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a)
\(\begin{array}{l}\left( {9x - {2^3}} \right):5 = 2\\9x - {2^3} = 2.5\\9x - 8 = 10\\9x = 18\\x = 2\end{array}\)
Vậy \(x = 2\)
b)
\(\begin{array}{l}\left[ {{3^4} - \left( {{8^2} + 14} \right):13} \right]x = {5^3} + {10^2}\\\left[ {81 - \left( {64 + 14} \right):13} \right]x = 125 + 100\\\left[ {81 - 78:13} \right]x = 125 + 100\\\left[ {81 - 6} \right]x = 225\\75x = 225\\x = 3\end{array}\)
Vậy \(x = 3\)
do y>x>0 => \(5^y>5\Rightarrow5^y⋮5\)
Mặt khác, \(2^x,2^x+1,2^x+2,2^x+3,2^x+4\)là 5 số tự nhiên liên tiếp và \(2^x\)không tận cùng bằng 0
=> \(2^x\)+1 hoặc \(2^x\)+3 chia hết cho 5
=> VT \(⋮\)5
Mà 11879 không chia hết cho 5
=> không tồn tại x,y thỏa mãn
\(\left(1\dfrac{3}{4}-\dfrac{4}{6}\right):\left(1\dfrac{1}{5}+2\dfrac{2}{5}+\dfrac{1}{5}\right)< x< 1\dfrac{1}{5}.1\dfrac{1}{4}+3\dfrac{2}{11}:2\dfrac{3}{121}\)
\(\Leftrightarrow\left(\dfrac{7}{4}-\dfrac{4}{6}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)< x< \dfrac{6}{5}.\dfrac{5}{4}+\dfrac{35}{11}:\dfrac{245}{121}\) \(\Leftrightarrow\left(\dfrac{21}{12}-\dfrac{8}{12}\right):\dfrac{19}{5}< x< \dfrac{3}{2}+\dfrac{35}{11}.\dfrac{121}{245}\) \(\Leftrightarrow\dfrac{13}{12}.\dfrac{5}{19}< x< \dfrac{3}{2}+\dfrac{2}{7}\) \(\Leftrightarrow\dfrac{65}{228}< x< \dfrac{21}{14}+\dfrac{4}{14}\) \(\Leftrightarrow\dfrac{65}{228}< x< \dfrac{25}{14}\) \(\Leftrightarrow x=1\)
1) \(3^x+3^{x+1}+3^{x+2}=351\)
\(\Rightarrow3^x\left(1+3^1+3^2\right)=351\)
\(\Rightarrow3^x.13=351\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
2) \(C=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)
\(\Rightarrow C=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)...+2^{96}\left(2+2^2+2^3+2^4\right)\)
\(\Rightarrow C=30+2^4.30...+2^{96}.30\)
\(\Rightarrow C=\left(1+2^4+...+2^{96}\right).30⋮30\)
mà \(30=5.6\)
\(\Rightarrow C⋮5\left(dpcm\right)\)
1,
Có \(3^x\)+ \(3^{x+1}\) + \(3^{x+2}\) = \(351\)
=> \(3^x\) + \(3^x\).\(3\) + \(3^x\).\(9\) = \(351\)
=> \(3^x\).\(13\) = \(351\)
=> \(3^x\) = \(27\)
=> \(x\) = \(3\)
2,
C = \(2\) + \(2^2\) + \(2^3\) + ... + \(2^{100}\)
2C = \(2^2\) + \(2^3\) + \(2^4\) + ... + \(2^{101}\)
2C - C = \(2^{101}\) - \(2\)
C = \(2^{101}\) - \(2\)
C = \(2\).\(\left(2^{100}-1\right)\)
C = 2.\(\left(\left(2^5\right)^{20}-1^{20}\right)\)
Có \(2^5\) \(-1\) \(⋮\) 5
=> \(\left(\left(2^5\right)^{20}-1^{20}\right)\) \(⋮\) 5
=> C \(⋮\) 5
3,
Xét \(\overline{abcdeg}\)
= \(\overline{ab}\).\(10000\) + \(\overline{cd}\).\(100\) + \(\overline{eg}\)
= \(\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\) + \(9.\left(1111.\overline{ab}+11.\overline{cd}\right)\)
Có\(\left\{{}\begin{matrix}9.\left(1111.\overline{ab}+11.\overline{cd}\right)⋮9\left(1111.\overline{ab}+11.\overline{cd}\inℕ^∗\right)\\\overline{ab}+\overline{cd}+\overline{eg}⋮9\end{matrix}\right.\)
=> \(\overline{abcdeg}⋮9\)
4,
S = \(3^0+3^2+3^4+...+3^{2002}\)
9S = \(3^2+3^4+3^6+...+3^{2004}\)
9S - S = \(3^2+3^4+3^6+...+3^{2004}\) - (\(3^0+3^2+3^4+...+3^{2002}\))
8S = \(3^{2004}-1\)
=> 8S \(< 3^{2004}\)
\(\Rightarrow\left(x+5\right)\left(y-3\right)=1\cdot15=3\cdot5\)
Ta có
| x+5 | 1 | 15 | 3 | 5 |
| y-3 | 15 | 1 | 5 | 3 |
| x | -4(ktm) | 10 | -2(ktm) | 0 |
| y | 18 | 4 | 8 | 6 |
Vậy \(\left(x;y\right)=\left\{\left(10;4\right);\left(0;6\right)\right\}\)
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
\(\left(x+5\right)^4=\left(x+5\right)^6\)
\(\Rightarrow\left(x+5\right)^4-\left(x+5\right)^6=0\)
\(\Rightarrow\left(x+5\right)-\left(x+5\right)^2=0\)
\(\Rightarrow\)\(x=-4\)
Câu 2 : không có yêu cầu đề bài thì làm kiểu gì?
\(\left(x+5\right)^4=\left(x+5\right)^6\)
\(\Rightarrow\left(x+5\right)^6-\left(x+5\right)^4=0\)
\(\Rightarrow\left(x+5\right)^4.\left[\left(x+5\right)^2-1\right]=0\)
\(\Rightarrow\left(x+5\right)^4=0\)hoặc \(\left(x+5\right)^2-1=0\)
=> x + 5 = 0 hoặc (x + 5)2 = 1
=> x + 5 = 0 hoặc x + 5 = 1 hoặc x + 5 = -1
=> x = -5 hoặc x = -4 hoặc x = -6