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![](https://rs.olm.vn/images/avt/0.png?1311)
(x - 13 + y)2 + (x - 6 - y)2 ≥ 0 + 0 = 0
Vì dấu "=" xảy ra nên x - 13 + y = 0 và x - 6 - y = 0
x + y = 13 và x - y = 6
x = (13 - 6) : 2 = 3,5
y = 13 - 3,5 = 9,5
Vậy x = 3,5 và y = 9,5
(\(x\) - 13 + y)2 + (\(x\) - 6 - y)2 = 0
(\(x\) - 13 + y)2 ≥ 0 ∀ \(x;y\)
(\(x-6-y\))2 ≥ 0 ∀ \(x;y\)
⇒(\(x-13+y\))2 + (\(x\) - 6- y)2 = 0
⇔ \(\left\{{}\begin{matrix}x-13+y=0\\x-6-y=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x-6-y=0\\x-13+y+x-6-y=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}y=x-6\\2x=19\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=\dfrac{19}{2}\\y=\dfrac{19}{2}-6\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=\dfrac{19}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(\(x\) -13 +y)2 + (\(x\) - 6 - y)2 = 0
(\(x-13+y\))2 ≥0; (\(x\) - 6 - y)2 ≥ 0∀ \(x;y\)
⇒(\(x-13+y\))2 + (\(x-6-y\))2 = 0
⇔ \(\left\{{}\begin{matrix}x-13+y=0\\x-6-y=0\end{matrix}\right.\)
⇒ -13 - 6 + 2\(x\) = 0 ⇒ \(x\) = \(\dfrac{19}{2}\) ⇒ y = \(\dfrac{19}{2}\) - 6 ⇒ y = \(\dfrac{7}{2}\)
Vậy (\(x\);y) = (\(\dfrac{19}{2}\); \(\dfrac{7}{2}\))
\(\left(x-13+y\right)^2+\left(x-6-y\right)^2=0\left(1\right)\)
Ta có :
\(\left\{{}\begin{matrix}\left(x-13+y\right)^2\ge0,\forall x;y\in R\\\left(x-6-y\right)^2\ge0,\forall x;y\in R\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x-13+y\right)^2=0\\\left(x-6-y\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-13+y=0\\x-6-y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=19\\y=x-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{2}\\y=\dfrac{19}{2}-6=\dfrac{7}{2}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{19}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\) thoả mãn đề bài
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\hept{\begin{cases}\left|x^2+y^2+z^2-1\right|=0\\\left(3y-4z\right)^4\ge0\\\left(3x-2y\right)^2\ge0\end{cases}}\Rightarrow\left|x^2+y^2+z^2-1\right|+\left(3y-4z\right)^4+\left(3x-2y\right)^2\ge0\)
dấu = xảy ra khi \(\hept{\begin{cases}\left|x^2+y^2+z^2-1\right|=0\\\left(3y-4z\right)^4=0\\\left(3x-2y\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x^2+y^2+z^2=1\\3y=4z\\3x-2y=0\end{cases}}\Rightarrow\hept{\begin{cases}x^2+y^2+z^2=1\\y=\frac{4z}{3}\\x=\frac{2y}{3}\end{cases}}\)
Vậy ...
p/s bài này chắc chỉ có dạng chung thôi bn :)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ Ta luôn có : \(\begin{cases}x^2\ge0\\\left(y-\frac{1}{10}\right)^4\ge0\end{cases}\)\(\Rightarrow x^2+\left(y-\frac{1}{10}\right)^4\ge0\)
Để dấu "=" xảy ra thì x = 0 , y = 1/10
b/ Tương tự.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left|3x+1\right|+\left|3x-5\right|=\left|3x+1\right|+\left|5-3x\right|\ge\left|3x+1+5-3x\right|=6\)(1)
\(\frac{12}{\left(y+3\right)^2+2}\le\frac{12}{2}=6\)(2)
\(\left(1\right);\left(2\right)\Rightarrow VT\ge VP."="\Leftrightarrow\hept{\begin{cases}-\frac{1}{3}\le x\le\frac{5}{3}\\y=-3\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài giải
b, \(x-5+\left|x-3\right|=4\)
\(\left|x-3\right|=4-x+5\)
\(\Rightarrow\orbr{\begin{cases}x-3=-4+x-5\\x-3=4-x+5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-x=-4-5+3\\x+x=4+5+3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x\ne-6\text{ ( loại ) }\\2x=12\end{cases}}\)\(\Rightarrow\text{ }x=6\)
c, \(\sqrt{\left(x+7\right)^2}+\left(x^2-49\right)^{2012}=0\)
\(\left(x+7\right)+\left(x^2-49\right)^{2012}=0\)
\(\Rightarrow\hept{\begin{cases}x+7=0\\\left(x^2-49\right)^{2012}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=-7\\x^2-49=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=-7\\x^2=49\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=-7\\x=\pm7\end{cases}}\)
\(\)\(\Rightarrow\text{ }x=-7\)
d, \(2\left|3-x\right|^{2017}+\left(y-x+1\right)^{2016}\le0\)
\(\text{Vì }\hept{\begin{cases}2\left|3-x\right|^{2017}\ge0\\\left(y-x+1\right)^{2016}\ge0\end{cases}}\) \(\Rightarrow\text{ Chỉ xảy ra trường hợp }2\left|3-x\right|^{2017}+\left(y-x+1\right)^{2016}=0\)
\(\Rightarrow\hept{\begin{cases}2\left|3-x\right|^{2017}=0\\\left(y-x+1\right)^{2016}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\left|3-x\right|^{2017}=0\\y-x+1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}3-x=0\\y-x+1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=3\\y-3+1=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\y-2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, |- \(x\) + 2| - |\(x\) + 7| = 0
|- \(x\) + 2| = | \(x\) + 7|
\(\left[{}\begin{matrix}-x+2=x+7\\-x+2=-x-7\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{5}{2}\\2=-7\left(loại\right)\end{matrix}\right.\)
vậy \(x\) = -\(\dfrac{5}{2}\)
b, |2\(x\) - 1| + |2 + y| ≥ 0
|2\(x\) - 1| ≥ 0 ∀ \(x\)
|2 + y| ≥ 0 ∀ y
⇒ |2\(x\) - 1| +|2 + y| ≥ 0 ∀\(x\) ; y
\(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\)=>(x−1)2+(y−2)2=0
Dấu "=" xảy ra khi (x−1)2=(y−2)2=0
(x-1)2=0=>x-1=0=>x=1
(y-2)2=0=>y-2=0=>y=2
Vậy x=1 và y=2