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

1)

\(\sqrt{0,36}.\sqrt{\dfrac{25}{16}}+\left(-\dfrac{1}{2}\right)^2\)

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

\(=\dfrac{3}{4}+\dfrac{1}{4}\)

\(=1\)

2)

\(\left(\dfrac{1}{4}-\dfrac{1}{\sqrt{64}}\right):\left(\dfrac{3}{2}\right)^2\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{8}\right):\dfrac{9}{4}\)

\(=\left(\dfrac{2}{8}-\dfrac{1}{8}\right):\dfrac{9}{4}\)

\(=\dfrac{1}{8}:\dfrac{9}{4}\)

\(=\dfrac{1}{8}.\dfrac{4}{9}\)

\(=\dfrac{1}{18}\)

3)

\(\sqrt{\dfrac{4}{81}}.\dfrac{27}{8}-\sqrt{0,49}.\dfrac{1}{7}\)

\(=\dfrac{2}{9}.\dfrac{27}{8}-\dfrac{7}{10}.\dfrac{1}{7}\)

\(=\dfrac{3}{4}-\dfrac{7}{10}.\dfrac{1}{7}\)

\(=\dfrac{3}{4}-\dfrac{1}{10}\)

\(=\dfrac{30}{40}-\dfrac{4}{40}\)

\(=\dfrac{13}{20}\)

d)

\(\sqrt{\dfrac{16}{81}}.\dfrac{27}{8}-\sqrt{0,25}.\left(-2\right)^2\)

\(=\dfrac{4}{9}.\dfrac{27}{8}-\dfrac{1}{2}.4\)

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

\(=\dfrac{3}{2}-2\)

\(=-\dfrac{1}{2}\)

\(#NguyễnNgọcKhánhLinh-VietNam\)

\(=\left(\dfrac{1}{2}-\dfrac{6}{5}\right):\dfrac{21}{20}-\dfrac{25}{4}+\dfrac{1}{2}=\dfrac{-7}{10}\cdot\dfrac{20}{21}-\dfrac{23}{4}\)

\(=\dfrac{-2}{3}-\dfrac{23}{4}=\dfrac{-8-69}{12}=-\dfrac{77}{12}\)

\(\sqrt{19}=4,3588....\approx4,36\)

\(\sqrt{19}=4,358898944\)

:v máy lag nãy giờ k zải đc r có người zải luôn

2x=căn 3

=>x=căn 3/2

`@` `\text {Answer}`

`\downarrow`

`x + 2/3 = 5x - 1/3`

`=> x + 2/3 - 5x + 1/3 = 0`

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

`=> -4x + 1 = 0`

`=> -4x = -1`

`=> 4x = 1`

`=> x = 1/4`

Vậy, `x = 1/4.`

Vận tốc của xe máy là:
\(\dfrac{2}{5}\cdot50=20\left(km/h\right)\) 

Quãng đường mà ô tô đi được:

\(s_1=v_1\cdot t=50t\)

Quãng đường mà xe máy đi được:

\(s_2=v_2\cdot t=20t\)

Mà tổng quãng đường hai xe đi được bằng quãng đường AB nên:

\(s_{AB}=s_1+s_2\)

\(\Rightarrow80=50t+20t\)

\(\Rightarrow70t=80\)

\(\Rightarrow t=\dfrac{8}{7}\left(h\right)\approx69\left(p\right)\)

Vậy hai xe gặp nhau 69 phút hai xe gặp nhau lúc:

9 giờ + 69 phút = 10 giờ 9 phút

giúp mình với ạ, helppp meeee

\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)

\(=\sqrt{2\left(1+2+3+...+\left(n-1\right)\right)+n}\)

\(=\sqrt{2\cdot\left(\dfrac{\left(n-1\right)\left(n-1+1\right)}{2}\right)+n}\)

\(=\sqrt{n\left(n-1\right)+n}=\sqrt{n^2}=n\)

Có nha bạn

\(0,\left(1\right)=\dfrac{1}{9}\)

\(0,\left(01\right)=\dfrac{1}{99}\)

\(1,\left(8\right)=\dfrac{187}{99}=\dfrac{187:11}{99:11}=\dfrac{17}{9}\)

\(0,\left(27\right)=\dfrac{27}{99}=\dfrac{27:9}{99:9}=\dfrac{3}{11}\)

\(\sqrt{\dfrac{3}{15}}=\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{5}{25}}=\dfrac{\sqrt{5}}{5}\)