Với mọi \(x\in R\) ta có:
\(\left(2x-1\right)^{2018}+\left|3y+x\right|\ge0\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{6}\end{matrix}\right.\)
Tìm các số hữu tỉ x, y thỏa mãn :
\(2016.\left|2x-27\right|^{2017}+2017.\left(3y+10\right)^{2018}=\left(-1\right)^{2019}+\left(-2020\right)^0\)
Giúp mk với , bài này khó quá
Tìm x biết
a) (5x-2)^10=(5x-2)^100
b)\(\left(\dfrac{2x-3}{4}\right)^{2016}\)+\(\left(\dfrac{3y+4}{5}\right)^{2018}\)=0
Bài 1: Tìm x,y,z biết:
a) \(\left|1-2x\right|+\left|2-3y\right|+\left|3-4z\right|=0\)
b) \(x+y=x:y=5\left(x-y\right)\)
cho x,y thỏa mãn \(\left(\dfrac{1}{3}-2x\right)^{2018}+\left(3y-x\right)^{2020}\le0\)
Chứng tỏ:\(\dfrac{1}{x}+\dfrac{1}{y}=24\)
Help Mn ơi lm giúp mk vs chiều nộp rồi
tim gia tri nho nhat , gia tri lon nhat
A=\(\left(4x-1\right)^4+\left|2x-3y\right|+25,6\)
B=\(\left(3x+2y\right)^2+\left|y-3\right|-10,5\)
C=\(40,5-\left(x-3\right)^2-\left|4x-3y\right|\)
D=\(-17.5-\left|y+2\right|-\left(3y+4\right)^4\)
giup minh nhe minh dang can gap
a) \(\left|2x-1\right|+\left|x+8\right|=4x\)
b) \(\left(\sqrt{x+1}\right)\left(\sqrt{x-3}\right)=0\)
c) \(3^x+3^{x+2}=2430\)
Tìm x :
a. \(\left|2x-1\right|=0\)
b. \(\left|2x-1\right|=\left|x\right|\)
Tìm x biết
a)\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\)
b)\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\)
c)\(\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
d)\(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)
e)\(2\left(x+1\right)-\dfrac{1}{3}.\left(x-1\right)=\dfrac{2}{3}\)
k)\(\left|4x-0,2\right|=0,2\)
Tim x,biet;
a/ \(\left(x-1\right)^2=0\)
b/ \(\left(x-2\right)^2-1=0\)
c/\(\left(2x-1\right)^3=-8\)
d/ \(\left(x+2\right)^2+1=0\)
