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câu A thiếu đề
B=\(x^2-2x+2017=\left(x-1\right)^2+2016>=2016\)
Min B=2016 khi x-1=0<=>x=1
+)D=\(-2x^2+4x+2017=-2\left(x^2-2x+1\right)+2019=-2\left(x-1\right)^2+2019< =2019\)
=>Max D=2019, dấu '=' xảy ra khi x-1=0<=>x=1
a) \(x^2y+2xy+y=y\left(x^2+2x+1\right)=y\left(x+1\right)^2\)
b) \(4x^2-4xy-6y^2+6xy=4x\left(x-y\right)+6y\left(x-y\right)=\left(x-y\right)\left(4x+6y\right)\)
\(=2\left(x-y\right)\left(2x+3y\right)\)
c) \(18x^5y+18x^3y-2x^3y^5-2xy^5=18x^3y\left(x^2+1\right)-2xy^5\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(18x^3y-2xy^5\right)=2xy\left(x^2+1\right)\left(9x^2-y^4\right)=2xy\left(x^2+1\right)\left(3x-y^2\right)\left(3x+y^2\right)\)
d)
d) \(-12x^5-12x^3y-3xy^2+36x^4+36x^2y+9y^2=-3x\left(4x^4+4x^2y+y^2\right)+9y\left(4x^4+4x^2y+y^2\right)\)\(=\left(4x^4+4x^2y+y^2\right)\left(9-3x\right)\)
a, \(2x^2+3\left(x+1\right)\left(x-1\right)-5x\left(x+1\right)\)
\(=2x^2+3\left(x^2-1\right)-5x^2-5x\)
\(=2x^2+3x^2-3-5x^2-5x\)
\(=\left(2x^2+3x^2-5x^2\right)-3-5x\)
\(=-\left(5x+3\right)\)
b, \(\left(4x+3y\right)\left(2x-5y\right)-\left(2x+6y\right)\left(3x-5y\right)\)
\(=8x^2-20xy+6xy-\left(15y^2-6x^2-10xy-18xy-30y^2\right)\)
\(=8x^2-20xy+6xy-15y^2+6x^2+10xy+18xy+30y^2\)
\(=\left(8x^2+6x^2\right)+\left(-20xy+6xy+10xy+18xy\right)+\left(-15y^2+30y^2\right)\)
\(=14x^2+14xy+15y^2\)
\(=14x.\left(x+y\right)+15y^2\)
Chúc bạn học tốt!!!
P = x2 - 2xy + 6y2 - 12x + 3y + 45
= x2 + y2 + 62 - 2xy - 12x + 12y + 5y2 - 9y + 4,05 + 4,95
= (y + 6 - x)2 + 5(y - 0,9)2 + 4,95 \(\ge\) 4,95
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}y+6-x=0\\y-0,9=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=6,9\\y=0,9\end{matrix}\right.\)
Answer:
\(2x^3+4x^2y+2xy^2\)
\(= 2 x ( x ² + 2 x y + y ² )\)
\(= 2 x ( x + y ) ² \)
\( − 3 x ^4 y − 6 x ^3 y ^2 − 3 x ^2 y ^3 \)
\(=-3x^2y(x^2+2xy+y^2)\)
\(=-3x^2y(x+y)^2\)
\(4x^5y^2+8x^4y^3+4x^3y^4\)
\(=4x^3y^2.x^2+4x^3y^2.2xy+4x^3y^2.y^2\)
\(=4x^3y^2.(x^2+2xy+y^2)\)
\(=4x^3y^2.(x+y)^2\)
\(a.4x^3-8x^2+4xy^3=4x\left(x^2-8x+y^3\right)\)
\(b.x^2+2xy+y^2-36=\left(x+y\right)^2-36=\left(x+y-6\right)\left(x+y+6\right)\) \(c.x^2-2xy+y^2-25=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\) \(d.x^2-5x+2xy-5y+y^2=\left(x+y\right)^2-5\left(x+y\right)=\left(x+y\right)\left(x+y-5\right)\) \(e.49+2xy-x^2-y^2=-\left(x^2-2xy+y^2-49\right)=-\left[\left(x-y\right)^2-49\right]=-\left(x-y-7\right)\left(x-y+7\right)\) \(f.3x^2-6x+3-3y^2=3\left(x^2-2x-y^2+1\right)\)
\(g.2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)\left(x+1\right)\)
\(h,\) giống câu f.
\(i.x^3-2x^2y+xy^2-64x=x\left(x^2-2xy+y^2-64\right)=x\left[\left(x-y\right)^2-64\right]=x\left(x-y-8\right)\left(x-y+8\right)\) \(k.3x+3y-x^2-2xy-y^2=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
1, \(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)\(=\frac{4y.y}{11x^2.x^2}.\frac{-3x^2}{2.4y}\)\(=\frac{y}{11x^2}.\frac{-3}{2}=\frac{-3y}{22x^2}\)
2, \(\frac{4x^2}{5y^2}:\frac{6x}{5y}:\frac{2x}{3y}\)\(=\frac{4x^2}{5y^2}.\frac{5y}{6x}.\frac{3y}{2x}\)\(=\frac{2x.2x}{5y.y}.\frac{5y}{3.2x}.\frac{3y}{2x}\)\(=\frac{2x}{y}.\frac{1}{3}.\frac{3y}{2x}\)
\(\frac{2x}{3y}.\frac{3y}{2x}=1\)
3, \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}\)\(=\frac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\frac{x+4}{2\left(x-2\right)}\)\(=\frac{\left(x+2\right)}{3}.\frac{1}{2}=\frac{x+2}{6}\)
4, \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\left(-\frac{2\left(x-2\right)}{x+2}\right)=\frac{5}{4}.\frac{-2}{1}=-\frac{5}{2}\)
5, \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{3}{-\left(x-6\right)}=\frac{x+6}{2\left(x+5\right)}.\frac{-3}{1}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
6, \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}=\frac{\left(x-3y\right)\left(x+3y\right)}{\left(xy\right)^2}.\frac{3xy}{2\left(x-3y\right)}=\frac{x+3y}{xy}.\frac{3}{2}=\frac{3\left(x+3y\right)}{2xy}\)
7, \(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}=\frac{3\left(x-y\right)\left(x+y\right)}{5xy}.\frac{5xy.3x}{-2\left(x-y\right)}=\frac{3\left(x+y\right)}{1}.\frac{3x}{-2}=\frac{-9x\left(x+y\right)}{2}\)
a)\(P=x^2+4x+2xy+3y^2+5y+2017\)
\(=x^2+2xy+y^2+4y+4+4x+2y^2+y+\dfrac{1}{8}+\dfrac{16103}{8}\)
\(=\left(x+y+2\right)^2+2\left(y^2+\dfrac{y}{2}+\dfrac{1}{16}\right)+\dfrac{16103}{8}\)
\(=\left(x+y+2\right)^2+2\left(y+\dfrac{1}{4}\right)^2+\dfrac{16103}{8}\ge\dfrac{16103}{8}\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{7}{4}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
b)\(Q=-x^2+4x-3y^2+6y+2017\)
\(=-x^2+4x-4-3y^2+6y+3+2024\)
\(=-\left(x^2-4x+4\right)-\left(3y^2-6y-3\right)+2024\)
\(=-\left(x-2\right)^2-3\left(y^2-2y-1\right)+2024\)
\(=-\left(x-2\right)^2-3\left(y-1\right)^2+2024\ge2024\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Ta có:
\(P=x^2+4x+2xy+3y^2+5y+2017\)
\(=x^2+2x\left(y+2\right)+\left(y+2\right)^2+2y^2+y+2013\)
\(=\left[x+\left(y+2\right)\right]^2+2\left(y^2+y+0,25\right)+2012,5\)
\(=\left(x+y+2\right)^2+2\left(y+0,5\right)^2+2012,5\ge2012,5\)
Dấu "=" xảy ra khi:
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2=0\\y+0,5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-0,5\\x=-1,5\end{matrix}\right.\)
Vậy \(minP=2012,5\) khi \(\left\{{}\begin{matrix}y=-0,5\\x=-1,5\end{matrix}\right.\)
Ta có:
\(Q=-x^2+4x-3y^2+6y+2017\)
\(=-\left(x^2-4x+4\right)-3\left(y^2-2y+1\right)+2024\)
\(=-\left(x-2\right)^2-3\left(y-1\right)^2+2024\le2024\)
Dấu "=" xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(maxQ=2024\) khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)