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\(\frac{1}{2}.\left(-\frac{11}{19}\right)-30\%\left(-\frac{1}{19}\right)+\frac{10}{19}.\frac{1111}{2222}=-\frac{11}{38}+\frac{3}{10}.\frac{1}{19}+\frac{10}{19}.\frac{1}{2}=-\frac{11}{38}+\frac{10}{38}+\frac{3}{190}=\frac{3}{190}-\frac{1}{38}=-\frac{1}{95}\)
\(23\left(x-1\right)+19=65\)
\(23\left(x-1\right)=65-19\)
\(23\left(x-1\right)=46\)
\(x-1=46:23\)
\(x-1=2\)
\(x=2+1\)
\(x=3\)
\(5x+3x=88\)
\(x\left(5+3\right)=88\)
\(x.8=88\)
\(x=88:8\)
\(x=11\)
\(x^3=64\)
\(x^3=4^3\)
\(\Rightarrow x=4\)
\(\left(5x-4\right):7-2=6\)
\(\left(5x-4\right):7=6+2\)
\(\left(5x-4\right):7=8\)
\(5x-4=8.7\)
\(5x-4=56\)
\(5x=56+4\)
\(5x=60\)
\(x=60:5\)
\(x=12\)
\(x^{50}=x\)
\(\Rightarrow x=1\)
\(4.2^x-3=125\)
\(4.2^x=125+3\)
\(4.2^x=128\)
\(2^x=128:4\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
k mk nha
#)Giải :
\(A=\frac{20^{18}+1}{20^{19}+1}\)và \(B=\frac{20^{17}+1}{20^{18}+1}\)
\(A=\frac{20^{18}+1}{20^{18+1}+1}\)và \(B=\frac{20^{17}+1}{20^{17+1}+1}\)
\(A=\frac{1}{20+1}\)và \(B=\frac{1}{20+1}\)
\(A=\frac{1}{21}\)và \(B=\frac{1}{21}\)
\(\Rightarrow A=B\)
#~Will~be~Pens~#
A>2018 +1+19/2019 +1+19
A>2018+20/2019+20
A>20(2017+1)/20(2018+1)
A>2017+1/2018+1
=>A>B
Chúc bạn học tốt
Ta có :
\(A=2+2^2+2^3+...+2^{19}+2^{20}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{18}+2^{19}+2^{20}\right)\)
\(=15+2^4\left(2+2^2+2^3\right)+2^7\left(2+2^2+2^3\right)+...+2^{18}\left(2+2^2+2^3\right)\)
\(=15+2^4.15+2^7.15+...+2^{18}.15\)
\(\Rightarrow\)\(A⋮15\)( vì xuất hiện thừa số 15 )
#Trang
#Fallen_Angel
Lời giải:
$A=(-1)+(-2)+(-3)+...+(-19)+20$
$=-(1+2+3+...+19)+20$
$=-19(19+1):2+20=-190+20=-170$