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\(M=1-2+3-4+5-6+...+2019-2020\)
\(\Rightarrow M=-1+\left(-1\right)+\left(-1\right)+...\left(-1\right)\)
\(\Rightarrow M=\left(-1\right).1010=-1010\)
M= 1-2+3-4+5-6+...+2019-2020
M= (-1)+(-1)+(-1)+...+(-1)
Tổng số cặp số có ở trên là:
2020:2=1010
M=(-1).1010
M=(-1010)
a: \(\Leftrightarrow3\cdot5^n\cdot25+4\cdot5^n\cdot\dfrac{1}{125}=19\cdot5^{10}\)
\(\Leftrightarrow5^n\cdot\left(75+\dfrac{4}{125}\right)=19\cdot5^{10}\)
\(\Leftrightarrow5^n\simeq2472903,228\)(vô lý)
b: \(\Leftrightarrow6^x\cdot11\cdot\dfrac{1}{6}+2\cdot6^x\cdot6=11\cdot6^{11}+2\cdot6^{11}\cdot36\)
\(\Leftrightarrow6^x\left(\dfrac{11}{6}+12\right)=6^{11}\cdot83\)
\(\Leftrightarrow6^x=6^{12}\)
hay x=12
=> \(3M=3^2+3^3+3^4+...+3^{101}\)
=> \(3M-M=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
=> \(2M=3^{101}-3\)
=> \(M=\frac{3^{101}-3}{2}\).
\(2N=2-2^2+2^3-2^4+...-2^{100}+2^{101}\)
=> \(2N-N=\left(2-2^2+2^3-2^4+...-2^{100}+2^{101}\right)-\left(1-2+2^2-2^3+...-2^{99}+2^{100}\right)\)
=> \(N=2^{101}-1\)
M = 3+3^2+3^3+....+3^100
3M = 3^2+3^3+...+3^101
3M - M = (3^2-3^2) + ... + (3^100 - 3^100) + 3^101 - 3
2M = 3^101 - 3
Vậy M = \(\frac{3^{101}-3}{2}\)
a) Ta có:
\(2n+1⋮n-3\)
\(\Rightarrow\left(2n-6\right)+7⋮n-3\)
\(\Rightarrow2\left(n-3\right)+7⋮n-3\)
\(\Rightarrow7⋮n-3\)
\(\Rightarrow n-3\in\left\{1;7\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n-3=1\Rightarrow n=4\\n-3=7\Rightarrow n=10\end{matrix}\right.\)
Vậy n=4 hoặc n=10
b) Ta có:
\(n^2+3n-13⋮n+3\)
\(\Rightarrow n\left(n+3\right)-13⋮n+3\)
\(\Rightarrow-13⋮n+3\)
\(\Rightarrow n+3\in\left\{1;13\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n+3=1\Rightarrow n=-2\left(loai\right)\\n+3=13\Rightarrow n=10\end{matrix}\right.\)
Vậy n=10
c) Ta có:
\(n^2+3⋮n-1\)
\(\Rightarrow n^2-1+4⋮n-1\)
\(\Rightarrow\left(n-1\right)\left(n+1\right)+4⋮n-1\)
\(\Rightarrow n+1+4⋮n-1\)
\(\Rightarrow n+5⋮n-1\)
\(\Rightarrow\left(n-1\right)+6⋮n-1\)
\(\Rightarrow6⋮n-1\)
\(\Rightarrow n-1\in\left\{1;2;3;6\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n-1=1\Rightarrow n=2\\n-1=2\Rightarrow n=3\\n-1=3\Rightarrow n=4\\n-1=6\Rightarrow n=7\end{matrix}\right.\)
Vậy n=2 hoặc n=3 hoặc n=4 hoặc n=7
a,\(2n+1=2n-6+7=2\left(n-3\right)+7\)
Do \(2\left(n-3\right)⋮n-3\)
\(\Rightarrow n-3\in\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}n-3=1\\n-3=-1\\n-3=7\\n-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=4\\n=2\\n=10\\n=-4\end{matrix}\right.\)
a, \(11^{25}\div11^{23}-3^5\div\left(1^{10}+2^3\right)-60\)
\(=11^{25}\div11^{23}-3^5\div\left(1+8\right)-60\)
\(=11^{25}\div11^{23}-3^5\div3^2-60\)
\(=11^2-3^3-60\)
\(=121-27-60\)
\(=94-60=34\)
b, \(2345-1000\div\left[19-2\left(21-18\right)^2\right]\)
\(=2345-1000\div\left[19-2.3^2\right]\)
\(=2345-1000\div\left[19-2.9\right]\)
\(=2345-1000\div1=2345-1000=1345\)
c, \(128-\left[68+8\left(37-35\right)^2\right]\div4\)
\(=128-\left[68+8.2^2\right]\div4\)
\(=128-\left[68+8.4\right]\div4\)
\(=128-\left[68+32\right]\div4\)
\(=128-100\div4=128-25=103\)
d, \(107-\left\{38+\left[7.3^2-24\div6+\left(9-7\right)^3\right]\right\}\div15\)
\(=107-\left\{38+\left[7.9-4+2^3\right]\right\}\div15\)
\(=107-\left\{38+\left[63-4+8\right]\right\}\div15\)
\(=107-\left\{38+\left[59+8\right]\right\}\div15\)
\(=107-\left\{38+67\right\}\div15\)
\(=107-105\div15\)
\(=107-7=100\)
e, \(50-\left[50-\left(2^3.5\right)\div2+3\right]\)
\(=50-\left[50-8.5\div2+3\right]\)
\(=50-\left[50-40\div2+3\right]\)
\(=50-\left[50-20+3\right]\)
\(=50-\left[30+3\right]\)
\(=50-33=17\)
Bài 2 :
a, \(5\left(x-9\right)=350-5^2\)
\(5\left(x-9\right)=350-25\)
\(5\left(x-9\right)=325\)
\(x-9=325\div5\)
\(x-9=65\)
\(\Rightarrow x=65+9\)
\(\Rightarrow x=74\)
Vậy x = 74
b, \(2\left(x-51\right)=2.2^3+20\)
\(2\left(x-51\right)=16+20\)
\(2\left(x-51\right)=36\)
\(x-51=36\div2\)
\(x-51=18\)
\(\Rightarrow x=18+51\)
\(\Rightarrow x=69\)
Vậy x = 69
c, \(2.3^x=162\)
\(\Rightarrow2.3^x=2.3^4\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Vậy x = 4
Có nhiều kết quả lắm :
Một là: m=7 ; n=1
Hai là :m=1; n=7
Ba là : m=5; n=5
Đúng thì k cho mình nhé
m=5
n=5