lm sao để viết dc cái phân số đó v

Mk nghĩ là ntn

cảm ơn bạn

giúp gì nói rõ ra chứ

giúp gì

\(BPT\Leftrightarrow1+\frac{1}{x+2}<1-\frac{1}{x+5}\)

=> \(\frac{1}{x+2}<-\frac{1}{x+5}\)

\(\Rightarrow\frac{1}{x+2}+\frac{1}{x+5}<0\)

\(\Rightarrow\frac{x+5+x+2}{\left(x+5\right)\left(x+2\right)}<0\)

=> \(\frac{2x+7}{x^2+7x+10}<0\)

\(\frac{x-2009-2010}{2008}+\frac{x-2008-2010}{2009}+\frac{x-2008-2009}{2010}=3\)

\(\Rightarrow\left(\frac{x-4019}{2008}-1\right)+\left(\frac{x-4018}{2009}-1\right)+\left(\frac{x-4017}{2010}-1\right)=0\)

\(\Rightarrow\frac{x-6027}{2008}+\frac{x-6027}{2009}+\frac{x-6027}{2010}=0\)

\(\Rightarrow\left(x-6027\right)\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}\right)=0\)

Mà \(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}\ne0\)

\(\Rightarrow x-6027=0\)

\(\Rightarrow x=6027\)

Vậy x = 6027

cảm ơn bn nhiều

\(\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{6}{2x}+\dfrac{6}{y}=\dfrac{1}{4}\)

\(\Leftrightarrow6\left(\dfrac{1}{2x}+\dfrac{1}{y}\right)=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{1}{2x}+\dfrac{1}{y}=\dfrac{1}{24}^{\left(1\right)}\)

Lại có: \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}^{\left(2\right)}\)

Lấy ⁽²⁾ trừ ⁽¹⁾ ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{2x}-\dfrac{1}{y}=\dfrac{1}{16}-\dfrac{1}{24}\)

\(\Leftrightarrow\dfrac{2-1}{2x}=\dfrac{1}{48}\)

\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{48}\)

=> 2x = 48

<=> x = 24

Thay x = 24 vào ⁽²⁾ ta có:

\(\dfrac{1}{24}+\dfrac{1}{y}=\dfrac{1}{16}\)

\(\Leftrightarrow\dfrac{1}{y}=\dfrac{1}{48}\)

=> y = 48

Vậy ...

Ta có: \(\dfrac{3}{x}\) + \(\dfrac{6}{y}\) = \(\dfrac{1}{4}\)

<=> 3(\(\dfrac{1}{x}\) + \(\dfrac{2}{y}\) ) = \(\dfrac{1}{4}\)

<=> \(\dfrac{1}{x}\) + \(\dfrac{2}{y}\) = \(\dfrac{1}{12}\) (1)

Mặt khác: \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\) = \(\dfrac{1}{16}\) (2)

Trừ (2) cho (1) vế theo vế ta được:

\(\dfrac{1}{x}\) + \(\dfrac{2}{y}\) - \(\dfrac{1}{x}\) - \(\dfrac{1}{y}\) = \(\dfrac{1}{12}\) - \(\dfrac{1}{16}\)

<=> \(\dfrac{1}{y}\) = \(\dfrac{1}{48}\) <=> y = 48

Thay y =48 vào (2) ta có: \(\dfrac{1}{x}\) + \(\dfrac{1}{48}\) = \(\dfrac{1}{16}\)

<=> \(\dfrac{1}{x}\) = \(\dfrac{1}{24}\) <=> x = 24

Vậy x =24 ; y =48

x³ + 2x² - 3x = x³ + 3x² - x² - 3x = x². (x +3) - x(x+3) = (x² - x).(x+3)

=> (ax² + bx + c).(x + 3) = (x² - x)(x + 3)

=>ax²  + bx + c = x² - x với mọi x

=> a = 1; b = -1; c = 0 

a) \(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}=\sqrt{3}-2\)

b) \(\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right).\frac{x+2\sqrt{x}}{\sqrt{x}}=\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{1}{\left(\sqrt{x}+2\right)^2}\right).\left(\sqrt{x}+2\right)\)

\(=\frac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}.\left(\sqrt{x}+2\right)=\frac{4}{x-4}\)

a, \(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}=\sqrt{3}-\sqrt{4}\)

b, Với x > 0 ; x \(\ne\)4

\(B=\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right).\frac{x+2\sqrt{x}}{\sqrt{x}}\)

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

\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}\pm2\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2}=\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}+2-\sqrt{x}+4}{\left(\sqrt{x}\pm2\right)}=\frac{6}{\left(\sqrt{x}\pm2\right)}\)

1.Với \(x-1\ge0\Rightarrow x\ge1\)

\(\Rightarrow x^2-3x+2+x-1=0\Rightarrow x^2-2x+1=0\)

\(\Rightarrow\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)

Với \(x-1< 0\Rightarrow x< 1\)

\(\Leftrightarrow x^2-3x+2-x+1=0\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}\left(l\right)}\)

Vậy x=1

2.\(\frac{x+2}{x-2}-\frac{1}{x}-\frac{2}{x\left(x-2\right)}=0\)

ĐK \(x\ne0\)và\(x\ne2\)

\(\Leftrightarrow\frac{x\left(x+2\right)-\left(x-2\right)-2}{x\left(x-2\right)}=0\Rightarrow x^2+2x-x+2-2=0\)

\(\Rightarrow x^2+x=0\Rightarrow x\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=-1\left(tm\right)\end{cases}}\)

Vậy x=-1