Cho : \(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=.......=\dfrac{a_{2017}}{a_{2018}}\) . Cm : \(\dfrac{a_1}{a_{2018}}=\left(\dfrac{a_1+a_2+....+a_{2017}}{a_2+a_3+.....+a_{2018}}\right)\)
Cho \(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=...=\dfrac{a_{n-1}}{an}=\dfrac{an}{a1}\)
\(a_1+a_2+...+an-1+an\ne0\)
Tính \(\dfrac{a_1^2+a_2^2+...+an^2}{\left(a_1+a_2+...+a_n\right)}\)
1)Tìm x, biết :
\(4.\left[3x-1\right]+\left[x\right]-2.\left[x-5\right]+7.\left[x-3\right]=12\)
\(\left[2\dfrac{1}{5}-x\right]+\left[x-\dfrac{1}{5}\right]+8\dfrac{1}{5}=1,2\)
\(3.\left[x+4\right]-\left[2.x+1\right]-5.\left[x-3\right]+\left[x-9\right]=5\)
\(2\left[x+3\dfrac{1}{2}\right]+\left[x\right]-3\dfrac{1}{2}=\left[2\dfrac{1}{5}-x\right]\)
Chứng minh rằng: \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{1}{2}\)
giúp mk nha !! ^_^
Bài 1:Tìm GTLN của các bểu thức:
a) A=5-3.(2x-1)2
b) B=\(\dfrac{1}{2.\left(x-1\right)^2+3}\)
c) C=\(\dfrac{x^2+8}{x^2+2},x\in Z\)
Bài 2: Tìm x sao cho:
a)(x-1)(x-2)>0
b) (x-2)2.(x+1).(x-4)<0
c) \(\dfrac{5}{x}< x\)
Làm hộ mình nhé
HElP ME!!
tính: a) 12.\(\left\{-\dfrac{2}{3}\right\}^2+\dfrac{4}{3}\) b) 12,5.\(\left\{\dfrac{-5}{7}\right\}+1,5.\left\{\dfrac{-5}{7}\right\}\)
Tìm x biết:
a) \(\left(-\dfrac{2}{3}\right)^2.x=\left(-\dfrac{2}{3}\right)^5\) ; b) \(\left(-\dfrac{1}{3}\right)^3.x=\dfrac{1}{81}\) ; c) (2x-3)\(^2\) ; d) (3x-2)\(^5\) =-243
bài 1 tìm x thuộc Q biết
a. |x|=\(^1_53\) và x<0
b.|x|=-2,1
c.|x-3,5|=5
d. |x+\(\dfrac{3}{4}\)|-\(\dfrac{1}{2}\)=0
e. |x-\(\dfrac{2}{5}\)|+\(\dfrac{1}{2}\)=\(\dfrac{3}{4}\)
f. \(\dfrac{5}{6}\)-|2-x|=\(\dfrac{1}{3}\)
g. (2x-5)^2=9
h. \(\sqrt{3-7x}\)=\(\dfrac{1}{4}\)
i. (\(\dfrac{2}{3}\))^x=\(\dfrac{8}{27}\)
k. (x+\(\dfrac{1}{2}\))^2=\(\dfrac{1}{9}\)
l. \(\dfrac{\left(-3^x\right)}{81}\)=-27
m.\(\left(x-2\right)\)^2x+3=(x-2)^2x+1(x thuộc N)
Câu 1:Thực hiện phép tính(tính một cách hợp lí nếu có thể):
a)\(\dfrac{1}{2}-\dfrac{-3}{6}=+\dfrac{5}{3}-\dfrac{9}{12}\)
b)\(\begin{matrix}&\left(\dfrac{-2}{3}\right)\end{matrix}.\dfrac{3}{11}+\left(\dfrac{-16}{9}\right):\dfrac{11}{3}\)
c)\(\begin{matrix}&\left(\dfrac{2}{3}\right)^0\end{matrix}-\sqrt{9+}\left(-\dfrac{^{ }1}{2}\right)^2\)