 4}}\end{array}\)
c)
\(\begin{array}{l}P.\frac{{x - 2}}{{x + 3}} = \frac{{{x^2} - 4{\rm{x}} + 4}}{{{x^2} - 9}}\\ \Rightarrow P = \frac{{{x^2} - 4{\rm{x}} + 4}}{{{x^2} - 9}}.\frac{{x + 3}}{{x - 2}}\\P = \frac{{{{(x - 2)}^2}(x + 3)}}{{(x - 3)(x + 3)(x - 2)}} = \frac{{x - 2}}{{x - 3}}\end{array}\)\(\)
d)
\(\begin{array}{l}P:\frac{{{x^2} - 9}}{{2{\rm{x}} + 4}} = \frac{{{x^2} - 4}}{{{x^2} + 3{\rm{x}}}}\\ \Rightarrow P = \frac{{{x^2} - 4}}{{{x^2} + 3{\rm{x}}}}.\frac{{{x^2} - 9}}{{2{\rm{x}} + 4}}\\P = \frac{{(x - 2)(x + 2)(x - 3)(x + 3)}}{{2{\rm{x}}(x + 3)(x + 2)}}\\P = \frac{{(x - 2)(x - 3)}}{{2{\rm{x}}}}\end{array}\)
a) P=\(\dfrac{4x-4}{x^3-8}\)( lấy VP-VT)
b)P=\(\dfrac{4x^2+48}{x^2-4}\) ( chuyển VT và thành VP+VT)
c) P=\(\dfrac{x-2}{x-3}\) ( chuyển VT thành VP.VT là ra)
d) \(\dfrac{\left(x-2\right)\left(x-3\right)}{2x}\)( lấy VP.VT)
b)(y-2)^3=y^3-8+12y-6y^2
c)8x^3+y^3=(2x+y)(4x^2+y^2-4xy)
2)
=(xy+2/3)^2
2.
pt <=> (x/2000 - 1) + (x+1/2001 - 1) + (x+2/2002 - 1) + (x+3/2003 - 1) + (x+4/2004 - 1 ) = 0
<=> x-2000/2000 + x-2000/2001 + x-2000/2002 + x-2000/2003 + x-2000/2004 = 0
<=> (x-2000).(1/2000 + 1/2001 + 1/2002 + 1/2003 + 1/2004) = 0
<=> x-2000=0 ( vì 1/2000 + 1/2001 + 1/2002 + 1/2003 + 1/2004 > 0 )
<=> x=2000
Tk mk nha
1.
a, = (2x-1)^2-2.(2x-1)+1-4
= (2x-1-1)^2-4
= (2x-2)^2-4
= (2x-2-2).(2x-2+2)
= 2x.(2x-4)
b, = [x.(x+3)].[(x+1).(x+2)]
= (x^2+3x).(x^2+3x+1)-8
= (x^2+3x+1)^2-1-8
= (x^2+3x+1)^2-9
= (x^2+3x+1-3).(x^2+3x+1+3)
= (x^2+3x-2).(x^2+3x+4)
= ((x+1).(x+3).(x^2+3x-2)
Tk mk nha
a) \(9x^2+6x+1=\left(3x\right)^2+2.3x.1+1^2=\left(3x+1\right)^2\)
b) \(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2=\left(x-0,5\right)^2\)
c) \(x^2y^4-2xy^2+1=\left(xy^2\right)^2-2.xy^2.1+1^2=\left(xy^2-1\right)^2\)
d) \(x^2+\dfrac{2}{3}x+\dfrac{1}{9}=x^2+2.x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2=\left(x+\dfrac{1}{3}\right)^2\)
a) \(9x^2+6x+1\)
\(=\left(3x\right)^2+2.3x.1+1^2\)
\(=\left(3x+1\right)^2\)
a) \(x^2+4y^2+4xy\)
\(=x^2+4xy+4y^2\)
\(=\left(x+2y\right)^2\)
\(a,x^2+4xy+4y^2=x^2+2.x.2y+\left(2y\right)^2.\)
\(=\left(x+2y\right)^2\)
\(b,\left(\frac{3}{2}x\right)^2-3xy+y^2\)
\(=\left(\frac{3}{2}x\right)^2-2.\frac{3}{2}x.y+y^2\)
\(=\left(\frac{3}{2}x-y\right)^2\)
\(c,\frac{x^2}{9}+\frac{x}{3}+\frac{1}{4}\)
\(=\left(\frac{x}{3}\right)^2+2.\frac{x}{3}.\frac{1}{2}+\left(\frac{1}{2}\right)^2\)
\(=\left(\frac{x}{3}+\frac{1}{2}\right)^2\)