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a/ A = 2x2 + y2 - 2xy - 2x + 3
= (x2 - 2xy + y2) + (x2 - 2x + 1) + 2
= (x - y)2 + (x - 1)2 + 2\(\ge2\)
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!!!
Bài 1. Rút gọn:
\(a, x\left(1-x\right)+6\left(x+3\right)\left(x+3\right)\)
\(=x-x^2+6\left(x^2+6x+9\right)\)
\(=x-x^2+6x^2+36x+54\)
\(=5x^2+37x+54\)
\(b, \left(2-3x\right)\left(2+3x\right)-\left(x+5\right)\left(x-5\right)\)
\(=\left(4-9x^2\right)-\left(x^2-25\right)\)
\(=-10x^2+29\)
\(c, \left(3x+1\right)\left(x+5\right)-\left(x-1\right)\left(x+1\right)\)
\(=3x^2+15x+x+5-x^2+1\)
\(=2x^2+16x+6\)
\(d,\left(2-3x\right)\left(2x+3\right)+6\left(x-1\right)^2\)
\(=\left(4x+6-6x^2-9x\right)+6\left(x^2-2x+1\right)\)
\(=4x+6-6x^2-9x+6x^2-12x+6\)
\(=-17x+12\)
\(e, x\left(5-x\right)-\left(2x+2\right)\left(3x+2\right)-\left(x-2\right)\left(x+2\right)\)
\(=5x-x^2-\left(6x^2+4x+6x+4\right)-\left(x^2-4\right)\)
\(=5x-x^2-6x^2-4x-6x-4-x^2+4\)
\(=-8x^2-5x\)
Bài 2:
a: VT\(=x^3-xy+x^2y^2-y^3-x^3+y^3-x^2y^2\)
=-xy
b: \(VT=x^2+6xy+9y^2-x^2+9y^2-6xy=18y^2=VP\)
a) \(4x\left(x-5\right)+3y\left(x-5\right)\)
\(=\left(x-5\right)\left(4x+3y\right)\)
b) \(x^2-2x-4y^2-4y\)
\(=\left[x^2-\left(2y\right)^2\right]-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
c) \(x^2+x-y^2+y\)
\(=\left(x^2-y^2\right)+\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+1\right)\)
d) \(3x^2+3y^2-6xy-12\)
\(=3\left(x^2+y^2-2xy-4\right)\)
\(=3\left[\left(x-y\right)^2-2^2\right]\)
\(=3\left(x-y-2\right)\left(x-y+2\right)\)
\(B=-2x^2-x+5\)
\(=-2\left(x^2+\dfrac{1}{2}x-\dfrac{5}{2}\right)\)
\(=-2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}-\dfrac{41}{16}\right)\)
\(=-2\left(x+\dfrac{1}{4}\right)^2+\dfrac{41}{8}\le\dfrac{41}{8}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{1}{4}=0\Leftrightarrow x=-\dfrac{1}{4}\)
Vậy Max B là : \(\dfrac{41}{8}\Leftrightarrow x=-\dfrac{1}{4}\)
\(A=-3x^2+x-2\)
\(=-3\left(x^2-\dfrac{1}{3}x+\dfrac{2}{3}\right)\)
\(=-3\left(x^2-2x.\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{23}{36}\right)\)
\(=-3\left[\left(x-\dfrac{1}{6}\right)^2+\dfrac{23}{36}\right]\)
\(=-3\left(x-\dfrac{1}{6}\right)^2-\dfrac{69}{26}\le-\dfrac{69}{26}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{6}=0\Leftrightarrow x=\dfrac{1}{6}\)
Vậy Max A là : \(\dfrac{-69}{26}\Leftrightarrow x=\dfrac{1}{6}\)
\(B=-2x^2-x+5\)
\(=-2\left(x^2-\dfrac{1}{2}x-\dfrac{5}{2}\right)\)
\(=-2\left(x^2-2x.\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{41}{16}\right)\)
\(=-2\left[\left(x-\dfrac{1}{4}\right)^2-\dfrac{41}{16}\right]\)
\(=-2\left(x-\dfrac{1}{4}\right)^2+\dfrac{41}{8}\le\dfrac{41}{8}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{4}\)
Vậy Max B là : \(\dfrac{41}{8}\Leftrightarrow x=\dfrac{1}{4}\)
\(C=-\left(x+1\right)^2-\left(2x-3\right)^2\)
\(=-x^2-2x-1-4x^2+12x-9\)
\(=-5x^2+10x-10\)
\(=-5\left(x^2-2x+1+1\right)\)
\(=-5\left[\left(x-1\right)^2+1\right]\)
\(=-5\left(x-1\right)^2-5\le-5\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy Max C là : \(-5\Leftrightarrow x=1\)
\(E=2-5x^2-y^2-4xy+2x\)
\(=-\left(4x^2+4xy+y^2\right)-\left(x^2-2x+1\right)+3\)
\(=-\left(2x+y\right)^2-\left(x-1\right)^2+3\le3\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=1\end{matrix}\right.\)
Vậy Max E là : \(3\Leftrightarrow x=1;y=-2\)
ns thật vs c tôi ms đọc đề bài thôi đã ko hiểu j rồi ns chi đến lm giúp c. Sr nhé
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 $B_{\max}=\frac{97}{22}$