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\(A=|2x+1|+|x-1|-|x-2|\)
Khi \(x< \frac{-1}{2}\) thì \(|2x+1|=-1-2x;|x-1|=1-x;|x-2|=2-x\)
\(\Rightarrow A=-2x-1+1-x+x-2\)
\(A=-2x-2\)
Khi \(\frac{-1}{2}\le x\le1\) thì \(|2x+1|=2x+1;|x-1|=1-x;|x-2|=2-x\)
\(\Rightarrow A=2x+1+1-x+x-2\)
\(A=2x\)
Khi \(1< x< 2\) thì \(|2x+1|=2x+1;|x-1|=x-1;|x-2|=2-x\)
\(\Rightarrow A=2x+1+x-1+x-2\)
\(A=4x-2\)
Khi \(x\ge2\) thì \(|2x+1|=2x+1;|x-1|=x-1;|x-2|=x-2\)
\(\Rightarrow A=2x+1+x-1+2-x\)
\(A=2x+2\)
a) \(D=x+\left|x\right|\)
- Nếu \(x\ge0\)thì: \(D=x+x=2x\)
- Nếu \(x< 0\) thì: \(D=x-x=0\)
b) \(E=\left|x-7\right|+6-x\)
- Nếu \(x\ge7\)thì: \(E=x-7+6-x=-1\)
- Nếu \(x< 7\)thì: \(E=7-x+6-x=13-2x\)
c) \(C=x+\frac{1}{2}-\left|x-\frac{2}{3}\right|\)
- Nếu \(x\ge\frac{2}{3}\)thì: \(C=x+\frac{1}{2}-\left(x-\frac{2}{3}\right)=x+\frac{1}{2}-x+\frac{2}{3}==\frac{7}{6}\)
- Nếu \(x< \frac{2}{3}\)thì: \(C=x+\frac{1}{2}-\left(\frac{2}{3}-x\right)=x+\frac{1}{2}-\frac{2}{3}+x=2x-\frac{1}{6}\)
a) \(|x|=5 \Leftrightarrow x=\pm 5\)
b) \(|x|<2 \Leftrightarrow -2<x<2\)
c) \(|x|=-1\)
Vì \(|x| ≥0 \forall x \Rightarrow |x|=-1 VN.\)
d) \(|x|=|-5| \Leftrightarrow |x|=5 \Leftrightarrow x= \pm 5\)
e) \(|x+3|=0 \Leftrightarrow x+3=0 \Leftrightarrow x=-3\).
f) \( |x-1|=4 \\ \Leftrightarrow x-1 = \pm 4 \\ \Leftrightarrow x=5 \vee x=-3\)
g) \( |x-5|=10 \\ \Leftrightarrow x-5=\pm10 \\ \Leftrightarrow x=15 vee x=-5\)
h) \(|x+1|+20=0 \Leftrightarrow |x+1|=-20 (VN) \).
g) \( |x-5|=10 \\ \Leftrightarrow x-5=\pm10 \\ \Leftrightarrow x=15 \vee x=-5\)
h) \(|x+1|+20=0 \Leftrightarrow |x+1|=-20 (VN) \).
\(|2x-8|+|-20|=|2x-8|+20\)
biểu thức nhỏ nhất khi x=0
\(Min=2.0-8+20=12\)
Đặt A = -(x+1)^2-/y-2/+11
Ta có: \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left|y-2\right|\ge0\end{cases}\Rightarrow\hept{\begin{cases}-\left(x+1\right)^2\le0\\-\left|y-2\right|\le0\end{cases}}}\)
\(\Rightarrow-\left(x+1\right)^2-\left|y-2\right|\le0\)
\(\Rightarrow A=-\left(x+1\right)^2-\left|y-2\right|+11\le11\)
Dấu "=" xảy ra khi x = -1, y = 2
Vậy GTLN của A = 11 khi x = -1, y = 2
Có: \(\left(x-2y+1\right)^2\ge0\forall x;y\)
\(\left|y+1\right|\ge0\forall y\)
\(\Rightarrow\left(x-2y+1\right)^2+\left|y+1\right|\ge0\forall x;y\)
\(\Rightarrow\left(x-2y+1\right)^2+\left|y+1\right|+17\ge17\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-2y+1\right)^2=0\\\left|y+1\right|=0\end{cases}}\)
\(\left|y+1\right|=0\Leftrightarrow y+1=0\Leftrightarrow y=-1\)
\(\left(x-2y+1\right)^2=0\Leftrightarrow x-2y+1=0\Leftrightarrow x-2.\left(-1\right)+1=0\Leftrightarrow x+2+1=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy GTNN của A = 17 \(\Leftrightarrow\left(x;y\right)=\left(-3;-1\right)\)
xy^2+xy-2x = x(y^2+y-2) = x ( y^2-y+2y-2) = x(y(y-1)+2(y-1) = x(y+2)(y-1)
OK!!!!!!!!!!!