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a) \(x^2+4y^2+4xy\)
\(=x^2+4xy+4y^2\)
\(=\left(x+2y\right)^2\)
\(a,x^2+4xy+4y^2=x^2+2.x.2y+\left(2y\right)^2.\)
\(=\left(x+2y\right)^2\)
\(b,\left(\frac{3}{2}x\right)^2-3xy+y^2\)
\(=\left(\frac{3}{2}x\right)^2-2.\frac{3}{2}x.y+y^2\)
\(=\left(\frac{3}{2}x-y\right)^2\)
\(c,\frac{x^2}{9}+\frac{x}{3}+\frac{1}{4}\)
\(=\left(\frac{x}{3}\right)^2+2.\frac{x}{3}.\frac{1}{2}+\left(\frac{1}{2}\right)^2\)
\(=\left(\frac{x}{3}+\frac{1}{2}\right)^2\)
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
\(=\left(\frac{3}{4}.\frac{x^m}{x}y\right)^2-2.\frac{3}{4}.\frac{x^my}{x}.\frac{4}{3}.\frac{y^m.x}{y}+\left(\frac{4}{3}.\frac{y^m}{y}x\right)^2\)
\(=\left(\frac{3}{4}.\frac{x^m}{x}y-\frac{4}{3}.\frac{y^m}{y}x\right)^2\)\(=\left(\frac{3}{4}.x^{m-1}.y-\frac{4}{3}.y^{m-1}.x\right)^2\)
\(\frac{a}{x-2}+\frac{b}{\left(x+1\right)^2}=\frac{a\left(x+1\right)^2+b\left(x-2\right)}{\left(x-2\right)\left(x+1\right)^2}=\frac{ax^2+\left(2a+b\right)x+\left(a-2b\right)}{x^3-3x-2}\)
\(\Rightarrow\frac{x^2+5}{x^3-3x-2}=\frac{ax^2+\left(2a+b\right)x+\left(a-2b\right)}{x^3-3x-2}\)
Đồng nhất hệ số, ta có :
\(\hept{\begin{cases}a=1\\2a+b=0\\a-2b=5\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=-2\end{cases}}}\)
\(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}=\left(x+\frac{1}{2}\right)^3\)
Bạn ghi sai đề nha
Hok tốt
\(x^3+\frac{3}{2}x^2+\frac{3}{2}x+\frac{1}{8}\)
\(=\left(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\right)+\frac{3}{2}x-\frac{3}{4}x\)
\(=\left(x+\frac{1}{2}\right)^3+\frac{3}{4}x\)
\(=\left(x+\frac{1}{2}\right)^3+\left(\sqrt[3]{\frac{3}{4}x}\right)^3\)
\(=\left(x+\frac{1}{2}+\sqrt[3]{\frac{3}{4}x}\right)\left[\left(x+\frac{1}{2}\right)^2-\left(x+\frac{1}{2}\right)\left(\sqrt[3]{\frac{3}{4}}\right)+\left(\sqrt[3]{\frac{3}{4}}\right)^2\right]\)
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
(x+3)2 =x2+6x+9
(x-1/2)(x+1/2)=x2-1/2
Câu 2 phải là thế này nhé: (x-1/2)(x+1/2)=x2-1/4