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\(a.3\times2^3+3\times4^2\)
\(=3\times2^3+3\times\left(2^2\right)^2\)
\(=3\times2^3+3\times2^4\)
\(=3\times\left(2^3+2^4\right)=3\times\left(8+16\right)=3\times24=72\)
\(520:\left\{\left[\left(16\cdot5+2^2\cdot5\right):5-5\right]+115\right\}+2017^0\)
\(520:\left\{\left[\left(80-20\right):5-5\right]+115\right\}+1\)
\(520:\left\{\left[60:5-5\right]+115\right\}+1\)
\(520:\left(7+115\right)+1\)
\(520:122+1\)
\(=5,2\)
520 : { [ (16 x 5 + 22 x 5 ) ) : 5 - 5 ] + 115 } + 2017 0
= 520 : { [ (80 + 20) : 5 - 5 ] + 115 } + 1
= 520 : { [ 100 : 5 - 5 ] + 115 } +1
= 520 : { [ 20 - 5 ] + 115 } +1
= 520 : { 15 + 115 } +1
= 520 : 130 +1
= 4 + 1
= 5
(x-2)+5=-x+3
x-2+5=-x+3
x+x=3+2-5
2x=0
x=0:2
x=0
Vậy x=0
c: \(=520:\left\{\left[\left(80+20\right):2-2\right]+82\right\}+1\)
\(=520:130+1\)
=4+1
=5
\(\left(2x-5\right)^5=\left(2x-5\right)^3\)
\(\left(2x-5\right)^5-\left(2x-5\right)^3=0\)
\(\left(2x-5\right)^3\left[\left(2x-5\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-5\right)^3=0\\\left(2x-5\right)^2-1=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x-5=0\\\left(2x-5\right)^2=1=\left(\pm1\right)^2\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x=5\\2x-5=-1\text{ hoặc }2x-5=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\2x=4\text{ hoặc }2x=6\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=2\text{ hoặc }x=3\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{5}{2}\text{ ; }2\text{ ; }3\right\}\)
\(\left(2x-5\right)^5=\left(2x-5\right)^3\)
=>\(\left(2x-5\right)^5-\left(2x-5\right)^3=0\)
=>\(\left(2x-5\right)^3.\left\{\left(2x-5\right)^2-1\right\}=0\)
=>\(\orbr{\begin{cases}\left(2x-5\right)^3=0\\\left(2x-5\right)^2-1=0\end{cases}}\)
=>\(\orbr{\begin{cases}2x-5=0\\\left(2x-5\right)^2=1\end{cases}}\)
=>\(x=\left\{\frac{5}{2};3;2\right\}\)\(\orbr{\begin{cases}x=\frac{5}{2}\\\orbr{\begin{cases}2x-5=1\\2x-5=-1\end{cases}}\end{cases}}\)
c: \(=520:\left\{\left[\left(80+20\right):2-2\right]+82\right\}+1\)
\(=520:\left\{48+82\right\}+1\)
=4+1
=5
a: \(=100\cdot104=10400\)
d: \(=2\cdot9+4\cdot27=18+108=126\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(A=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{3}{2015}\)
\(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)\)
Xét :
\(\frac{1}{2016}< \frac{1}{2015}\)\(;\)\(\frac{1}{2017}< \frac{1}{2015}\)\(;\)\(\frac{1}{2018}< \frac{1}{2015}\)
\(\Rightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{2015}+\frac{1}{2015}+\frac{1}{2015}\)
\(\Leftrightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}< 0\)
Suy ra : \(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)>4-0=4\) ( đpcm )
...
a,12^8:12^4=122 ( sai )
b, 5^9=15 ( sai )
c, 5^3.5^2=5^5 ( đúng )
=> có câu c, đúng còn câu a, và b, là sai .
cần gấp nha mn