\(x=\frac{2^{2018}+2^{2019}}{2^{2017}+2^{2018}}=\frac{2^{2019}}{2^{2017}}=\frac{2^2}{2}=\frac{4}{2}=2.\)
\(y=\sqrt{10^2-8^2}-\sqrt{3^2+4^2}\)
\(=\sqrt{100-64}-\sqrt{9+16}\)
\(=\sqrt{36}-\sqrt{25}\)
\(=6-5=1\)
Cho a,b,c,d khác 0, thỏa mãn :
\(\frac{x^{2018}+y^{2018}+z^{2018}+t^{2018}}{a^2+b^2+c^2+d^2}\) =\(\frac{x^{2018}}{a^2}\)+\(\frac{y^{2018}}{b^2}\)
Tính A=x2019+y2019+z2019+t2019
Cho các số \(a,b,c,d\ne0\). Tính
\(T=x^{2019}+y^{2019}+z^{2019}+t^{2019}\)
Biết \(x,y,z,t\)thoả mãn: \(\frac{x^{2018}+y^{2018}+z^{2018}+t^{2018}}{a^2+b^2+c^2+d^2}=\frac{x^{2018}}{a^2}+\frac{y^{2018}}{b^2}+\frac{z^{2018}}{c^2}+\frac{t^{2018}}{d^2}\)
Tìm x, y
| x - 2017 | + | y - 2018 | ≤ 0
3| x - y |5 + 10| y + 2/3 |7 ≤ 0
1/2(3/4x - 1/2)2018 + 2017/2018|4/5 y+ 6/25| ≤ 0
2017 |2x - y | 2018 + 2018 | y - 4 |2017 ≤ 0
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/2)+(2019/3)+(2019/4)+(2019/5)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/1)+(2019/2)+(2019/3)+(2019/4)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Bài 1: Tính giá trị của biểu thức sau
A=1-\(\dfrac{50-\dfrac{4}{2018}+\dfrac{2}{2019}-\dfrac{2}{2020}}{100-\dfrac{8}{2018} +\dfrac{4}{2019}-\dfrac{4}{2020}}\)
B=\(\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
C=\(x^{2020}\)-\(y^{2020}\)+\(xy^{2019}\)-\(x^{2019}\).y+2019 biết x-y=0
Mong mn giúp đỡ
cho x^2018+y^2018+z^20018+t^2018/a^2+b^2+c^2+d^2
=x^2018/a^2+y^2018/b^2+z^2018/c^2+t^2018/d^2tính T=x^2019+y^2019+z^2019+t^2019
giúp mik nha mn ơi.
mik cần gấp bâgiowf
Tìm x , y thỏa mãn :
a) \(\frac{1}{2}\times(\frac{3}{4}x-\frac{1}{2})^{2018}+\frac{2017}{2018}\times/\frac{4}{5}y+\frac{6}{25}/\le0\)0
b) \(2017\times/2x-y/+2018\times(y-4)^{2017}\le0\)
cm \(\frac{2_{ }!+\sqrt{1}}{2!}\)+\(\frac{3!+\sqrt{4}}{3!}\)+...+\(\frac{2018!+\sqrt{2017^2}}{2018!}\)
