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2.
\(a,x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
\(b,x^2-3y^2=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
\(c,\left(3x-2y\right)^2-\left(2x-3y\right)^2\\ =\left(3x-2y-2x+3y\right)\left(3x-2y+2x-3y\right)\\ =\left(x+y\right)\left(5x-5y\right)=5\left(x-y\right)\left(x+y\right)\)
\(d,9\left(x-y\right)^2-4\left(x+y\right)^2\\ =\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\)
\(e,\left(4x^2-4x+1\right)-\left(x+1\right)^2\\ =\left(2x-1\right)^2-\left(x+1\right)^2\\ =\left(2x-1-x-1\right)\left(2x-1+x+1\right)\\ =3x\left(x-2\right)\)
\(f,x^3+27=\left(x+3\right)\left(x^2+3x+9\right)\)
\(g,27x^3-0,001=\left(3x-0,1\right)\left(9x^2+0,027x+0,01\right)\)
\(h,125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
Bài 3 :
a) \(x^4+2x^2+1=\left(x^2+1\right)^2\)
b) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
c) \(-x^2-2xy-y^2=-\left(x+y\right)^2\)
e) \(\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
f) \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
g) \(x^3+6x^2+12x+8=\left(x+2\right)^3\)
h) \(x^3+1-x^2-x=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)=\left(x+1\right)\left(x^2-2x+1\right)=\left(x+1\right)\left(x-1\right)^2\)
l) \(\left(x+y\right)^2-x^3-y^3=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)=3xy\left(x+y\right)\)
e rất mún giúp chị nhưng e mới lớp 6 à , tiếc quá !! e hỉu giờ chị như thế naò mà !! mong là sẽ có ai giúp chị !! chị đừng lo nhé !!
\(3x^2-5x+2\)
\(=3x^2-3x-2x+2\)
\(=3x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-2\right)\)
Đề sai rồi bạn phải + 2 chứ
(oh) hóa trị 1 mà zn hóa trị 2=> cthh la zn(oh)2
với lại ko có oh2 dau chi co OH hoac la H2O
\(\left(X^2+2x+1\right)+\left(4y^2+\frac{4.1y}{4}+\frac{1}{16}\right)+2-\frac{1}{16}.\)
\(\left(x+1\right)^2+\left(2y+\frac{1}{4}\right)^2+\frac{15}{16}\ge\frac{15}{16}\)
\(x^2+4y^2+2x-y+2\)
\(=\left(x^2+2x+1\right)+\left[\left(2y\right)^2-2.2y.\frac{1}{4}+\left(\frac{1}{4}\right)^2\right]+\frac{15}{16}\)
\(=\left(x+1\right)^2+\left(2y-\frac{1}{4}\right)+\frac{15}{16}\)
Ta có: \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(2y-\frac{1}{4}\right)\ge0\forall y\end{cases}\Rightarrow\left(x+1\right)^2+\left(2y-\frac{1}{4}\right)+\frac{15}{16}\ge\frac{15}{16}}\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(2y-\frac{1}{4}\right)=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1=0\\2y-\frac{1}{4}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=\frac{1}{8}\end{cases}}}\)
Vậy GTNN của \(x^2+4y^2+2x-y+2=\frac{15}{16}\Leftrightarrow\hept{\begin{cases}x=-1\\y=\frac{1}{8}\end{cases}}\)
Tham khảo nhé~
\(A=\left(x^2-2x+1\right)+2013=\left(x-1\right)^2+2013\ge2013\\ A_{min}=2013\Leftrightarrow x=1\\ B=-\left(x^2-5x+\dfrac{25}{4}\right)+1993,25=-\left(x-\dfrac{5}{2}\right)^2+1993,25\le1993,25\\ B_{max}=1993,25\Leftrightarrow x=\dfrac{5}{2}\\ C=-\left(x^2-4x+4\right)+4=-\left(x-2\right)^2+4\le4\\ C_{max}=4\Leftrightarrow x=2\\ D=\left(x^2-4xy+4y^2\right)+\left(y^2-6y+9\right)+8\\ D=\left(x-2y\right)^2+\left(y-3\right)^2+8\ge8\\ D_{min}=8\Leftrightarrow\left\{{}\begin{matrix}x=2y=6\\y=3\end{matrix}\right.\)
\(E=\dfrac{6x-2}{3x^2+1}=\dfrac{3x^2+1-3x^2+6x-3}{3x^2+1}=1-\dfrac{3\left(x-1\right)^2}{3x^2+1}\le1\\ E_{max}=1\Leftrightarrow3\left(x-1\right)^2=0\Leftrightarrow x=1\\ F=1+\dfrac{10}{3x^2+9x+7}\\ \text{Có }3x^2+9x+7=3\left(x+\dfrac{3}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\\ \Leftrightarrow F\le1+\dfrac{10}{\dfrac{1}{4}}=41\\ F_{max}=41\Leftrightarrow x=-\dfrac{3}{2}\)
\(G=\dfrac{x^2+100x+196}{x}=x+100+\dfrac{196}{x}\\ \Leftrightarrow G\ge2\sqrt{x\cdot\dfrac{196}{x}}+100=2\cdot14+100=128\\ \Leftrightarrow G_{max}=128\Leftrightarrow x^2=196\Leftrightarrow x=14\left(x>0\right)\\ H=\dfrac{x}{\left(x+2012\right)^2}=\dfrac{x}{x^2+4024x+2012^2}\\ \Leftrightarrow Hx^2+\left(4024\cdot H-1\right)x+2012^2=0\\ \Delta\ge0\Leftrightarrow\left(4024\cdot H-1\right)^2-4\cdot2012^2\cdot H\ge0\\ \Leftrightarrow4024^2H^2-8048\cdot H+1-4024^2\ge0\\ \Leftrightarrow\left[{}\begin{matrix}H\le\dfrac{4025}{4024}\\H\ge-\dfrac{4023}{4024}\end{matrix}\right.\\ \text{Vậy H ko có min hay max}\)