(2x-1)2016>(=)0
l3y-1l>(=)0
=>(2x-1)2016+l3y-1l>(=)0
mà (2x-1)2016+l3y-1l<(=)0
=>(2x-1)2016=l3y-1l=0
=>2x-1=3y-1=0
=>x=1/2;y=1/3
vậy x=1/2;y=1/3
Tìm \(x;y\)biết: \(\left(2x-1\right)^{2016}+\left(3y+6\right)^{2014}+\left(z-1\right)^{2012}\)
\(C=2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\frac{2015}{2016}\right)^0\)
a,\(\left|y^2-\frac{1}{4}\right|\)+\(\left(2x^2-32\right)^8\)= 0
b,\(\left(3y^2-27\right)^{30}+\left|x^2-16\right|\le\)0
Biết \(\left(2x-1\right)^{2016}+\left(3y+6\right)^{2014}+\left(z-1\right)=0\)
Tính \(4x+y-3z\)
C=\(x\)\(\left[x^2-y\right]\)x\(\left[x^3-2y^2\right]\)x\(\left[x^4-3y^3\right]\)x\(\left[x^5-4y^4\right]\)tại \(x=2,y=-2\)
D=\(x^2\left[x+y\right]\)-\(y^2\)\(\left[x+y\right]\)+\(\left[x^2-y^2\right]\)+2\(\left[x+y\right]\)+3 biết rằng x+y+1=0
M=\(\left[x+y\right]\)x\(\left[y+z\right]\)x\(\left[x+z\right]\)biết ranhwfx+y+z=xyz=2
Tìm x , y biết:
\(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}\le0\)
Bài 1 :Rút gọn
\(\left(4x^2-3y\right)a2y-\left(3x^2-4y\right)3y\)
\(4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)
\(2ax^2-a\left(1+2x^2\right)-\left\{a-x\left(x+a\right)\right\}\)
Bài 2:Tìm x
a)\(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+1=0\)
b)\(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
Bài 3:Rút gọn
\(x\left(1+x+x^2+...+x^9\right)-\left(1+x+x^2+...+x^9\right)\)
tìm tất cả các x,y thỏa mãn biết: a)\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
b)\(\left|2x-27\right|^{2015}+\left(3y+10\right)^{2016}=0\)
c)sao cho (2007ab) à bình phương của một số tự nhiên
