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(3x-72).809=8010
(3x-72)=810:809
(3x-72)=801
(3x-72)=80
3x-49=80
3x=80+49
3x=129
x=129:3
x=43
80-(4.25-3.8)=1024-x+4
80-(100-24)=1028-x
80-76=1028-x
4=1028-x
x=1028-4
x=1024
k cho mk nha
\(80-\left(4.5^2-3.2^3\right)=2^{10}-\left(x-4\right)\) )
=> \(80-\left[2^2\left(5^2-3.2\right)\right]=2^{10}-x+4\)
=> \(80-\left[2^2\left(25-6\right)\right]=2^{10}-x+4\)
=> \(80-\left(4.19\right)=2^{10}-x+4\)
=> \(80-76-4=2^{10}-x\)
=> \(2^{10}-x=0\)
=> \(x=1024\)
80 - (4.52 - 3.23) = 210 - (x - 4)
<=> 80 - (100 - 24) = 210 - (x - 4)
<=> 80 - 76 = 210 - (x - 4)
<=> 4 + x - 4 = 210
<=> x = 210
@Đức Đạt Đỗ (Đạt 301 Channel)
80-[2^2(5^2-6)]=2^10-(x-4)
80-(2^2.19)=2^10-x+4
80-76=2^10-x+4
\(\Rightarrow\)2^10-x+4=4
x+4=1024-4=1020
x=1020-4=1016
90-[4.52-3.23]=210-(x-40
90-[100-24]=210-(x-4)
90-76=210-(x-4)
14=1024-(x-4)
x-4=1024-14
x-4=1010
x=1010+4
x=1014
\(2^{100}\)và \(1024^9\)
Ta có: \(1024^9=\left(2^{10}\right)^9=2^{90}\)
Vì \(2^{100}>2^{90}\)
\(\Rightarrow2^{100}>1024^9\)
a) Ta có : 10249 = (210)9 = 290
2100 > 290 => 2100 > 10249
b) Ta có : 912 = (32)12 = 324 ; 277 = (33)7 = 321
324 > 321 => 912 > 277
c) Ta có : 12580 = (53)80 = 5240 ; 25118 = (52)118 = 5236
5240 > 5236 => 12580 > 25118
d) tự làm
a/ \(3^{x+2}+3^x=10\)
\(\Leftrightarrow3^x.\left(3^2+1\right)=10\)
\(\Leftrightarrow3^x.10=10\)
\(\Leftrightarrow3^x=1\)
\(\Leftrightarrow3^x=3^0\)
\(\Leftrightarrow x=0\)
Vậy...
b/ \(4.2^x+2^x=80\)
\(\Leftrightarrow2^x\left(4+1\right)=80\)
\(\Leftrightarrow2^x.5=80\)
\(\Leftrightarrow2^x=16\)
\(\Leftrightarrow2^x=2^4\)
\(\Leftrightarrow x=4\)
Vậy..
c/ \(\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
Vậy..