3x10^2-[1200-(4^2-2x3)^3]
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a) \(3.5^2-27:3^2-5^2.4-18:3^2\)
\(=3.\left(5^2-5^2\right).27:\left(3^2-3^2\right)\)
\(=15.0.27.0\)
\(=0.0=0\)
b)
2x-1 là bội của x+3
=> 2x-1 chia hết cho x+3
hay [2(x+3)-7] chia hết ho x+ 3
=> 7 chia hết cho x+ 3
x+3 εεƯ(7)={1,-1,7,-7}
x+3=1 x+3=-1 x+3=7 x+3= -7
x = 1-3 x = -1-3 x = 7-3 x = -7-3
x = -2 x = -4 x =4 x = -10
Vậy x= -2, x=-4,x= 4, x= -10
c) \(205-\left[1200-\left(4^2-2.3\right)^3\right]:40\)
\(=205-\left[1200-16-6^3\right]:40\)
\(=205-\left[1200-10^3:40\right]\)
\(=205-1200-1000:40\)
\(=205-200:40\)
\(=205-5\)
\(=200\)
\(3^2.5+2^3.10-3^4:3\)
\(=5\left(3^2+2^3.2\right)-3^{4-1}\)
\(=5\left(9+16\right)-3^3\)
\(=5.25-27\)
\(=125-27=98\)
\(3^2\times5+2^3\times10-3^4:3\\ =9\times5+8\times10-27\\ =45+80-27\\ =98.\)
\(\frac{2^4.2^6}{\left(2^5\right)^2}-\frac{2^5.15^3}{6^3.10^2}=\frac{2^{10}}{2^{10}}-\frac{2^5.\left(3.5\right)^3}{\left(2.3\right)^3.\left(2.5\right)^2}=1-\frac{2^5.3^3.5^3}{2^3.3^3.2^2.5^2}\)
\(=1-\frac{2^5.3^3.5^3}{\left(2^3.2^2\right).3^3.5^2}=1-\frac{2^5.3^3.5^3}{2^5.3^3.5^2}=\frac{1.1.5}{1.1.1}=5\)
d: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
3.102 - [1200 - (42 - 2.3)3 ]
= 3.100 - [1200 - (16 - 2.3)3]
= 300 - [ 1200 - 103]
= 300 - 200
= 100