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a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
Chứng tỏ rằng :
a) 1+5+52+53+.......+5501 \(⋮\)6
b) 2+22 +23 +.. + 2100 vừa \(⋮\)31, vừa \(⋮\) cho 5
a/ \(1+5+5^2+..........+5^{501}\)
\(=\left(1+5\right)+\left(5^2+5^3\right)+............+\left(5^{500}+5^{501}\right)\)
\(=1\left(1+5\right)+5^2\left(1+5\right)+...........+5^{500}\left(1+5\right)\)
\(=1.6+5^2.6+.............+5^{500}.6\)
\(=6\left(1+5^2+..........+5^{500}\right)⋮6\left(đpcm\right)\)
b/ \(2+2^2+2^3+............+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+............+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+............+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+..........+2^{96}.31\)
\(=31\left(2+........+2^{96}\right)⋮31\left(đpcm\right)\)
a)1+5+5^2+5^3+........+5^501
= 6+(5^2+5^3)+(5^4+5^5)......+(5^500+5^501)
=6+150+150(5^2+5^3)+150(5^4+5^5).......150(5^499+5^500)
=6+150(5^2+5^3+.......+5^500)
mà 6 chia hết cho 6
150(5^2+5^3+.......+5^500) chia hết cho 6
=> 6+150(5^2+5^3+.......+5^500) chia hết cho 6
=> 6+150+150(5^2+5^3)+150(5^4+5^5).......150(5^499+5^500) chia hết cho 6
=> 6+(5^2+5^3)+(5^4+5^5)......+(5^500+5^501) chia hết cho 6
=> 1+5+5^2+5^3+........+5^501 chia hết cho 6
Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)
Bài 1:
a) Ta có: \(\frac{-5}{7}+\frac{2}{7}+\frac{4}{-9}+\frac{4}{9}\)
\(=-\frac{3}{7}+\frac{-4}{9}+\frac{4}{9}\)
\(=-\frac{3}{7}\)
b) Ta có: \(\left(\frac{1}{2}:\frac{3}{4}\right)^2\)
\(=\left(\frac{1}{2}\cdot\frac{4}{3}\right)^2\)
\(=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c) Ta có: \(\frac{1}{2}+\frac{3}{4}-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{4}{5}-\frac{3}{4}\)
\(=\frac{1}{2}-\frac{4}{5}\)
\(=\frac{5}{10}-\frac{8}{10}=\frac{-3}{10}\)
d) Ta có: \(5^6:5^4+2^3\cdot2^2-225:15^2\)
\(=5^2+2^5-\frac{15^2}{15^2}\)
\(=25+32-1\)
\(=56\)
e) Ta có: \(\frac{7}{23}+\frac{4}{17}-\frac{7}{23}+\frac{13}{17}\)
\(=\frac{4}{17}+\frac{13}{17}\)
\(=\frac{17}{17}=1\)
g) Ta có: \(19\frac{1}{4}\cdot\frac{7}{12}-15\frac{1}{4}\cdot\frac{7}{12}\)
\(=\frac{7}{12}\left(19+\frac{1}{4}-15-\frac{1}{4}\right)\)
\(=\frac{7}{12}\cdot4=\frac{7}{3}\)
\(\left(2^{10}+2^9\right)+\left(2^8+2^7\right)+....+\left(2^2+2\right)\)
\(=2^9.\left(2+1\right)+2^7.\left(2+1\right)+...+2.\left(2+1\right)\)
\(=2^9.3+2^7.3+...+2.3\)
\(=3.\left(2^9+2^7+...+2\right)⋮3\)
P/S: mấy bài khác tương tự
\(a,2^{10}+2^9+2^8+...+2\)
\(=\left(2^{10}+2^9\right)+\left(2^8+2^7\right)+...+\left(2^2+2\right)\)
\(=2^9\left(2+1\right)+2^7\left(2+1\right)+...+2\left(2+1\right)\)
\(=2^9.3+2^7.3+...+2.3\)
\(=3\left(2^9+2^7+...+2\right)⋮3\left(đpcm\right)\)
\(b,1+3+3^2+3^3+...+3^{99}\)
\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{98}+3^{99}\right)\)
\(=4+3^2\left(1+3\right)+...+3^{98}\left(1+3\right)\)
\(=4+3^2.4+...+3^{98}.4\)
\(=4\left(1+3^2+...+3^{98}\right)⋮4\left(đpcm\right)\)
\(c,1+5+5^2+5^3+...+5^{1975}\)
\(=\left(1+5\right)+\left(5^2+5^3\right)+...+\left(5^{1974}+5^{1975}\right)\)
\(=6+5^2\left(1+5\right)+...+5^{1974}\left(1+5\right)\)
\(=6+5^2.6+...+5^{1974}.6\)
\(=6\left(1+5^2+...+5^{1974}\right)⋮6\left(đpcm\right)\)
a) \(A=1+3+3^2+.....+3^{10}⋮4\)
\(=\left(1+3\right)+\left(3^2+3^3\right)+.......+\left(3^9+3^{10}\right)\)
\(=\left(1+3\right)+\left(3^2\cdot1+3^2\cdot3\right)+.....+\left(3^9\cdot1+3^9\cdot3\right)\)
\(=\left(1+3\right)+3^2\left(1+3\right)+....+3^9\left(1+3\right)\)
\(=4\cdot1+3^2\cdot4+.......+3^9\cdot4\)
\(=4\cdot\left(1+3^2+.....+3^9\right)⋮4\)
Do đó A \(⋮\) 4
b) \(B=16^5+2^{15}⋮33\)
Ta có \(B=16^5+2^{15}\)
\(=\left(2^4\right)^5+2^{15}\)
\(=2^{20}+2^{15}\)
\(=2^{15}\cdot2^5+2^{15}\cdot1\)
\(=2^{15}\cdot\left(2^5+1\right)\)
\(=2^5\cdot\left(32+1\right)\)
\(=2^{15}\cdot33⋮33\)
Do đó \(B⋮33\)
bài 1 : thực hiện phép tính
a) 3.52+15.22-26:2
= 3.25 + 15.4 - 26 : 2
= 75 + 60 - 13
= 135 - 13
= 122
b) 20:22+59:58
= 20:4 + 5
= 5 + 5
= 10
c) 100:52+7.32
= 100:25 + 7.9
= 4 + 63
= 67
d) 295-(31-22.5)2
= 295-(31-4.5)2
= 295 - 112
= 295 - 121
= 174
e) (-47)-[(45.24-52.12):14]
= (-47)-[(45.16-25.12):14]
= (-47)-[(720-300):14]
= (-47)-( 420:14 )
= (-47) - 30
= -77
f) (-2011)+5.[300-(17-7)2]
= (-2011)+5.(300-102)
= (-2011)+5.(300-100)
= (-2011)+5.200
= (-2011)+1000
= -1011
g) 5.[29-(6-1)2]-129
= 5.(29-52)-129
= 5.(29-25)-129
= 5.4-129
= 20-129
= -109
Đúng thì tik cái nha ! Thanks nhiều !
