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Bài 1 :
\(e,x^2+2xy+y^2-2x-2y+1\)
\(=\left(x+y-1\right)^2\)
Bài 2:
\(b,2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\left(x^2+1>0\right)\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
a,\(x^3+2x^2y+xy^2-9x\)
=x(\(x^2+2xy+y^2\)-9)
=x[(\(x^2+2xy+y^2\))-9]
=x[\(\left(x+y\right)^2\)-9]
b,2x-2y-\(x^2+2xy-y^2\)
=(2x-2y)-(\(x^2-2xy+y^2\))
=2(x-y)-\(\left(x-y\right)^2\)
=(x-y)(2-x+y)
c,\(x^4-2x^2\)
=\(x^2\left(x^2-2\right)\)
d,\(x^2-4x+3\)
=\(x^2-4x+4-1\)
=\(\left(x^2-4x+2^2\right)\)-1
=\(\left(x-2\right)^2\)-1
=(x-2-1)(x-2+1)
thông cảm mk chỉ làm đc từng này thôi
à..mà bạn xem lại ý e, cho mk đc k
k) \(x^3-x+3x^2+3xt^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
h) \(a^3-a^2x-ay+xy\)
\(=a^2\left(a-x\right)-y\left(a-x\right)\)
\(=\left(a^2-y\right)\left(a-x\right)\)
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
Lời giải:
\(3x^2+y^2+z^2+2x-2y+2xy+3=0\)
\(\Leftrightarrow (x^2+y^2+1+2xy-2y-2x)+2(x^2+2x+1)+z^2=0\)
\(\Leftrightarrow (x+y-1)^2+2(x+1)^2+z^2=0\)
Vì \(\left\{\begin{matrix} (x+y-1)^2\geq 0\\ (x+1)^2\geq 0\\ z^2\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{R}\)
Do đó: \((x+y-1)^2+2(x+1)^2+z^2\geq 0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} (x+y-1)^2=0\\ (x+1)^2=0\\ z^2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-1\\ y=2\\ z=0\end{matrix}\right.\)