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1/ \(\frac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)
\(=\frac{5^{20}.3^{29}-3^{31}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{19}}=\frac{3^{29}.5^{18}.\left(25-9\right)}{3^{29}.5^{18}.\left(7-5\right)}=\frac{16}{2}=8\)
CÁC BÀI CÒN LẠI TƯƠNG TỰ HẾT NHÉ E
Bài 1:
a. https://olm.vn/hoi-dap/detail/100987610050.html
b. Giống nhau hoàn toàn => P=Q
Chỉ biết thế thôi
có : Q = [ 2 + 2^2 ] + [ 2^3 +2^4] + ... + [2^9 + 2^10]
Q = 2 [1+2] +2^3[1 +2]+ ...+ 2^9 [1+2]
Q = 2 . 3+2^3 .3 +... + 2^9 .3
Q = 3. [ 2 + 2^3 +... + 2^9]
Vậy Q chia hết cho 3
S=30+32+34+36+...+3200
6S=32+34+36+...+3202
6S-S=(32+34+36+...+3202)-(1+32+34+...+3200)
5S=1+(32-32)+(34-34)+...+(3200-3200)+3202
S=(3200+1):5\(\frac{ }{ }\)
Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)
Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)
\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)
\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)
\(A=\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^{14}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{5.\left(2^2\right)^9.\left(3^2\right)^9.\left(2^2\right)^6-2.\left(2^2\right)^{14}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{18}}{5.2^{28}.3^{18}-7.2.2^{28}.3^{18}}\)
\(A=5.\frac{\left(2^{18}.2^{12}\right).3^{18}.2^{29}.3^{18}}{2^{28}.3^{18}.\left(5-7-2\right)}\)
\(A=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{5.2.2^{29}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{2^{20}.2^{18}.\left(5.2-1\right)}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{2^{29}.3^{18}.9}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{2.11}{11.\left(-2\right)}\)
\(A=-2\)
Vậy : A = -2
Ta có : \(A=4+2^2+2^3+.....+2^{30}\)
\(\Rightarrow2A=8+2^3+2^4+.....+2^{31}\)
=> \(2A-A=2^{31}+8-4-2^2\)
<=> \(2A=2^{31}\)
Mk ghi nhầm nhé :
Kết quả cuối cùng là : \(A=2^{31}\)