\(\frac{-150.\left(1352-41\right)}{\left(1352-41\right).\left(150-15\right)}< x< \frac{\left(2400:48\right)-250}{350-\left(3600:12\right)}\)
Tìm số nguyên x biết:
a)\(\frac{-150\left(1352-41\right)}{\left(1352-41\right)\left(50-45\right)}\)<x<\(\frac{\left(2400:48\right)-250}{350-\left(3600:12\right)}\)
b)10,07-[3,927-(3,63-\(\frac{3}{20}\))\(\le\)x\(\le\)1,5 \(\left(9,2-\frac{2}{5}\right)\)+(3,2-3,12).(\(4\frac{1}{10}\)-2,25)
tim x thuocZ :-150(1352-41) / (1352-41)(150-15) < x < (2400:48)-250 / 350-(3600-12)
Rút gọn:
a,\(\frac{15.\left(1352-41\right)}{\left(1352-41\right)\left(150-15\right)}\)
b,\(\frac{2^{10}.6^{15}+3^{14}.15.4^{13}}{2^{18}.18^7.3^3+3^{15}.2^{25}}\)
a/\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
b/ \(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)
Tìm x :
\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+..+\frac{1}{\left(5x+1\right)\left(5x+6\right)}=\frac{10}{41}\)
Bài 1 : Thưc Hiên Phep tính
a, \(A=\frac{5}{3}+\frac{5}{7}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
b, \(B=\frac{5}{7}.\frac{2}{11}+\frac{5}{7}.\frac{12}{11}-\frac{5}{7}.\frac{7}{11}\)
c, \(G=\frac{\left(-2\right)}{3}+\frac{\left(-5\right)}{7}+\frac{2}{3}+\frac{\left(-2\right)}{7}\)
e, \(H=\frac{\left(-5\right)}{7}.\frac{2}{11}+\frac{\left(-5\right)}{7}.\frac{9}{11}\)
g, \(N=\frac{-5}{13}+\frac{5}{7}+\frac{20}{41}+\frac{-8}{13}+\frac{21}{41}\)
j, \(E=\frac{5}{7}.\frac{12}{11}+\frac{5}{7}.\frac{12}{11}-\frac{5}{7}.\frac{17}{11}\)
Tính
a)\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
b) \(\frac{10}{-3}+\frac{13}{10}-\frac{1}{6}+\frac{7}{10}\)
Tìm GTLN ( max )
\(i=\frac{30}{\left|x\right|+2}\) ; \(m=\frac{48}{\left|x\right|+\left|y-5\right|+12}\)