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a) \(B=3+3^2+3^3+...+3^{120}\)
\(B=3\cdot1+3\cdot3+3\cdot3^2+...+3\cdot3^{119}\)
\(B=3\cdot\left(1+3+3^2+...+3^{119}\right)\)
Suy ra B chia hết cho 3 (đpcm)
b) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+...+\left(3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3\right)+\left(1\cdot3^3+3\cdot3^3\right)+\left(1\cdot3^5+3\cdot3^5\right)+...+\left(1\cdot3^{119}+3\cdot3^{119}\right)\)
\(B=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+3^5\cdot\left(1+3\right)+...+3^{119}\cdot\left(1+3\right)\)
\(B=3\cdot4+3^3\cdot4+3^5\cdot4+...+3^{119}\cdot4\)
\(B=4\cdot\left(3+3^3+3^5+...+3^{119}\right)\)
Suy ra B chia hết cho 4 (đpcm)
c) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3+3^2\cdot3\right)+\left(1\cdot3^4+3\cdot3^4+3^2\cdot3^4\right)+...+\left(1\cdot3^{118}+3\cdot3^{118}+3^2\cdot3^{118}\right)\)
\(B=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+3^7\cdot\left(1+3+9\right)+...+3^{118}\cdot\left(1+3+9\right)\)
\(B=3\cdot13+3^4\cdot13+3^7\cdot13+...+3^{118}\cdot13\)
\(B=13\cdot\left(3+3^4+3^7+...+3^{118}\right)\)
Suy ra B chia hết cho 13 (đpcm)
(-4;-3;-2;-1;0;1;2;3;4)
Ko có dấu ngoặc nhọn nên mik xài ngoặc tròn nha
a)\(\hept{\begin{cases}x⋮18\\x⋮24\end{cases}\Rightarrow x\in BC\left(18,24\right)}\)
Ta có
\(18=3^2.2\)
\(24=2^3.3\)
\(\Rightarrow BCNN\left(18,24\right)=3^2.2^3=72\)
\(\Rightarrow BC\left(18,24\right)=\left\{0;72;144;216;...\right\}\)
Mà \(100< x< 150\)
\(\Rightarrow x=144\)
b)\(\hept{\begin{cases}126⋮x\\36⋮x\end{cases}\Rightarrow x\inƯC\left(126,36\right)}\)
Ta có
\(126=2.3^2.7\)
\(36=2^2.3^2\)
\(\RightarrowƯCLN\left(126,36\right)=2.3^2=18\)
\(\RightarrowƯC\left(126,36\right)=\left\{1;2;3;6;9;18\right\}\)
Mà \(x>10\)
\(\Rightarrow x=18\)
c)\(\hept{\begin{cases}48⋮x\\32⋮x\end{cases}\Rightarrow x\inƯC\left(48,32\right)}\)
Mà x lớn nhất \(\Rightarrow x=ƯCLN\left(48,32\right)\)
Ta có
\(48=2^4.3\)
\(32=2^5\)
\(\RightarrowƯCLN\left(48,32\right)=2^4=16\)
Vậy \(x=16\)
d)\(\hept{\begin{cases}x⋮18\\x⋮24\\x⋮54\end{cases}\Rightarrow x\in BC\left(18,24,54\right)}\)
Mà x nhỏ nhất khác 0 \(\Rightarrow x=BCNN\left(18,24,54\right)\)
Ta có
\(18=2.3^2\)
\(24=2^3.3\)
\(54=2.3^3\)
\(\Rightarrow BCNN\left(18,24,54\right)=2^3.3^3=216\)
Vậy \(x=216\)
\(\frac{x}{-7}=\frac{5}{-35}\)
\(\frac{x.5}{-35}=\frac{5}{-35}\)
=> x . 5 = 5
x = 5 : 5
x = 1
a, \(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow\left(x-1\right).3=8.9\)
\(\Rightarrow\left(x-1\right).3=72\)
\(\Rightarrow x-1=72:3\)
\(\Rightarrow x-1=24\)
\(\Rightarrow x=24+1\)
\(\Rightarrow x=25\)
b, \(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow-x.x=-9.4\)
\(\Rightarrow-\left(x^2\right)=-36\)
\(\Rightarrow x^2=36\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
c, \(\frac{x}{4}=\frac{18}{x+1}\)
\(\Rightarrow x\left(x+1\right)=4.18\)
\(\Rightarrow x.x+x.1=72\)
\(\Rightarrow x^2+x=72\)
\(\Rightarrow x^2+x-72=0\)
\(\Rightarrow x^2+x-8^2+8=0\)
\(\Rightarrow x=8\)
a) 169 . ( 3x - 9.17 ) + 24 : 3 = 30
169 . ( 3x - 153 ) + 8 = 30
169 . ( 3x - 153 ) = 30 - 8
169 . ( 3x - 153 ) = 22
3x - 153 = 22 : 169
3x - 153 = \(\frac{22}{169}\)
3x = \(\frac{22}{169}+153\)
3x = \(\frac{25879}{169}\)
x = \(\frac{25879}{169}:3\)
x = \(\frac{25879}{507}\)
Vậy \(x=\frac{25879}{507}\)
b) \(\left(\frac{4}{5}:\frac{6}{5}+\frac{1}{5}:\frac{1}{x}\right).30-26=54\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right).30=54+26\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right).30=80\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right)=80:30\)
\(\frac{2}{3}+\frac{1}{5}.x=\frac{8}{3}\)
\(\frac{1}{5}.x=\frac{8}{3}-\frac{2}{3}\)
\(\frac{1}{5}.x=2\)
\(x=2:\frac{1}{5}\)
\(x=10\)
Vậy \(x=10\)
c) \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{18}\right):12\frac{1}{9}=0\)
\(\frac{1}{2}-\left(\frac{59}{9}+x-\frac{117}{18}\right):\frac{109}{9}=0\)
\(\frac{1}{2}.\left(\frac{59}{9}-\frac{117}{18}+x\right).\frac{9}{109}=0\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right).\frac{9}{109}=0\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right)=0:\frac{9}{109}\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right)=0\)
\(\frac{1}{18}+x=0:\frac{1}{2}\)
\(\frac{1}{18}+x=0\)
\(x=0-\frac{1}{18}\)
\(x=\frac{-1}{18}\)
Vậy \(x=\frac{-1}{18}\)
d) 720 : [ 41 - ( 2x - 5 ) ] = 210
41 - 2x + 5 = 720 : 210
41 + 5 - 2x = \(\frac{24}{7}\)
46 - 2x = \(\frac{24}{7}\)
2x = \(46-\frac{24}{7}\)
2x = \(\frac{298}{7}\)
x = \(\frac{298}{7}:2\)
x = \(\frac{149}{7}\)
Vậy \(x=\frac{149}{7}\)
\(a,-\frac{x}{4}=-\frac{9}{x}\)
\(x^2=36\)
\(x=\pm6\)
\(c,\frac{x}{4}=\frac{18}{x+1}\)
\(x^2+x=72\)
\(\left(x-8\right)\left(x+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-9\end{cases}}\)
a) x = 100.
b) x = 28.
c) x = 0.
d) x = 0.
Bài cô t vừa giao nè