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1. a)
Cách 1 : \(A=\left\{10;11;12;13;14\right\}\)
Cách 2 : \(A=\left\{x\in N\left|9< x< 15\right|\right\}\)
b) Từ 240 đến 460 có tất cả số số lẻ là :
\(\left(460-240\right)\div2+1=111\) ( số lẻ )
2.
\(a)4.13.25\)
= ( 25 . 4 ) . 13
= 100 . 13
= 1300
\(b)113+25+887\)
= ( 113 + 887 ) + 25
= 1000 + 25
= 1025
\(c)\left(5^{29}\div5^{25}\right).5\)
= \(5^4.5\)
= \(5^5\)
3.
\(a)x+4=44\)
\(x=44-4\)
\(x=40\)
\(b)60-4.\left(x+5\right)=12\)
\(4.\left(x+5\right)=60-12\)
\(4.\left(x+5\right)=48\)
\(x+5=48\div4\)
\(x+5=12\)
\(x=12-5\)
\(x=7\)
4.
\(\left[36.4-4.\left(82-7.11\right)^2\right]\div4.2016^0\)
\(=\left[36.4-4\left(82-77\right)^2\right]\div4.1\)
\(=\left[36.4-4.5^2\right]\div4\)
\(=\left[36.4-4.25\right]\div4\)
\(=\left(144-100\right)\div4\)
\(=44\div4\)
\(=11\)
4n+2 + 4n+3 + 4n+4 + 4n+5 = 85. ( 22016 : 2 2012)
=> 4n.16 + 4n.64 + 4n.256 + 4n.1024 = 85.24
=> 4n.(16 + 64 + 256 + 1024) = 85.16
=> 4n.1360 = 1360
=> 4n = 1 = 40
=> n = 0
Ta có:\(\left(x-2\right)^{10}=\left(x-2\right)^{12}\)
\(\Leftrightarrow\left(x-2\right)^{12}-\left(x-2\right)^{10}=0\)
\(\Leftrightarrow\left(x-2\right)^{10}\left[\left(x-2\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^{10}=0\\\left(x-2\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x\in\left\{3;1\right\}\end{cases}}\)
Vậy \(x\in\left\{-1,2,3\right\}\)
b) Ta có: \(4^{x+2}+4^{x+3}+4^{x+4}+4^{x+5}=85.\left(2^{2016}:2^{2012}\right)\)
\(\Leftrightarrow4^{x+2}\left(1+4+4^2+4^3\right)=85.2^4\)
\(\Leftrightarrow4^{x+2}.85=1360\)
\(\Leftrightarrow4^x=16\)
\(\Leftrightarrow4^x=4^2\)
\(\Leftrightarrow x=2\)
Vậy x=2
\(A=15+22+29+...+127+134\)
\(=\frac{\left(134+15\right)\left[\left(134-15\right):7+1\right]}{2}\)
\(=\frac{149.18}{2}\)
\(=2682:2\)
\(=1341\)
\(b,4^{n+2}+4^{n+3}+4^{n+4}+4^{n+5}=85.\left(2^{2016}:2^{2012}\right)\)
\(\Rightarrow4^n\left(4^2+4^3+4^4+4^5\right)=85.2^4\)
\(\Rightarrow4^n.1360=1360\)
\(\Rightarrow4^n=1=4^0\)
\(\Rightarrow n=0\)