Tìm x , y biết :
a) \(x^2+\left(y-\frac{1}{10}\right)^4=0\)
b) \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)
Tìm x,y biết:
a/\(x^2+\left(y-\frac{1}{10}\right)=0\)
b/\(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)10\le0\)
Tìm x,y biết
\(\left\{\frac{1}{2}.x-5\right\}^{10}+\left\{y^2-\frac{1}{4}\right\}^{20}\le0\)
Vì \(\left(\frac{1}{2}x-5\right)^{10}\ge0\)và \(\left(y^2-\frac{1}{4}\right)^{20}\ge0\)
nên \(\left(\frac{1}{2}x-5\right)^{10}+\left(y^2-\frac{1}{4}\right)^{20}=0\)
<=>\(\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\)<=>\(\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}\)
Ta có:\(\hept{\begin{cases}\left\{\frac{1}{2}x-5\right\}^{10}\ge0\forall x\\\left\{y^2-\frac{1}{4}\right\}^{20}\ge0\forall y\end{cases}}\)
Mà \(\left\{\frac{1}{2}x-5\right\}^{10}+\left\{y^2-\frac{1}{4}\right\}^{20}\le0\)
\(\Rightarrow\left\{\frac{1}{2}x-5\right\}^{10}+\left\{y^2-\frac{1}{4}\right\}^{20}=0\)
\(\Leftrightarrow\hept{\begin{cases}\left\{\frac{1}{2}x-5\right\}^{10}=0\\\left\{y^2-\frac{1}{4}\right\}^{20}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}}\)
Vậy \(x=10;y=\pm\frac{1}{2}\)
Tìm x, y biết :
\(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)
Tìm x.y , biết
a )\(\left(x-1\right)^2+\left(y-3\right)^2=0\)
b) \(\left(2x-\frac{1}{2}\right)^4+\left(y+\frac{3}{2}\right)^8=0\)
c) \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)
\(\left(x-1\right)^2+\left(y-3\right)^2=0\)
mà \(\left(x-1\right)^2\ge0;\left(y-3\right)^2\ge0\)
nên để: \(\left(x-1\right)^2+\left(y-3\right)^2=0\) thì:
\(x-1=y-3=0\Rightarrow x=1;y=3\)
a)x-1=y-3=0
x=1 va y=3
b)2x-1/2=y+3/2=0
x=1/4 va y=-3/2
c)1/2x-5=y2-1/4=0
1/2.x=5 va y2=1/4
x=10 va y=1/2 hoac x=10 va y=-1/2
a) x = 1 và y = 3
b) x = \(\frac{1}{4}\) và y = \(-\frac{3}{2}\)
c) x = 10 và y = \(\frac{1}{2}\)
Tìm x biết : a) \(x^2+\left(y-\frac{1}{10}\right)4=0\)
b) \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)
Tìm x:
\(a,\)\(x^2+\left(9+\frac{1}{10}\right)^2=0\)
\(b,\)\(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)
nhầm a, \(x^2+\left(9-\frac{1}{10}\right)^2=0\)
\(a;x^2+\left(9-\frac{1}{10}\right)^2=0\)
\(\Leftrightarrow x^2+\frac{89^2}{100}=0\)
\(\Leftrightarrow x^2=-\frac{7921}{100}\)
Mà\(x^2\ge0\Rightarrow x\in\varnothing\)
\(a,x^2+\left(9+\frac{1}{10}\right)^2=0\)
\(\Rightarrow x^2+\left(\frac{91}{10}\right)^2=0\)
\(\Rightarrow x^2+\frac{91^2}{10^2}=0\)
\(\Rightarrow x^2+\frac{8281}{100}=0\)
\(\Rightarrow x^2=-\frac{8281}{100}\)
Mà x2 \(\ge\)0 với mọi x
=> \(x\in\varnothing\)
b) \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)
\(\text{Ta có : }\left(\frac{1}{2}x-5\right)^{20}\ge0\forall x\)
\(\left(y^2-\frac{1}{4}\right)^{10}\ge0\forall y\)
\(\Rightarrow\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\ge0\forall x;y\)
\(\text{mà }\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)
\(\Rightarrow\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}=0\)
\(\Rightarrow\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}=0\\\left(y^2-\frac{1}{4}\right)^{10}=0\end{cases}\Rightarrow\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}\Rightarrow}\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}}\)
Vậy các cặp (x;y) thỏa mãn là : \(\left(10;\frac{1}{2}\right);\left(10;\frac{-1}{2}\right)\)
Tìm x, y
\(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)
Xét \(\left(\frac{1}{2}x-5\right)^{20}\ge0\)
\(\left(y^2-\frac{1}{4}\right)^{10}\ge0\)
\(\Rightarrow\) \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\ge0\)
mà \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}\)
x = 10
y = \(\frac{1}{2}\)
nha
..........................
\(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)
Tìm x và y
Vì \(\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\frac{1}{4}\right)^{10}\ge0\end{cases}\Rightarrow\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}}\ge0\)
Theo đề bài:
\(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)
=> \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}=0\)
<=>\(\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}=0\\\left(y^2-\frac{1}{4}\right)^{10}=0\end{cases}}\)
<=>\(\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\)
<=>\(\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}}\)
<=>\(x=10\) và \(y=-\frac{1}{4}\) hoặc \(y=\frac{1}{4}\)
Vậy ...
Tìm x biết
a)\(\frac{x+1}{x-4}>0\)
b)\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
c)\(\left(x+2\right)\left(x-3\right)< 0\)
d)\(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|\le0\)
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
\(\Rightarrow\frac{x-4}{x-4}+\frac{5}{x-4}>0\)
\(\Rightarrow1+\frac{5}{x-4}>0\)
\(\Rightarrow\frac{5}{x-4}>-1\)
\(\Rightarrow\frac{-5}{-x+4}>-\frac{5}{5}\)
\(\Rightarrow-x+4< -5\)
\(\Rightarrow-x< -9\)
\(\Rightarrow x>9\)