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\(a,4^{72}v\text{à}8^{48}\)
TA CÓ:\(4^{72}=\left(2^2\right)^{72}=2^{144}\)
\(8^{48}=\left(2^3\right)^{48}=2^{144}\)
\(\Rightarrow4^{72}=8^{48}\)
\(b,5^{127}v\text{à}2^{254}\)
TA CÓ:\(2^{252}2^{2\times127}=\left(2^2\right)^{127}=4^{127}\)
\(5^{127}>4^{127}\left(v\text{ì5>4}\right)\)\(5^{127}>4^{127}\left(v\text{ì}5>4\right)\)
\(\Rightarrow5^{127}>2^{254}\)
a) Ta có : 472 = 43.24 = (43)24 = 6424
848 = 82.24 = (82)24 = 6424
Ta thấy : 6424 = 6424 => 472 = 848
b) Ta có : 2254 = 22.127 = (22)127 = 4127
Vì 5 > 4 => 5127 > 2254
\(\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{2018.2023}\)
Ta có : \(\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{2018.2023}\)
\(=\frac{1}{5}.\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{2018.2023}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{2018}-\frac{1}{2023}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{3}-\frac{1}{2023}\right)\)
\(=\frac{1}{5}.\frac{2020}{6069}=\frac{404}{6069}\)
Bài 1:
a) \(\left|2y+1\right|=7\)
\(\Rightarrow2y+1=7\) hoặc \(2y+1=-7\)
\(\Rightarrow2y=6\) hoặc \(2y=-8\)
\(\Rightarrow y=3\) hoặc \(y=-4\)
Vậy................
b) \(\left|y-8\right|-15=22\)
\(\left|y-8\right|=37\)
\(\Rightarrow y-8=37\) hoặc \(y-8=-37\)
\(\Rightarrow y=45\) hoặc \(y=-29\)
Vậy \(y\in\left\{45;-29\right\}\)
\(2^2+2^2.2^2\)
\(=2^2+2^4\)
\(=4+16=20\)
\(2^2+2^2.2^2=2^2.\left(1+2^2\right)=4.5=20\)
\(2^2+2^2\cdot2^2\)
\(=2^2+2^{2+2}\)
\(=2^2+2^4\)
\(=2^{2+4}\)
\(=2^6\)
\(=64\)
\(2^2+2^2.2^2\)
\(=2^2+2^{2+2}\)
\(2^2+2^4\)
\(=4+16\)
\(=20\)
TK MÌNH NHÉ
\(2^2+2^2.2^2=4+4.4=4+16=20\)
\(2^2+2^2.2^2\)
\(=2^2+2^{2+2}\)
\(=2^2+2^4\)
\(=2^{2+4}\)
\(=2^6\)
\(^{2^2}\)+ \(^{2^2}\). \(^{2^2}\)= 4 + 4 . 4 = 4 + 16 = 20
Tk mik nha
2.2+2.2.2.2=4+16=20
2^2 + 2^2 .2^2 = 2^ 2 + (2.2)^2= 2^2+ 4^ 2= 4+ 16= 20
=20