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chứ kiểu vậy thì ko có ai giải cho bạn đâu
a) Ta có \(\left(2^{17}+17^2\right)\cdot\left(9^{15}-15^9\right)\cdot\left(4^2-2^4\right)\)
=\(\left(2^{17}+17^2\right)\cdot\left(9^{15}-15^9\right)\cdot\left(16-16\right)\)
=\(\left(2^{17}+17^2\right)\cdot\left(9^{15}-15^9\right)\cdot0\)=0
b) \(\left(7^{1997}-7^{1995}\right):\left(7^{1994}\cdot7\right)\)
=\(\left(7^{1995}\left(7^2-1\right)\right):7^{1995}\)
=\(7^2-1\)=\(49-1\)=\(48\)
c Giống câu a
1.
\(\left(x+2\right)^3=\frac{1}{8}\)
\(\Rightarrow\left(x+2\right)^3=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow x+2=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}-2\)
\(\Rightarrow x=-\frac{3}{2}\)
Vậy \(x=-\frac{3}{2}.\)
2.
b) Ta có:
\(5^5-5^4+5^3\)
\(=5^3.\left(5^2-5+1\right)\)
\(=5^3.\left(25-5+1\right)\)
\(=5^3.21\)
Vì \(21⋮7\) nên \(5^3.21⋮7.\)
\(\Rightarrow5^5-5^4+5^3⋮7\left(đpcm\right).\)
c) Ta có:
\(2^{19}+2^{21}+2^{22}\)
\(=2^{19}.\left(1+2^2+2^3\right)\)
\(=2^{19}.\left(1+4+8\right)\)
\(=2^{19}.13\)
Vì \(13⋮13\) nên \(2^{19}.13⋮13.\)
\(\Rightarrow2^{19}+2^{21}+2^{22}⋮13\left(đpcm\right).\)
Chúc bạn học tốt!
\(a,\left[2^{17}+16^2\right]\cdot\left[9^{15}-3^{15}\right]\cdot\left[2^4-4^2\right]\)
\(=\left[2^{17}+16^2\right]\cdot\left[9^{15}-3^{15}\right]\cdot\left[16-16\right]\)
\(=\left[2^{17}+16^2\right]\left[9^{15}-3^{15}\right]\cdot0=0\)
\(b,\left[8^{2017}-8^{2015}\right]\cdot\left[8^{2014}\cdot8\right]\)
\(=8^{2015}\left[8^2-1\right]\cdot8^{2015}\)
\(=8^{2015}\cdot63\cdot8^{2015}=8^{4030}\cdot63\)sửa lại câu b , có vấn đề rồi
\(c,\frac{2^8+8^3}{2^5\cdot2^3}=\frac{2^8+\left[2^3\right]^3}{2^5\cdot2^3}=\frac{2^8+2^9}{2^8}=\frac{2^8\left[1+2\right]}{2^8}=3\)
2.a, \(2^6=\left[2^3\right]^2=8^2\)
Mà 8 = 8 nên 82 = 82 hay 26 = 82
b, \(5^3=5\cdot5\cdot5=125\)
\(3^5=3\cdot3\cdot3\cdot3\cdot3=243\)
Mà 125 < 243 nên 53 < 35
c, 26 = [23 ]2 = 82
Mà 8 > 6 nên 82 > 62 hay 26 > 62
d, 7200 = [72 ]100 = 49100
6300 = \(\left[6^3\right]^{100}\)= 216100
Mà 49 < 216 nên 49100 < 216100 hay 7200 < 6300
\(a,347\cdot2^2-2^2\cdot\left(216+184\right):8\)
\(< =>1388-4\cdot400:8\)
\(< =>1388-1600:8\)
\(< =>1388-200\)
\(< =>1188\)
\(b,132-\left[116-\left(132-128\right)^2\right]\)
\(< =>132-\left[116-\left(4^2\right)\right]\)
\(< =>132-100\)
\(< =>32\)
\(c,\left[184:\left(96-124:31\right)-2\right]\cdot3651\)
\(< =>\left[184:\left(96-4\right)-2\right]\cdot3651\)
\(< =>\left[184:92-2\right]\cdot3651\)
\(< =>\left[2-2\right]\cdot3651\)
\(< =>0\cdot3651\)
\(< =>0\)
\(e,\left(2+4+6+8+...+2020\right)\cdot\left(36\cdot333-108\cdot111\right)\)
\(< =>\left[\left(2020-2\right):2+1\right]\cdot\left(36\cdot333-108\cdot111\right)\)
\(< =>1010\cdot\left(11988-11988\right)\)
\(< =>1010\cdot0\)
\(< =>0\)
\(g,1024:2^4+140:\left(38+2^5\right)-7^{23}:7^{21}\)
\(< =>64+140:70-49\)
\(< =>64+2-49\)
\(< =>66-49\)
\(< =>17\)
\(h,\left(44\cdot52\cdot60\right):\left(11\cdot13\cdot15\right)\)
\(< =>137280:2145\)
\(< =>64\)
\(j,\left(2^{17}+15^4\right)\cdot\left(3^{19}-2^{17}\right)\cdot\left(2^4-4^2\right)\)
\(< =>\left(131072+50625\right)\cdot\left(1162261467-131072\right)\cdot\left(16-16\right)\)
\(< =>181697\cdot1162130395\cdot0\)
\(< =>0\)
mk chuc ban hoc tot nhe :))