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Vì \(\left(x-8\right)^{2016}\ge;\sqrt{y-10}\ge0\)
\(\Rightarrow\left(x-8\right)^{2016}+\sqrt{y-10}\ge0\)
Mà \(\left(x-8\right)^{2016}+\sqrt{y-10}=0\) \(\Rightarrow\left(x-8\right)^{2016}=0;\sqrt{y-10}=0\)
\(\Rightarrow x-8=0;y-10=0\)
\(\Rightarrow x=8;y=10\)
Ta có :(x - 8)2016 + \(\sqrt{y}-10\) = 0
( x- 8 )2016 >=0 ; \(\sqrt{y-10}>=0\)
=> ( x- 8 ) = 0 => x= 8
=> (y - 10 ) =0 => y = 10
=> x+y = 8+10
=> x+y = 18
\(\left(x-8\right)^{2016}+\sqrt{y-10}=0\)
Mà \(\left(x-8\right)^{2016}+\sqrt{y-10}\ge0\)
\(\Rightarrow\left[\begin{matrix}\left(x-8\right)^{2016}=0\\\sqrt{y-10}=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x-8=0\\y-10=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=8\\y=10\end{matrix}\right.\)
\(\Rightarrow x+y=8+10=18\)
Vậy x + y = 18
\(\left(x-8\right)^{2016}+\sqrt{y-10}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-8\right)^{2016}=0\\\sqrt{y-10}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
\(\Rightarrow x+y=8+10=18\)
Vậy.........................................
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, \(2\left|3-x\right|^{2017}+\left(y-x+1\right)^{2016}\le0\)
\(\text{Vì }\hept{\begin{cases}2\left|3-x\right|^{2017}\ge0\\\left(y-x+1\right)^{2016}\ge0\end{cases}}\) \(\Rightarrow\text{ Chỉ xảy ra trường hợp }2\left|3-x\right|^{2017}+\left(y-x+1\right)^{2016}=0\)
\(\Rightarrow\hept{\begin{cases}2\left|3-x\right|^{2017}=0\\\left(y-x+1\right)^{2016}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\left|3-x\right|^{2017}=0\\y-x+1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}3-x=0\\y-x+1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=3\\y-3+1=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\y-2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}\)
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3
Vì \(\hept{\begin{cases}\left|x-2\right|\ge0\\\sqrt{\left(y+1\right)^{2015}}\ge0\end{cases}\Rightarrow\left|x-2\right|+\sqrt{\left(y+1\right)^{2015}}\ge}0\)
Dấu "=" của đẳng thức xảy ra khi \(\left|x-2\right|=\sqrt{\left(y+1\right)^{2015}}=0\)
\(\left|x-2\right|=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
\(\sqrt{\left(y+1\right)^{2015}}=0\Leftrightarrow\left(y+1\right)^{2015}=0\Leftrightarrow y+1=0\Leftrightarrow y=-1\)
Thay x=2 và y=-1 vào biểu thức P ta có:
\(P=2x^3+15y^3+2016=2.2^3+15.\left(-1\right)^3+2016=16+\left(-15\right)+2016=2017\)
Vậy ................
Ta có : \(\left(x-y\right)^{2018}=\left(x-y\right)^{2016}\)
\(\Leftrightarrow\left(x-y\right)^{2018}-\left(x-y\right)^{2016}=0\)
\(\Leftrightarrow\left(x-y\right)^{2016}\left[\left(x-y\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-y\right)^{2016}=0\left(1\right)\\\left(x-y\right)^2-1=0\left(2\right)\end{cases}}\)
+) Từ (1) \(\Rightarrow x-y=0\) kết hợp với giả thiết : \(x+y=0\)
\(\Rightarrow x=y=0\)
+) Từ (2) \(\Rightarrow\orbr{\begin{cases}x-y=1\\x-y=-1\end{cases}}\)
*) Với \(x-y=1\) kết hợp với giả thiết \(x+y=0\)
\(\Rightarrow y=-\frac{1}{2},x=\frac{1}{2}\)
*) Với \(x-y=-1\) kết hợp với giả thiết \(x+y=0\)
\(\Rightarrow y=\frac{1}{2},x=-\frac{1}{2}\)
Vậy : \(\left(x,y\right)\in\left\{\left(0,0\right);\left(\frac{1}{2},-\frac{1}{2}\right);\left(-\frac{1}{2},\frac{1}{2}\right)\right\}\)
Xét : \(\left\{\begin{matrix}\left(x-8\right)^{2016}\ge0\\\sqrt{y-10}\ge0\end{matrix}\right.\)
=> Để \(\left(x-8\right)^{2016}+\sqrt{y-10}=0\)
Thì ( x- 8)2016= \(\sqrt{y-10}\)= 0
\(\Rightarrow\left\{\begin{matrix}\left(x-8\right)^{2016}=0\\\sqrt{y-10}=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=8\\y=10\end{matrix}\right.\)
=> x+ y= 8+ 10= 18
Vậy x+ y= 18
Ta có 2 trường hợp:
Th1: (x-8)2016 và \(\sqrt{y-10}\) là 2 số trài dấu.
Nhưng \(\left(x-8\right)^{2016}\ge0\) \(\forall x\)
\(\sqrt{y-10}\ge0\) \(\forall y\)
\(\Rightarrow\)(x-8)2016 và \(\sqrt{y-10}\) ko thể trái dấu
Th2: \(\left(x-8\right)^{2016}=\sqrt{y-10}=0\)
\(\Rightarrow\left\{\begin{matrix}\left(x-8\right)=0\\y-10=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=8\\y=10\end{matrix}\right.\)
Vậy x+y=8+10=18