Question

a) 0,1 . √9 + 0,2 . √16

b) (-1/3)² - 3/8 : (0,5)^{3 +} 1234⁰

c) (x - 3/5)² - 0,2 = -1/25

b) | x - 2| = 5

Answer 1

`a, 0,1 . sqrt(9) + 0,2 . sqrt(16)`

`= 0,1 . 3 + 0,2 . 4`

`= 0,3 +0,8`

`= 1,1`

`b, (-1/3)^2 - 3/8 : (0,5)^3 +1234^0`

`= 1/9 - 3/8 : 1/8 +1`

`= 1/9 - 3+1`

`= 1/9 -2`

`= -17/9`

`c, (x-3/5)^2 - 0.2 = -1/25`

`=> (x-3/5)^2 = -1/25+0,2`

`=> (x-3/5)^2=4/25`

`=> (x-3/5)^2 = (+-2/5)^2`

`=>[(x-3/5 = 2/5),(x -3/5 = -2/5):}`

`=> [(x= 1),(x = 1/5):}`

`d, |x-2|=5`

`=> x-2=+-5`

`=> [(x-2=5),(x-2=-5):}`

`=> [(x=7),(x=-3):}`

 

Answer 2

`0,1 . sqrt{9} + 0,2 . sqrt{16}`

`= 0,1 . 3 + 0,2 . 4`

`= 0,3 + 0,8`

`= 1,1`

`(-1/3)^2 - 3/8 : (0,5)^3 + 1234^0`

`= 1/9 - 3/8 : 1/8 + 1`

`= 1/9 - 1 + 1`

`= 1/9 - (1-1) `

`= 1/9`

Answer 3

`(x-3/5)^2 - 0,2 = -1/25`

`=> (x-3/5)^2 = 1/5 -1/25`

`=> (x-3/5)^2 = 4/25`

`=> (x - 3/5)^2 = (2/5)^2`

`=> x - 3/5 = 2/5` hoặc `x - 3/5 = -2/5`

`=> x = 1` hoặc `x = 1/5`

Vậy ...

`|x-2| = 5`

`=> x - 2 = 5` hoặc `x - 2=  -5`

`=> x = 7` hoặc ` x = -3`

Vậy ...

Answer 4

\(a,0,1.\sqrt{9}+0,2.\sqrt{16}\)
\(=0,1.3+0,2.14\)
\(=0,3+2,8\)
\(=3,1\)
\(b,\left(\dfrac{-1}{3}\right)^2-\dfrac{3}{8}:\left(0,5\right)^3+1234^0\)
\(=\dfrac{1}{9}-\dfrac{3}{8}:\dfrac{1}{8}+1\)
\(=\dfrac{1}{9}-\dfrac{3}{8}.8+1\)
\(=\dfrac{1}{9}-3+1\)
\(=\dfrac{-17}{9}\)
\(c,\left(x-\dfrac{3}{5}\right)^2-0,2=\dfrac{-1}{25}\)
\(\left(x-\dfrac{3}{5}\right)^2=\dfrac{-1}{25}+0,2\)
\(\left(x-\dfrac{3}{5}\right)^2=\dfrac{4}{25}\)
\(\left(x-\dfrac{3}{5}\right)^2=\left(\dfrac{2}{5}\right)^2\) or \(\left(x-\dfrac{3}{5}\right)^2=\left(\dfrac{-2}{5}\right)^2\)
\(x-\dfrac{3}{5}=\dfrac{2}{5}\) or \(x-\dfrac{3}{5}=\dfrac{-2}{5}\)
\(x=\dfrac{2}{5}+\dfrac{3}{5}\) or \(x=\dfrac{-2}{5}+\dfrac{3}{5}\)
\(x=\dfrac{5}{5}\) or \(x=\dfrac{1}{5}\)
\(d,\left|x-2\right|=5\)
\(x-2=5\) or \(x-2=-5\)
\(x=5+2\) or \(\left(-5\right)+2\)
\(x=7\) or \(x=-3\)