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Bài 1:
520 : [ 515 . 6 + 515 . 19 ]
= 520 : [ 515 . ( 6 + 19 ) ]
= 520 : [ 515 . 25 ]
= 520 : [ 515 . 52 ]
= 520 : 517
= 53
Bài 2
a) 5x + x = 39 - 311 : 39
=> 6x = 39 - 32
=> 6x = 39 - 9
=> 6x = 30
=> x = 5
b) 7x - x = 521 : 519 + 3 . 22 - 70
=> 6x = 52 + 3 . 4 - 1
=> 6x = 25 + 12 - 1
=> 6x = 36
=> x = 6
c) 7x - 2x = 617 : 615 + 44 : 11
=> ( 7 - 2 )x = 62 + 4
=> 5x = 36 + 4
=> 5x = 40
=> x = 8
1. \(5^{20}:\left(5^{15}.6+5^{15}.19\right)=5^{20}:\left\{5^{15}\left(6+19\right)\right\}\)
\(=5^{20}:\left\{5^{15}.25\right\}=5^{20}:\left\{5^{15}.5^2\right\}\)
\(=5^{20}:5^{17}=5^3=125\)
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\(=\frac{7.6^{11}.2^{19}.3^5-2^{19}.6^{15}}{3^{21}.2^{28}-3^{17}.2^{28}}=\frac{2^{19}.6^{11}.\left(7.3^5-6^4\right)}{3^{17}.2^{28}.\left(3^4-1\right)}=\frac{2^{30}.3^{11}.3^4.\left(7.3-2^4\right)}{3^{17}.2^{28}.\left(3^4-1\right)}=\frac{2^2.13}{3^2.80}=\frac{13}{9.20}=\frac{13}{180}\)
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3x + 42 = 196 : ( 193 . 192 ) - 3 . 12018
3x + 42 = 196 : 195 - 3 . 1
3x + 42 = 19 - 3
3x + 42 = 17
3x + 16 = 17
3x = 17 - 16
3x = 1
\(\Rightarrow\) x = 0.
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a)=0 vì 24-42=0 số nào nhân vs 0 cũng =0
b) = 100+(98-97)+(96-95)+....+(2-1)
=100+1+1+....+1(có 46 số 1 )
=100+46
=146
a)(217 + 154).(319 - 217).(24 - 42) = 0
b)100+98+96+...+4+2-97-95-...-3-1
= 100 + (98 - 97) + (96 -95) + .... + (4 - 3) + (2 - 1)
= 100 + 1 + 1 + .... + 1 + 1 (98 : 2 = 49 số 1)
= 100 + 49
= 149
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Ta có :
\(\left(44.52.60\right):\left(11.13.15\right)=\frac{44.52.60}{11.13.15}=\frac{4.4.4}{1.1.1}=64\)
Ta có :
\(\left(2^{17}+15^4\right)\left(3^{19}-2^{17}\right)\left(2^4-4^2\right)\)
\(=\)\(\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).\left[2^4-\left(2^2\right)^2\right]\)
\(=\)\(\left(2^{17}+15^4\right)\left(3^{19}-2^{17}\right)\left(2^4-2^4\right)\)
\(=\)\(\left(2^{17}+15^4\right)\left(3^{19}-2^{17}\right).0\)
\(=\)\(0\)
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a) Không thể vì: \(\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}=1+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}>1\)
b) Ta có: \(\dfrac{a}{b}< 1\) thì \(\dfrac{a}{b}>\dfrac{a-m}{b-m}\)
CM: \(\dfrac{a}{b}=\dfrac{a\cdot\left(b-m\right)}{b\cdot\left(b-m\right)}=\dfrac{ab-am}{b^2-bm}\left(1\right)\\ \dfrac{a-m}{b-m}=\dfrac{\left(a-m\right)\cdot b}{\left(b-m\right)\cdot b}=\dfrac{ab-am}{b^2-bm}\left(2\right)\)
Vì \(\dfrac{a}{b}< 1\Rightarrow a< b\Rightarrow am< bm\Rightarrow ab-am>ab-bm\left(3\right)\)
Từ (1), (2), (3) ta có \(\dfrac{a}{b}>\dfrac{a-m}{b-m}\)
Vậy
\(B=\dfrac{17^{19}-1}{17^{20}-1}>\dfrac{17^{19}-1-16}{17^{20}-1-16}=\dfrac{17^{19}-17}{17^{20}-17}=\dfrac{17\cdot\left(17^{18}-1\right)}{17\cdot\left(17^{19}-1\right)}=\dfrac{17^{18}-1}{17^{19}-1}=A\)
Vậy B > A
Sai đề rồi bạn, phải là:
\(15:(3^{19}:3^{18})-2^{19}:2^{17}\)
\(=15:3-2^2=5-4=1\)