\(64^x\cdot9^2=16^x\cdot2^{2x+9}-431\)
Tính hợp lý:
\(H=\frac{\left(3\cdot4\cdot2^{16}\right)}{11\cdot2^{13}\cdot4^{11}-16^9}\)
\(I=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)
\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)
\(=\frac{2^9.3^9.-17}{1}\)
Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)
\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3.2^{18}}{2^{35}.3^2}\)
\(=\frac{1}{2^{17}.3}\)
a)\(\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}\)
b\(\left(1\cdot2\cdot3....\cdot9-1\cdot2\cdot3.....\cdot8-1\cdot2\cdot3....7\cdot8^2\right)\)
c)1152-(374+1152)+(-65+374)
d)(10^2+11^2+12^2):(13^2+14^2)
\(|x+\frac{2017}{2018}|+|2y-6,4|+|z-\frac{2^{19}\cdot3^9\cdot15.2^{18}\cdot9^4}{2^9\cdot3^5\cdot2^{10}\cdot12^{10}}|=0\)
Vì |x + | >= 0 Với mọi x
|2y – 6,4| >= 0 Với mọi y
|z - | >= 0 Với mọi z
ð |x + | + |2y – 6,4| + |z - | >= 0 Với mọi x,y,z
Mà |x + | + |2y – 6,4| + |z - | = 0
Dấu ‘‘=’’ xảy ra khi x + = 0
2y – 6,4 = 0
z - =0
=> x = -
y = 3,2
z = = 33,75
\(\frac{12^2\cdot4^{16}}{11\cdot2^{13}\cdot4^{11}-16^9}+\frac{7\cdot9+14\cdot27+21\cdot36}{21\cdot27+42\cdot81+63\cdot108}\)
tính\(\frac{9^{4^{ }}\cdot27^5\cdot3^6\cdot4^4}{3^8\cdot8^{14}\cdot24\cdot3\cdot8^2}\)
\(\frac{5\cdot415\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot2^{19}-7\cdot2^{19}.27^6}\)
\(\frac{8^5\cdot24^4\cdot72^2}{16^{12}\cdot125^2\cdot94^4}\)
Câu 1 : \(1,321338308x10^{-4}\)
Câu 2 : \(1316,572106\)
Câu 3 : \(1,641302619x10^{-13}\)
Ủng hộ nhé,tớ đang âm.
Tính:
\(B=\frac{18^6\cdot2^{12}\cdot4^3\cdot9^3}{16^3\cdot6^9\cdot27^3}\)
\(B=\frac{18^6\cdot2^{12}\cdot4^3\cdot9^3}{16^3\cdot6^9\cdot27^3}\)
\(=>B=\frac{\left(3^2\cdot2\right)^6\cdot2^{12}\cdot\left(2^2\right)^3\cdot\left(3^2\right)^3}{\left(2^4\right)^3\cdot\left(2\cdot3\right)^9\cdot\left(3^3\right)^3}\)
\(=>B=\frac{3^{12}\cdot2^6\cdot2^{12}\cdot2^6\cdot3^6}{2^{12}\cdot2^9\cdot3^9\cdot3^9}\)
\(=>B=\frac{\left(3^{12}\cdot3^6\right)\cdot\left(2^6\cdot2^{12}\cdot2^6\right)}{\left(2^{12}\cdot2^9\right)\cdot\left(3^9\cdot3^9\right)}\)
\(=>B=\frac{3^{18}\cdot2^{24}}{2^{21}\cdot3^{18}}\)
\(=>B=\frac{2^{24}}{2^{21}}\)
\(=>B=2^{24-21}\)
\(=>B=2^3\)
\(=>B=8\)
\(B=\frac{18^6.2^{12}.4^3.9^3}{16^3.6^9.27^3}\)
\(=\frac{2^6.3^{12}.2^{12}.2^6.3^6}{2^{12}.2^9.3^9.3^9}\)
\(=\frac{2^{24}.3^{18}}{2^{21}.3^{18}}\)\(=2^3=8\)
\(A\frac{6^{10}-3^9\cdot2^8\cdot5}{27^3\cdot4^5+16^3\cdot9^4}\)
\(A=\frac{6^{10}-3^9.2^8.5}{27^3.4^5+16^3.9^4}\)
\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{\left(3^3\right)^3.\left(2^2\right)^5+\left(2^4\right)^3.\left(3^2\right)^4}\)
\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{3^9.2^{10}+2^{12}.3^8}\)
\(=\frac{3^9.2^8.\left(3.2^2-1.1.5\right)}{3^8.2^{10}.\left(3.1+2^2\right)}\)
\(=\frac{3^9.2^8.7}{3^8.2^{10}.7}\)
\(=\frac{3}{2^2}=\frac{3}{4}\)
Bài làm :
\(A=\frac{6^{10}-3^9.2^8.5}{27^3.4^5+16^3.9^4}\)
\(=\frac{\left(2.3\right)^{10}-3^9.2^8.5}{\left(3^3\right)^3.\left(2^2\right)^5+\left(2^4\right)^3.\left(3^2\right)^4}\)
\(=\frac{2^{10}.3^{10}-3^9.2^8.5}{3^9.2^{10}+2^{12}.3^8}\)
\(=\frac{2^8.3^9.\left(2^2.3-5\right)}{3^8.2^{10}.\left(3+2^2\right)}\)
\(=\frac{3.7}{2^2.7}\)
\(=\frac{3}{4}\)
Học tốt
Tính \(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(B=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó
A=\(\frac{15\cdot3^{11}+4.27^4}{9^7}\)
B=\(\frac{2^{19}\cdot2^{73}+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
C=\(\frac{5\cdot12^3\cdot4^{11}-16^8}{\left(3\cdot2^{17}\right)^2}\)
D=\(\frac{4^7\cdot2^8}{3\cdot2^{15}\cdot16^2-5\cdot2^2\cdot\left(2^{10}\right)^2}\)