Thực hiện phép tính:
a) (-2).(-2).(-2)
b) (-0,5).(-0,5);
c) \(\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\)
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a: \(=3^2+\left(3^4\right)^{-0.75}-5^{2\cdot0.5}\)
\(=9-5+3^{-3}=4+\dfrac{1}{27}=\dfrac{109}{27}\)
b: \(=2^{4-6\sqrt{7}}\cdot2^{6\sqrt{7}}=2^{4-6\sqrt{7}+6\sqrt{7}}=2^4=16\)
Bài 1: Thực hiện phép tính:
\(\text{a, 1,3 + 2,5 – 4,7 + 5,6 – 4,3=}0,4\)
\(\text{b, - 5,7 + 4,2 – 8,2 + 11,7}=2\\ \)
\(\text{c, 25.(- 0,8).4.(-0,5).0,224}=8,96\)
Bài 1:
a) 35.43 + 35.56 + 35
= 35. (43 + 56 + 1)
= 35. (99 + 1)
= 35.100
= 3500
b) 40 + (139 – 172 + 99) – (139 + 199 – 172)
= 40 + 139 – 172 + 99 – 139 – 199 + 172
= 40 + (139 – 139) + (172 – 172) + (99 – 199)
= 40 + 0 + 0 + (-100)
= -60
Bài 5:
Theo đề bài, ta có :
n + 6 chia hết cho n , n cũng chia hết cho n
Mặt khác :
[(n + 6) - n] chia hết cho n \(\leftrightarrow\) (n + 6 - n) chia heet cho n
Vậy N là ước của 6 nên:
\(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
N là số nguyên dương : \(n\in\left\{1;2;3;6\right\}\)
Vậy......
5:
n+6 chia hết cho n
=>6 chia hết cho n
mà n là số tự nhiên
nên n thuộc {1;2;3;6}
1:
a: =35(43+56+1)=35*100=3500
b: =40+139+99-172-139-199+172
=40-40=0
`@` `\text {Ans}`
`\downarrow`
`a.`
\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-0,4+1\)
`= -0,1 + 1`
`= 0,9`
`b.`
\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)
`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)
`=`\(1+4\cdot\left(-2,25\right)\)
`= 1+ (-9) = -8`
`c.`
\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)
`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)
`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)
`=`\(\dfrac{13}{6}\div2\)
`=`\(\dfrac{13}{12}\)
`d.`
\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)
`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)
`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)
`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)
`e.`
\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)
`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)
`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)
`f.`
\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)
`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)
`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)
`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)
`= 0,8 \div (-2/15)`
`=-6`
`@` `yHGiangg.`
1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)
\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)
\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)
\(=5-\dfrac{4}{2}\)
\(=5-2\)
\(=3\)
b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)
\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)
\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)
\(=0+0+0+2022\)
\(=2022\)
2) \(0,7^2\cdot x=0,49^2\)
\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)
\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)
\(\Rightarrow x=\left(0,7\right)^2\)
\(\Rightarrow x=0,49\)
b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)
\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)
\(\Rightarrow x=\left(-0,5\right)^5\)
\(\Rightarrow x=-\dfrac{1}{32}\)
2:
a: =>x*0,49=0,49^2
=>x=0,49
b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5
\(=\left(3.5-0.5\right)\cdot\dfrac{1}{27}+\dfrac{4}{3}=\dfrac{1}{9}+\dfrac{4}{3}=\dfrac{1}{9}+\dfrac{12}{9}=\dfrac{13}{9}\)
a) \((-2).(-2).(-2) =4.(-2) = -8\)
b) \((-0,5).(-0,5) = 0,25\)
c)
\(\begin{array}{l}\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\\ = \frac{{1.1.1.1}}{{2.2.2.2}}\\ = \frac{1}{{16}}\end{array}\)