273 + [ 34 + 27 + ( - 273 ) ]
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a)
\(\left(-8537\right)+\left(1975+8537\right)\)
\(=\left(-8537\right)+1975+8537\)
\(=1975\)
b)
\(\left(35-17\right)+\left(17+20-35\right)\)
\(=35-17+17+20-35\)
\(=20\)
c)
\(273+\left[-34+27+\left(-273\right)\right]\)
\(=273-34+27-273\)
\(=27-34\)
\(=-7\)
\(B=\left|157\dfrac{13}{27}-273\dfrac{7}{19}\right|-96\dfrac{14}{27}+15\dfrac{12}{19}\)
\(=273\dfrac{7}{19}-153\dfrac{13}{27}-96\dfrac{14}{27}+15\dfrac{12}{19}\)
\(=\left(273+15+\dfrac{7}{19}+\dfrac{12}{19}\right)-\left(153+96+\dfrac{13}{27}+\dfrac{14}{27}\right)\)
\(=289-250=39\)
1: \(\dfrac{f\left(x\right)}{x-3}=\dfrac{2x^2-6x+\left(a+6\right)x-3a-18+3a+19}{x-3}\)
=2x^2+(a+6)+3a+19/x-3
Để f(x)/x-3 dư 4 thì 3a+19=4
=>3a=-15
=>a=-5
2: \(\dfrac{f\left(x\right)}{x-5}=\dfrac{3x^2-15x+\left(a+15\right)x-5a-75+5a+102}{x-5}\)
\(=3x+a+15+\dfrac{5a+102}{x-5}\)
Để dư là 27 thì 5a+102=27
=>5a=-75
=>a=-15
A*27=4*273
A*27=1092
A=1092:27
A=40,44444444
bn nên xem lại đề bài đi
\(A=8^2\cdot324=8^2\cdot18^2=144^2=\left(12^2\right)^2=12^4\)
\(B=27^3\cdot9^4\cdot243=\left(3^3\right)^3\cdot\left(3^2\right)^4\cdot3^5=3^9\cdot3^8\cdot3^5=3^{22}\)
\(C=6^2\cdot36^4\cdot216^2=6^2\cdot\left(6^2\right)^4\cdot\left(6^3\right)^2=6^2\cdot6^8\cdot6^6=6^{16}\)
\(D=3^4\cdot81^2\cdot9^5=3^4\cdot\left(3^4\right)^2\cdot\left(3^2\right)^5=3^4\cdot3^8\cdot3^{10}=3^{22}\)
\(273+\left[34+27+\left(-273\right)\right].\)
\(=273+34+27-273\)
\(=0+61\)
\(=61\)
TL:
Ta có: \(273+\left[34+27+\left(-273\right)\right]\)
\(=273+34+27-273\)
\(=\left[273+\left(-273\right)\right]+\left(34+27\right)\)
\(=0+61\)
\(=61\)
\(HT\)