Bài 11: Số vô tỉ. Khái niệm về căn bậc hai

\(\sqrt{7}< \sqrt{9}=3\)

can bac 7<3

\(\sqrt{5}< \sqrt{7}\)

`sqrt{x+32} =41`

`Đk : x>=-32`

`x+32 = 41^2`

`x+32=1618`

`x=1628-32`

`x=1649`

Vậy `x=1649`

Lời giải:
$\sqrt{x+32}=41$

$x+32=41^2=1681$

$x=1681-32=1649$

sqrt{x+4} =12`

`đk :x >=-4`

`<=> x+4 =12^2`

`x+4 =144`

`x=144-4`

`x=140`

Vậy `x=140(t//m)`

`sqrt{25} - 3sqrt{4/9+1}`

`=sqrt{5^2} - 3 sqrt{ 4/9 +9/9}`

`=5 -3sqrt{13/9}`

`=5 - 3 * (sqrt{13})/3`

`=5-sqrt{13}`

\(=3-1+3:2=2+1,5=3,5\)

Sửa đề: \(\sqrt{25}-3\sqrt{\dfrac{4}{9}}+1\)

=5-3*2/3+1

=5-2+1=4

\(=5-3.\dfrac{2}{3}+1=5-2+1=4\)

\(=\left(0.125\cdot8\right)^{100}=1\)

\(=\left(0,125.\sqrt{64}\right)^{100}=1^{100}=1\)

sửa : \(\sqrt{\dfrac{16}{9}}+\left(\dfrac{2}{3}\right)^9:\left(-\dfrac{2}{3}\right)^8-2020\)

\(=\dfrac{4}{3}+\dfrac{2}{3}-2020=2-2020=-2018\)