Ta có: \(\left(2x-\frac{1}{6}\right)^2\ge0\forall x\)
\(\left|3y+12\right|\ge0\forall y\)
=> \(\left(2x-\frac{1}{6}\right)^2+\left|3y+12\right|\ge0\forall x;y\)
=> \(\hept{\begin{cases}2x-\frac{1}{6}=0\\3y+12=0\end{cases}}\)
=> \(\hept{\begin{cases}2x=\frac{1}{6}\\3y=-12\end{cases}}\)
=> \(\hept{\begin{cases}x=\frac{1}{12}\\y=-4\end{cases}}\)
\(\left(2x-\frac{1}{6^{ }}\right)^2+\left|3y+12\right|\le0\)
Tìm x, biết:
\(\left(2x-5\right)^4+\left(3y+1\right)^6\le0\)
tìm x,y biết:
a)\(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)
b)\(x=6:\frac{1}{3}-0,8:\frac{1,5}{\frac{3}{2}.0,4.\frac{50}{1:\frac{1}{2}}}+\frac{1}{4}+\frac{1+\frac{1}{2}.\frac{1}{0,25}}{6-\frac{46}{1+2,2.10}}\)
c)\(y=77^{-1}.7^4.11^2.\left(1.11\right)^4.\left(7^2\right)^{-8}.\left(7^{-8}\right)^{-3}.\frac{1}{\left(-11\right)^{-3}}\)
1.Tính
A= \(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\)
B=\(1-\frac{1}{1-2^{-1}}+\frac{1}{1+\frac{1}{2^{-1}}}\)
2. Tìm x, biết:
a) \(\left(2x-3\right)^2=36\)
b) \(\left(2x-1\right)^5=243\)
c) \(\left(7x+2\right)^{-1}=3^{-2}\)
d) \(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)
Tìm x , y biết:
\(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}\le0\)
Tìm x, y, z, biết :
a) \(\left|\frac{1}{2}+x\right|+\left|x+y+z\right|+\left|\frac{1}{3}+y\right|=0\)
b) \(\left|\frac{1}{12}-x\right|+\left|\frac{1}{25}-y\right|+\left|z-\frac{14}{3}\right|\le0\)
Tìm x,y biết \(\left(2x-5\right)^{2000}+\left(3y-4\right)^{2002}\le0\)
Tìm x,y biết:
a) \(x\left(x-y\right)=\frac{3}{10}\) và \(y\left(x-y\right)=\frac{-3}{50}\)
b) \(\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}\le0\)
Tính giá trị của biểu thức \(A=4x^3y^2-\frac{1}{4}x+2y-5\), biết: \(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\le0\)
