(x-3)+(x-2)+...+(x+10)+11=11-72
=>14x+49=-49
=>14x=-98
hay x=-7
(x-3)+(x-2)+...+(x+10)+11=11-72
=>14x+49=-49
=>14x=-98
hay x=-7
Tìm x \(\in\) Z :
a \(\left(\dfrac{3}{4}\right)^{2x}=\dfrac{12.5^9.\left(-7\right)^{10}+3.7^9.\left(-5\right)^{11}}{35^9.2^4}\)
b, \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
NHANH NHÁ GIÚP MINK VỚI GIẢI RÕ CHI TIẾT MAI MIK NỘP RỒI ! THANKS TRC^^
1)Tìm số hạng thứ 100 của dãy
a) \(1;4;-7;-10;13;16;-19;-22\);....
b)\(3;-9;15;-21;27;....\)
2)Tìm x
a)\(4+6+8+10+...+x=990\)
b)\(\left(-1\right).1+\left(-1^2\right).2+\left(-1^3\right).3+...\left(-1^{2017}\right).2017-x=0\)
Chứng minh rằng :
a)\(\dfrac{1}{x}\)-\(\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\)
b)\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\)
c)\(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
Tìm x;y \(\in\) Z
\(\left|3x+2y\right|\) +\(\left|4y-1\right|\) \(\le\) 0
\(\left|x+y-7\right|\) \(\left|xy-10\right|\) \(\le\) 0
\(\left|x-y-2\right|\) +\(\left|y+3\right|\) =0
1:rút gọn
\(\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}\)
2: tìm x
\(\dfrac{3\cdot\left(x-2\right)}{4}-\dfrac{2\cdot\left(1+2x\right)}{3}=1\dfrac{1}{4}-5\cdot\dfrac{\left(1+3x\right)}{6}-\dfrac{x-2}{12}\)
Tìm STN x biết rằng:
a)\(720:[41-\left(2x-5\right)]=2^3\cdot5\)
b)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5+50\)
d)\(\left(x+3\right)^{10}=\left(x-3\right)^{17}\)
Tìm x:
\(\dfrac{5}{6}.|\dfrac{3}{8}-x|-\left(\dfrac{-7}{8}+\dfrac{11}{12}-\dfrac{5}{6}\right)=-\left(\dfrac{-1}{2}\right)^2\)
1: rút gọn rồi tính
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right)\) : \(\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
2: tìm x: \(3\cdot\left(4-x\right)+\left(x+2\right)\cdot\left(1+2x\right)=7\cdot\left(1+x\right)-2x\cdot\left(2-x\right)\)
3: tìm x: \(\dfrac{2\cdot\left(1+x\right)}{3}-\dfrac{5\cdot\left(2-x\right)}{6}=1\dfrac{1}{3}-\dfrac{3\cdot\left(2x+3\right)}{4}-1\dfrac{1}{2}\cdot\left(x+1\right)\)
4: cho a= \(3+3^{2^3}+3^3+3^4+...+3^{360}\)