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2C=4x^2+2x-10=((2x)^2+4x\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\))-\(\dfrac{41}{4}\)
=\(\left(2x+\dfrac{1}{2}\right)^2\)-41/4\(\ge\dfrac{-41}{4}\)
=> C\(\ge\dfrac{-41}{8}\)
Vậy min C = \(\dfrac{-41}{8}\)khi x=\(\dfrac{-1}{4}\)
\(4A=4x^2+44y^2+24xy-8y+20=\left(2x\right)^2+2.2x.6y+\left(6y\right)^2+8y^2-8y+20=\left(2x+6y\right)^2+2\left(4y^2-4y+1\right)+18=\left(2x+6y\right)^2+2\left(2y-1\right)^2+18\ge18\)
Bài 1:
a)
\(A=x^2+y^2-xy-3y+2016=(x^2-xy+\frac{y^2}{4})+(\frac{3y^2}{4}-3y+3)+2013\)
\(=(x-\frac{y}{2})^2+3(\frac{y}{2}-1)^2+2013\)
\(\geq 2013\)
Vậy GTNN của $A$ là $2013$. Giá trị này đạt được khi \(\left\{\begin{matrix} x-\frac{y}{2}=0\\ \frac{y}{2}-1=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} y=2\\ x=1\end{matrix}\right.\)
b)
\(B=2x^2+5y^2+4xy-6+5x-9\)
\(=5(y^2+\frac{4}{5}xy+\frac{4}{25}x^2)+\frac{6}{5}x^2+5x-15\)
\(=5(y+\frac{2}{5}x)^2+\frac{6}{5}(x^2+\frac{25}{6}x+\frac{25^2}{12^2})-\frac{485}{24}\)
\(=5(y+\frac{2}{5}x)^2+\frac{6}{5}(x+\frac{25}{12})^2-\frac{485}{24}\geq \frac{-485}{24}\)
Vậy GTNN của $B$ là $\frac{-485}{24}$
Giá trị này đạt được khi \(\left\{\begin{matrix} y+\frac{2}{5}x=0\\ x+\frac{25}{12}=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-\frac{25}{12}\\ y=\frac{5}{6}\end{matrix}\right.\)
c)
\(C=x^2+xy+y^2-3x-3y+2018\)
\(=\frac{4x^2+4xy+4y^2-12x-12y+8072}{4}=\frac{(4x^2+4xy+y^2)+3y^2-12x-12y+8072}{4}\)
\(=\frac{(2x+y)^2-6(2x+y)+3y^2-6y+8072}{4}\)
\(=\frac{(2x+y)^2-6(2x+y)+9+3(y^2-2y+1)+8060}{4}=\frac{(2x+y-3)^2+3(y-1)^2+8060}{4}\)
\(\geq \frac{8060}{4}=2015\)
Vậy $C_{\min}=2015$. Giá trị đạt được khi \(\left\{\begin{matrix} 2x+y-3=0\\ y-1=0\end{matrix}\right.\Leftrightarrow x=y=1\)
Bài 2:
a)
\(-A=x^2+4y^2-2x+4y-5=(x^2-2x+1)+(4y^2+4y+1)-7\)
\(=(x-1)^2+(2y+1)^2-7\geq -7\)
\(\Rightarrow A\leq 7\)
Vậy GTLN của $A$ là $7$.
Giá trị này đạt được khi \(\left\{\begin{matrix} x-1=0\\ 2y+1=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=\frac{-1}{2}\end{matrix}\right.\)
b)
ĐKĐB \(\Leftrightarrow B+2x^2+10y^2-6xy-4x+3y-2=0\)
\(\Leftrightarrow 2x^2-2x(3y+2)+(10y^2+3y-2+B)=0\)
Coi đây là PT bậc 2 ẩn $x$. Vì dấu "=" tồn tại nên PT luôn có nghiệm
\(\Rightarrow \Delta'=(3y+2)^2-2(10y^2+3y-2+B)\geq 0\)
\(\Leftrightarrow B\leq \frac{-11y^2+6y+8}{2}=\frac{\frac{97}{11}-11(y-\frac{3}{11})^2}{2}\leq \frac{97}{22}\)
Vậy $B_{\max}=\frac{97}{22}$
a) \(\frac{6xy+4y}{4x^2y^2}+\frac{2xy-4y}{4x^2y^2}\)
\(=\frac{6xy+4y+2xy-4y}{4x^2y^2}\)
\(=\frac{8xy}{4x^2y^2}\)
\(=\frac{2}{xy}\)
b) \(\frac{5}{x+3}-\frac{3}{x-3}+\frac{30}{x^2-9}\)
\(=\frac{5\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{3\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{30}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{5x-15-3x-9+30}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{2x+6}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{2}{x-3}\)
c) \(\frac{2x+8}{\left(x+2\right)^2}:\frac{x+4}{x+2}\)
\(=\frac{2\left(x+4\right)}{\left(x+2\right)^2}\cdot\frac{x+2}{x+4}\)
\(=\frac{2\left(x+4\right)\left(x+2\right)}{\left(x+2\right)\left(x+2\right)\left(x+4\right)}\)
\(=\frac{2}{x+2}\)
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
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) \(-12x^5-12x^3y-3xy^2+36x^4+36x^2y+9y^2=-3x\left(4x^4+4x^2y+y^2\right)+9y\left(4x^4+4x^2y+y^2\right)\)\(=\left(4x^4+4x^2y+y^2\right)\left(9-3x\right)\)
\(a,A-B+C\)
\(=\left(5xy^2-4x^2y-6xy^2\right)-\left(8yx^2-4y^2x+3y^2\right)+\left(-2xy^2+3yx^2+5x^2\right)\)
\(=5xy^2-4x^2y-6xy^2-8yx^2+4y^2x-3y^2-2xy^2+3yx^2+5x^2\)
\(=\left(5xy^2-6xy^2-2xy^2+4xy^2\right)+\left(-4x^2y-8x^2y+3x^2y\right)+\left(-3y^2+5x^2\right)\)
\(=xy^2-9x^2y-3y^2+5x^2\)
\(b,2\left(A+B\right)+C=2A+2B+C\)
\(=2\left(5xy^2-4x^2y-6xy^2\right)+2\left(8yx^2-4y^2x+3y^2\right)-2xy^2+3yx^2+5x^2\)
\(=10xy^2-8x^2y-12xy^2+16x^2y-8xy^2+6y^2-2xy^2+3x^2y+5x^2\)
\(=-12xy^2+11x^2y+5x^2+6y^2\)