6. Tìm GTNN P= |2x-8|+|2x-10| Q= |x-3|+|2y-5|
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\(A=\left|2x+8\right|+6\ge6\Rightarrow Min_A=6\)
\(B=\left|2y+4\right|+7+\left|4x+3\right|\ge7\Rightarrow Min_B=7\)
\(C=x^2+2x+5=\left(x+1\right)^2+4\ge4\Rightarrow Min_C=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left|4x-3\right|+\left|5y+7,5\right|+10\)
Mà \(\left|4x-3\right|\ge0\)với mọi x
\(\left|5y+7,5\right|\ge0\)với mọi y
\(\Rightarrow A\)có GTNN là 10
Để A có GTNN thì :
\(4x-3=0\) \(5y+7,5=0\)
\(4x=3\) \(5y=-7,5\)
\(x=\frac{3}{4}\) \(y=-1,5\)
\(B=\frac{5,8}{\left|2,5-x\right|+5,8}\)
Mà \(\left|2,5-x\right|\ge0\)
\(\Rightarrow\)GTNN \(\left|2,5-x\right|+5,8=5,8\)
Để B có GTLN \(\Rightarrow2,5-x=0\)
\(\Rightarrow x=2,5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhận xét : Lũy thừa bậc chẵn hay giá trị tuyệt đối của 1 số hữu tỉ luôn lớn hơn hoặc bằng 0(bằng 0 khi số hữu tỉ đó là 0)
1)\(\left(2x+\frac{1}{3}\right)^4\ge0\Rightarrow\left(2x+\frac{1}{3}\right)^4-10\ge-10\).Vậy GTNN của A là -10 khi :
\(\left(2x+\frac{1}{3}\right)^4=0\Rightarrow2x+\frac{1}{3}=0\Rightarrow2x=\frac{-1}{3}\Rightarrow x=\frac{-1}{6}\)
\(|2x-\frac{2}{3}|\ge0;\left(y+\frac{1}{4}\right)^4\ge0\Rightarrow|2x-\frac{2}{3}|+\left(y+\frac{1}{4}\right)^4-1\ge-1\).Vậy GTNN của B là -1 khi :
\(\hept{\begin{cases}|2x-\frac{2}{3}|=0\Rightarrow2x-\frac{2}{3}=0\Rightarrow2x=\frac{2}{3}\Rightarrow x=\frac{1}{3}\\\left(y+\frac{1}{4}\right)^4=0\Rightarrow y+\frac{1}{4}=0\Rightarrow y=\frac{-1}{4}\end{cases}}\)
2)\(\left(\frac{3}{7}x-\frac{4}{15}\right)^6\ge0\Rightarrow-\left(\frac{3}{7}x-\frac{4}{15}\right)^6\le0\Rightarrow-\left(\frac{3}{7}x-\frac{4}{15}\right)+3\le3\).Vậy GTLN của C là 3 khi :
\(\left(\frac{3}{7}x-\frac{4}{15}\right)^6=0\Rightarrow\frac{3}{7}x-\frac{4}{15}=0\Rightarrow\frac{3}{7}x=\frac{4}{15}\Rightarrow x=\frac{4}{15}:\frac{3}{7}=\frac{28}{45}\)
\(|x-3|\ge0;|2y+1|\ge0\Rightarrow-|x-3|\le0;-|2y+1|\le0\Rightarrow-|x-3|-|2y+1|+15\le15\)
Vậy GTLN của D là 15 khi :\(\hept{\begin{cases}|x-3|=0\Rightarrow x-3=0\Rightarrow x=3\\|2y+1|=0\Rightarrow2y+1=0\Rightarrow2y=-1\Rightarrow y=\frac{-1}{2}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=x^2+2y^2+2xy+2x+4y-1\)
\(A=\left(x^2+2xy+y^2\right)+\left(y^2+2y\right)+\left(2x+2y\right)-1\)
\(A=\left[\left(x+y\right)^2+2\left(x+y\right)+1\right]+\left(y^2+2y+1\right)-3\)
\(A=\left(x+y+1\right)^2+\left(y+1\right)^2-3\ge-3\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}}\)
Vậy GTNN của \(A\) là \(-3\) khi \(x=0\) và \(y=-1\)
Chúc bạn học tốt ~
Đặt \(B=-x^2-2x-y^2-8y-10\)
\(-B=\left(x^2+2x+1\right)+\left(y^2+8y+16\right)-7\)
\(-B=\left(x+1\right)^2+\left(y+4\right)^2-17\ge-17\)
\(B=-\left(x+1\right)^2-\left(y+4\right)^2+17\le17\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}-\left(x+1\right)^2=0\\-\left(y+4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-4\end{cases}}}\)
Vậy GTLN của \(B\) là \(17\) khi \(x=-1\) và \(y=-4\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
|3x-7|+|3x-2|+8 >= 5+8 = 13
Dấu "=" xảy ra <=> 3/2 <= x <= 7/3
k mk nha
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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, \(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}=\frac{3\left(x-y\right)\left(x+y\right)}{5xy}.\frac{5xy.3x}{-2\left(x-y\right)}=\frac{3\left(x+y\right)}{1}.\frac{3x}{-2}=\frac{-9x\left(x+y\right)}{2}\)