Ta có \(1^2+2^2+3^2+...+49^2=\frac{49\left(49+1\right)\left(2.49+1\right)}{6}=40425\)
\(\Rightarrow40425\left(2-x\right)=-1\frac{1}{5}\Rightarrow2-x=-\frac{2}{67375}\Rightarrow x=\frac{134752}{67375}\)
Ta có \(1^2+2^2+3^2+...+49^2=\frac{49\left(49+1\right)\left(2.49+1\right)}{6}=40425\)
\(\Rightarrow40425\left(2-x\right)=-1\frac{1}{5}\Rightarrow2-x=-\frac{2}{67375}\Rightarrow x=\frac{134752}{67375}\)
Tìm x : \(\frac{\left(x-3\right)^2}{2}-1\frac{1}{3}\left(x+2\right)^2-\frac{5}{4}\left(x-1\right)\left(x+1\right)=1\frac{1}{2}x\left(x-2\right)-x-4\)
Tìm x : \(\frac{\left(x-3\right)^2}{2}-1\frac{1}{3}\left(x+2\right)^2-\frac{5}{4}\left(x-1\right)\left(x+1\right)=1\frac{1}{2}x\left(x-2\right)-x-4\)
Tìm x,y
a)\(\left(\frac{2}{3}:x+\frac{1}{4}\right)^2=\frac{49}{64}\)
b)\(\frac{2}{5}:\)/x-2/-1/3=1/2
c)\(\frac{1}{8}-\left(\frac{1}{12}-\frac{1}{3}x\right)^3=0\)
1. Tìm x ϵ Q sao cho:
a) (2x-3). (x+1) < 0.
b) \(\left(x-\frac{1}{2}\right).\left(x+3\right)\)> 0.
2. Tính:
S=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{999.1001}\)
3. Tìm x: Biết x không thuộc{-2; -5; -10; -17}
\(\frac{3}{\left(x+2\right).\left(x+5\right)}+\frac{5}{\left(x+5\right).\left(x+10\right)}+\frac{7}{\left(x+10\right).\left(x+17\right)}=\frac{x}{\left(x+2\right).\left(x+17\right)}\)
\(b.\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)
Tìm x biết:
a,\(\left(5x+1\right)^2=\frac{36}{49}\)
b,\(\left(2x-\frac{1}{2}\right)^2=\frac{1}{4}\)
Tìm x, biết
|x2-25|+|x2-100|=75
b,\(\left|\frac{x}{2}-2\right|\in Z;\left|\frac{x}{2}-2\right|< 2\)
c, \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
tim xEz biet:
a)\(x^2+\left(y-\frac{1}{4}\right)^4=6\)
b)\(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)
c)\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
Tìm x :
\(-5\left(x+\frac{1}{5}\right)+\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}_{ }-\frac{5}{6}\)