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Bài 1:
Tao có:
\(81^7mod\left(405\right)\)
\(81^3\equiv81mod\left(405\right)\)
\(81^6\equiv81^2\equiv81mod\left(405\right)\)
\(81^7\equiv81^2.81\equiv81mod\left(405\right)\)
Ta có:
\(27^9mod\left(405\right)\)
\(27^3\equiv243mod\left(405\right)\)
\(27^9\equiv243^3\equiv162mod\left(405\right)\)
Ta có:
\(9^{13}mod\left(405\right)\)
\(9\equiv9mod\left(405\right)\)
\(9^3\equiv324mod\left(405\right)\)
\(9^9\equiv324^3\equiv324mod\left(405\right)\)
\(9^{10}\equiv324.9\equiv81mod\left(405\right)\)
\(9^{13}\equiv81.324\equiv324mod\left(405\right)\)
\(81^7+27^9-9^{13}:405=81+162-324:405=-0,2\)
\(\Rightarrow81^7+27^9-9^{13}⋮405\left(đpcm\right)\)
Casio không biết có áp dụng ntn vào bài này được không nữa? Nhưng mình ôn hổm rày thấy có bài gần giống vậy, nên mình làm thử bạn tham khảo nha chúc bạn học tốt! ^^
a: =18x941+18x59
=18(941+59)
=18x1000=18000
b: \(=81:27-16:8=3-2=1\)
c: =30-40+25=-10+25=15
d: =17(85+15)-150=1700-150=1550
e: =-150-180-200=-530
f: =17+15+40=72
1. a, 3x + 2 \(⋮2x-1\)
Có 3(2x - 1) \(⋮2x-1\)
Và 2(3x - 2) \(⋮2x-1\)
=> 6x - 4 - 6x + 3 \(⋮2x-1\)
<=> -1 \(⋮2x-1\)
=> 2x - 1 \(\inƯ\left(1\right)=\left\{\pm1\right\}\)
=> 2x = 2; 0
=> x = 1; 0 (thỏa mãn)
@Lớp 6B Đoàn Kết
1. b, x2 - 2x + 3 \(⋮x-1\)
<=> x(x - 2) + 3 \(⋮x-1\)
<=> x(x - 1) - x + 3 \(⋮x-1\)
<=> x(x - 1) - (x - 1) - 2 \(⋮x-1\)
<=> (x - 1)2 - 2 \(⋮x-1\)
<=> -2 \(⋮x-1\)
=> x - 1 \(\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> x = 2; 0; 3; -1 (thỏa mãn)
@Lớp 6B Đoàn Kết
Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
a) \(2^x=32\)
Ta có: \(2^5=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
b) Sửa đề tí: \(9< 3^x< 81\)
\(\Rightarrow3^2< 3^x< 3^4\)
\(\Rightarrow2< x< 4\)
\(\Rightarrow x=\left\{3\right\}\)
Vậy x = 3
c) Ta có: \(25\le5^x\le125\)
\(\Rightarrow5^2\le5^x\le5^3\)
\(\Rightarrow2\le x\le3\)
\(\Rightarrow x=\left\{2;3\right\}\)
Vậy x = 2 hoặc x = 3
d) \(\left(x-2\right)^3\times5=40\)
\(\Rightarrow\left(x-2\right)^3=8\)
Mà \(8=2^3\Rightarrow\left(x-2\right)^3=2^3\)
Suy ra: x - 2 = 2
Vậy x = 4
Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
5x+1 = 125 = 53
x + 1 = 3 => x = 2
b) 3x-1 = 81 = 34
x - 1 = 4 => x= 5
c) (x - 6)2 = 9 = 32
x - 6 = 3 => x = 9
(2x - 7)2 = 169 = 132
2x - 7 = 13 => x = 10
d) x3 = 216 = 23.3
x3 = (2.3)3 = 63
Vậy x = 6
3x = 81
<=> x=4
b) x2=81
<=> x = 9;-9
c) (2x+3)3=125
<=> (2x+3)3=53
<=> 2x+3 = 5
<=> 2x=2
<=> x=1
d) (2x-3)4 = 625
<=>(2x-3)4=54
<=> 2x-3=5
<=> 2x=8
<=> x=4
bài 1 là:817+279-913⋮405
sorry, mk viết sai đề
bài 1 vs 3 đâu bn
Bạn ơi bài 1 nó đâu có chia hết đâu mà chứng minh?
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vt lung tung