Question

h) \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)

i) \(x^2-\dfrac{7}{6}x+\dfrac{1}{3}=0\)

k) \(\dfrac{13}{x-1}+\dfrac{5}{2x-2}-\dfrac{6}{3x-3}\)

Answer 1

`h)x/2-1/x=1/12(x ne 0)`

`<=>6x^2-12=x`

`<=>6x^2-x-12=0`

`<=>6x^2-9x+8x-12=0`

`<=>3x(2x-3)+4(2x-3)=0`

`<=>(2x-3)(3x+4)=0`

`<=>` \(\left[ \begin{array}{l}x=\dfrac32\\x=-\dfrac43\end{array} \right.\) 

`i)x^2-7/6x+1/3=0`

`<=>6x^2-7x+2=0`

`<=>6x^2-3x-4x+2=0`

`<=>3x(2x-1)-2(2x-1)=0`

`<=>(2x-1)(3x-2)=0`

`<=>` \(\left[ \begin{array}{l}x=\dfrac12\\x=\dfrac23\end{array} \right.\) 

Câu cuối không có dấu "=" nên không tìm được x :v

Answer 2

- Hai câu h, i bấm nốt đáp án để đẹp nha ;-; câu k thiếu đề :v

Answer 3

h) Ta có: \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)

\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)

\(\Leftrightarrow12\left(x^2-2\right)-2x=0\)

\(\Leftrightarrow12x^2-2x-24=0\)

\(\Delta=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{2+34}{12}=\dfrac{36}{12}=3\\x_2=\dfrac{2-34}{12}=\dfrac{-32}{12}=-\dfrac{8}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{3;-\dfrac{8}{3}\right\}\)

i) Ta có: \(x^2-\dfrac{7}{6}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow6x^2-7x+2=0\)

\(\Leftrightarrow6x^2-3x-4x+2=0\)

\(\Leftrightarrow3x\left(2x-1\right)-2\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)