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T = 5.5 + 6.6 + .... + 30.30
T = 5.(6 - 1) + 6.(7-1) + ... + 30.(31 - 1)
T = 5.6 - 5 + 6.7 - 6 + ... + 30.31 - 30
T = (5.6 + 6.7 + ... + 30.31) - (5 + 6 + ... + 30)
Đặt A = 5.6+ 6.7 + ... + 30.31
B = 5 + 6 + ... + 30
Ta có :
3A = 5.6.3 + 6.7.3 + ... + 30.31 . 3
3A = 5.6.(7-4) + 6.7.(8-5) + ... + 30.31.(32-29)
3A = 5.6.7 - 4.5.6 + 6.7.8 - 5.6.7 + ... + 30.31.32 - 29.30.31
3A = (5.6.7 + 6.7.8 + ... + 30.31.32) - (4.5.6 + 5.6.7 + ... + 29.30.31)
3A = 30.31.32 - 4.5.6
3A = 29640
A = 29640 : 3
A = 9880
SSH của B là : (30 - 5) : 1 + 1 = 26 (số hạng)
Tổng B là : (30 + 5) . 26 : 2 =455
=> T = A - B = 9880 - 455 = 9425
c, S = 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561
S = (3 + 2187) + (9 + 6561) + (27 + 243) + (81 + 729) + 1
S = 2190 + 6570 + 270 + 810 + 1
S = (2190 + 810) + 6570 + 270 + 1
S = 3000 + 6570 + 270 + 1
S = 9570 + 270 + 1
S = 9840 + 1
S = 9841
Vậy S = 9841
\(C=\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{45.47}\)
\(\Rightarrow C=\frac{6}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\)
\(\Rightarrow C=3.\left(\frac{1}{3}-\frac{1}{47}\right)\)
\(\Rightarrow C=3.\frac{44}{141}\)
\(\Rightarrow C=\frac{44}{47}\)
\(C=\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{45.47}=3.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{45.47}\right)=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\\ \)
\(=3.\left(\frac{1}{3}-\frac{1}{47}\right)=\frac{3.44}{141}=\frac{44}{47}\)
a.
\(A=5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9=5.2^{30}.3^{18}-2^2.3^{20}.2^{27}\)
\(=5.2^{30}.3^{18}-3^{20}.2^{29}=2^{29}.3^{18}.\left(5.2-3^2\right)=2^{29}.3^{18}\)
\(B=5.2^9.\left(2.3\right)^{19}-7.2^{29}.\left(3^3\right)^6=5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}=5.2^{28}.3^{19}-7.2^{29}.3^{18}\)
\(=2^{28}.3^{18}.\left(5.3-7.2\right)=2^{28}.3^{18}\)
=> \(A:B=\left(2^{29}.3^{18}\right):\left(2^{28}.3^{18}\right)=\frac{\left(2^{29}.3^{18}\right)}{\left(2^{28}.3^{18}\right)}=2\)
b. kiểm tra lại đề bài nhé
\(a)\) \(427-98=329\)
\(b)\) \(2\cdot19\cdot15+3\cdot43\cdot10+62\cdot80\)
\(=\left(2\cdot15\right)\cdot19+\left(3\cdot10\right)\cdot43+62\cdot80\)
\(=30\cdot19+30\cdot43+62\cdot80\)
\(=30\cdot\left(19+43\right)+62\cdot80\)
\(=30\cdot62+62\cdot80\)
\(=62\cdot\left(30+80\right)\)
\(=62\cdot110=6820\)
\(c)\) Đặt \(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)
\(\Rightarrow2M=1-\frac{1}{3^6}\)
\(\Rightarrow M=\frac{728}{2\cdot729}=\frac{364}{729}\)
Vậy \(M=\frac{364}{729}\)