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 3: Tính nhanh:

\(\text{a, -13,45 – 7,98 – 8,55}=-29,98\)

\(\text{b, 9,72 + 8,38 + 3,62}=21,72\)

\(c,\)\(\)\(\left| {45,37} \right| - \left| { - 29,73} \right| + \left( { - 12,27} \right)\)\(=45,37-29,73-12,27=3,38\)

\(\text{d, 31,71 – 7,41.15 – 2,59.15}=-118,29\)

Bài 1:

a) \(1,3+2,5-4,7+5,6-4,3\) 

\(=\left(1,3+2,5+5,6\right)+\left(-4,7-4,3\right)\) 

\(=9,4+\left(-9\right)\) 

\(=0,4\) 

b) \(-5,7+4,2-8,2+11,7\) 

\(=\left(-5,7+11,7\right)+\left(4,2-8.2\right)\) 

\(=6+\left(-4\right)\) 

\(=2\) 

c) \(25.\left(-0,8\right).4.\left(-0,5\right).0,224\) 

\(=\left(25.4\right).\left(-0,8.-0,5\right).0,224\) 

\(=100.0,4.0,224\) 

\(=40.0,224\) 

\(=8,96\)

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

Mình sửa lại bài 1 một chút nha bạn

1:

a: =35(43+56+1)=35*100=3500

b: =40+139+99-172-139-199+172

=40+99-139+139-199+172-172

=139-199=-60

`@` `\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.`

0,2:5+2:0,5 > 0,2:0,5+2:5

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

=-2x^3+6x^2-x

`-2x.(x^2-3x+0,5)`

`=(-2x.x^2)+[-2x.(-3x)]+(-2x.0,5)`

`= -2x^3 + 6x^2 -x`

\(=\left(3.5-0.5\right)\cdot\dfrac{1}{27}+\dfrac{4}{3}=\dfrac{1}{9}+\dfrac{4}{3}=\dfrac{1}{9}+\dfrac{12}{9}=\dfrac{13}{9}\)