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\(S=3^1+3^2+3^3+.....+3^{100}\) \(=\left(3^1+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=120+3^5.\left(3^1+3^2+3^3+3^4\right)+....+3^{97}.\left(3^1+3^2+3^3+3^4\right)\)
\(=1.120+3^5.120+...+3^{97}.120\)
\(=\left(1+3^5+...+3^{97}\right).120\)
\(\Rightarrow S⋮120\)
Vậy ........
\(a,A=\frac{-251489}{15750}\)
b,\(B=\frac{47}{12}\)
\(c,C=\frac{11}{25}\)
\(d,D=\frac{10}{39}\)
Vì \(2x+3y⋮17\Rightarrow4.\left(2x+3y\right)⋮17\)\(=\left(8x+12y\right)\)
Vì \(\left(8x+12y\right)⋮17\)và \(9x+5y⋮17\)\(\Rightarrow\left(8x+12y\right)+\left(9x+5y\right)⋮17\)\(\Rightarrow17x+17y⋮17\)
\(\Rightarrow17\left(x+y\right)⋮17\)vì do \(17⋮17\)nên\(17\left(x+y\right)⋮17\)
=> Nếu \(2x+3y⋮17\)thì \(9x+5y⋮17\)
k mình nhé.
CHÚC BẠN HỌC GIỎI.
Lời giải:
$(-55).(-17)-55.(-2+17)=(-55)(-17)-55.15=(-55)(-17)+(-55).15$
$=(-55)(-17+15)=(-55).(-2)=55.2=110$
\(a)\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+21\)
\(=\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{2}{9}+21\)
\(=\dfrac{109}{136}+\dfrac{191}{9}\)
\(=\dfrac{26957}{1224}\)
\(b)\dfrac{25}{20}-\dfrac{18}{13}+5+\dfrac{5}{13}\)
\(=\dfrac{5}{4}-\dfrac{18}{13}+5+\dfrac{5}{13}\)
\(=\dfrac{-7}{52}+\dfrac{70}{13}\)
\(=\dfrac{21}{4}\)
\(4x:17=0\)
\(4x=0:17\)
\(\Rightarrow x=0\)
\(7x-8=713\)
\(7x=705\)
\(\Rightarrow x=100\frac{5}{7}\)
\(8\left(x-3\right)=0\)
\(8.x-8.3=0\)
\(8x=0+8.3\)
\(8x=24\)
\(\Rightarrow x=3\)
( -17 ) . 75 + (-25 ). 17 – 120
=17.(-1).75+(-25).17-120
=17.(-75-25)-120
=17.(-100)-120
=-1700-120
=-1820