Mk nghĩ là ntn

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

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

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

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

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

\(\Rightarrow x-6027=0\)

\(\Rightarrow x=6027\)

Vậy x = 6027

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giúp gì nói rõ ra chứ

giúp gì

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

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

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

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

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

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

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

+) Với : x ≥ 1 , ta có :

( 1) ⇔ x² - 3x + 2 + x - 1 = 0

⇔ x² - 2x + 1 = 0

⇔ ( x - 1)² = 0

⇔ x = 1 ( TM ĐK )

+) Với : x < 1 , ta có :

( 1) ⇔ x² - 3x + 2 + 1 - x = 0

⇔ x² - 4x + 3 = 0

⇔ x² - x - 3x + 3 = 0

⇔ x( x - 1) - 3( x - 1) = 0

⇔ ( x - 1)( x - 3) = 0

⇔ x = 1 ( KTM ) hoặc : x = 3 ( KTM )

KL.......

3. \(\dfrac{x+2}{x-2}-\dfrac{1}{x}-\dfrac{2}{x\left(x-2\right)}=0\) ( x # 2 ; x # 0)

⇔ \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=0\)

⇔ x² + 2x + 2 - x - 2 = 0

⇔ x² + x = 0

⇔ x( x + 1) = 0

⇔ x = 0 ( KTM) hoặc : x = -1 ( TM )

KL....

Ta có :

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+....+\frac{1}{\left(x+5\right)\left(x+6\right)}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+....+\frac{1}{x+5}-\frac{1}{x+6}\)

\(=\frac{1}{x}-\frac{1}{x+6}\)

\(=\frac{6}{x\left(x+6\right)}\)