a,

\(97.13+130.0,3=13.(97+10.0,3)=13.100=1300\)

b,

\(86.153-530.86= -377.86=-32422\)

a) Ta có: \(\dfrac{6x^2-8xy}{9xy-12y^2}\)

\(=\dfrac{2x\left(3x-4y\right)}{3y\left(3x-4y\right)}=\dfrac{2x}{3y}\)
b) \(\dfrac{2a^3-18a}{a^4-81}\)

\(=\dfrac{2a\left(a^2-9\right)}{\left(a^2-9\right)\left(a^2+9\right)}=\dfrac{2a}{a^2+9}\)

Em bấm vào biểu tượng \(\sum\) trên thanh công cụ và gõ phân số để mn dễ hỗ trợ nhé!

`(x^2+x-6)/(x^2+4x+3):(x^2-10x+25)/(x^2-4x-5)(x ne -1,x ne 5,x ne -3)`

`=((x-2)(x+3))/((x+1)(x+3)):(x-5)^2/((x+1)(x-5))`

`=(x-2)/(x+1):(x-5)/(x+1)`

`=(x-2)/(x-5)`

\(\dfrac{x^2+x-6}{x^2+4x+3}=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x+3\right)\left(x+1\right)}=\dfrac{x-2}{x+1}\)

\(\dfrac{x^2-10x+25}{x^2-4x-5}=\dfrac{\left(x-5\right)^2}{\left(x-5\right)\left(x+1\right)}=\dfrac{x-5}{x+1}\)

1: \(\dfrac{-7}{4}\cdot\dfrac{2}{9}=\dfrac{-14}{36}=\dfrac{-7}{18}\)

2: \(=\dfrac{12}{13}\cdot\dfrac{26}{5}=\dfrac{24}{5}\)

3: \(=\dfrac{20}{11}\cdot\dfrac{55}{21}=\dfrac{100}{21}\)

4: \(=\dfrac{-40}{240}=\dfrac{-1}{6}\)

a) \(97\cdot13+130\cdot0,3\)

\(=97\cdot13+13\cdot3\)

\(=13\cdot\left(97+3\right)\)

\(=13\cdot100\)

\(=1300\)

b) \(86\cdot153-530\cdot8,6\)

\(=86\cdot153-53\cdot86\)

\(=86\cdot\left(153-53\right)\)

\(=86\cdot100\)

\(=8600\)

a) \(97.13+130.0,3\)

\(=97.13+13.3\)

\(=13\left(97+3\right)=13.100=1300\)

b) \(86.153-530.8,6\)

\(=86.153-53.86\)

\(=86.\left(153-53\right)=86.100=8600\)

a) 97 x 13 + 130 x 0,3

= 97 x 13 + 13 x 10 x 0,3

= 97 x 13 + 13 x 3

= 13 x ( 97 + 3)

= 13 x 100

= 1300

b) 86 x 153 - 530 x 8,6

= 86 x 153 - 53 x 10 x 8,6

= 86 x 153 - 53 x 86

= 86 x ( 153 - 53 )

= 86 x 100

= 8600

-Chia nhỏ ra bạn ơi để nhận được câu tl sớm nhất.

-Bạn đặt không mất gì nên cứ đặt thoải mái đuyyy.

-Để dài như này khum ai làm đouuu.

a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{x-3\sqrt{x}}\right):\dfrac{2}{\sqrt{x}-3}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{2}\)

\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)

b) Thay \(x=3-2\sqrt{2}\) vào A, ta được:

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

c) Để \(A< \dfrac{2}{3}\) thì \(\dfrac{\sqrt{x}+1}{2\sqrt{x}}-\dfrac{2}{3}< 0\)

\(\Leftrightarrow\dfrac{3\left(\sqrt{x}+1\right)-4\sqrt{x}}{6\sqrt{x}}< 0\)

\(\Leftrightarrow-\sqrt{x}+3< 0\)

\(\Leftrightarrow-\sqrt{x}< -3\)

\(\Leftrightarrow\sqrt{x}>3\)

hay x>9

Vậy: Để \(A< \dfrac{2}{3}\) thì x>9