a,5+x/3x-6-2x-3/2x-4=1/2 b,x(x+1).(x+3/4)=0 c,x^2/x+2=x-3-x/x+2 d,(x+1/2).(3/4-2x)=0 e,x+3/x-3-x-3/x+3=9/x^2-9

0 bình luận về “a,5+x/3x-6-2x-3/2x-4=1/2 b,x(x+1).(x+3/4)=0 c,x^2/x+2=x-3-x/x+2 d,(x+1/2).(3/4-2x)=0 e,x+3/x-3-x-3/x+3=9/x^2-9”

1. Đáp án + Giải thích các bước giải:

   `a)`

   `(5+x)/(3x-6)-(2x-3)/(2x-4)=1/2` `(x ne 2)`

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

   `<=> (2(5+x)-3(2x-3))/(6(x-2))=(3(x-2))/(6(x-2))`

   `=> 2(5+x)-3(2x-3)=3(x-2)`

   `<=> 10+2x-6x+9=3x-6`

   `<=> 19-4x=3x-6`

   `<=> -4x-3x=-6-19`

   `<=> -7x=-25`

   `<=> x=25/7 (TM)`

   `=> S={25/7}`

   $b) x(x+1).(x+\dfrac{3}{4})=0 \\ ⇔ \left[ \begin{array}{l}x=0\\x+1=0\\x+\dfrac{3}{4}=0\end{array} \right.\\⇔\left[ \begin{array}{l}x=0\\x=-1\\x=-\dfrac{3}{4}\end{array} \right.\\ $ `⇒S={0;-1;-\frac{3}{4}}`

   `c)`

   `(x^2)/(x+2)=x-(3-x)/(x+2)` `(x ne -2)`

   `<=> (x^2)/(x+2)=(x(x+2)-3+x)/(x+2)`

   `=> x^2=x^2+2x-3+x`

   `<=> x^2+3x-3=x^2`

   `<=> 3x=3`

   `<=> x=1 (TM)`

   `=> S={1}`

   `d)`

   `(x+1/2).(3/4-2x)=0`

   `<=>` \(\left[ \begin{array}{l}x+\dfrac{1}{2}=0\\\dfrac{3}{4}-2x=0\end{array}\right.\) `<=>` \(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=\dfrac{3}{8}\end{array} \right.\)`=> S={1/2;3/8}`

   `e)`

   `(x+3)/(x-3)-(x-3)/(x+3)=9/(x^2-9)` `(x ne +-3)`

   `<=> ((x+3)^2-(x-3)^2)/(x^2-9)=9/(x^2-9)`

   `=> (x+3)^2-(x-3)^2=9`

   `<=> 12x=9`

   `<=> x=3/4 (TM)`

   `=> S={3/4}`

   Trả lời

2. Đáp án:

    

   Giải thích các bước giải:

   `a)`

   `(5+x)/(3x-6)-(2x-3)/(2x-4)=1/2` (đk: `x\ne2`)

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

   `<=>(2(x+5)-3(2x-3))/(6(x-2))=(3(x-2))/(6(x-2))`

   `=>2x+10-6x+9=3x-6`

   `<=>-4x+19=3x-6`

   `<=>-7x=-25`

   `<=>x=25/7` (tmđk)

   Vậy `S={25/7}`

   `b)`

   `x(x+1)(x+3/4)=0`

   `<=>` \(\left[ \begin{array}{l}x=0\\x+1=0\\x+\dfrac{3}{4}=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=0\\x=-1\\x=-\dfrac{4}{3}\end{array} \right.\) 

   Vậy `S={0;-1;-4/3}`

   `c)`

   `(x^2)/(x+2)=x-(3-x)/(x+2)` (đk: `x\ne-2`)

   `<=>(x^2)/(x+2)=(x(x+2)-3+x)/(x+2)`

   `=>x^2=x^2+2x-3+x`

   `<=>x^2+3x-3=x^2`

   `<=>3x=3`

   `<=>x=1` (tmđk)

   Vậy `S={1}`

   `d)`

   `(x+1/2)(3/4-2x)=0`

   `<=>` \(\left[ \begin{array}{l}x+\dfrac{1}{2}=0\\\dfrac{3}{4}-2x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=\dfrac{3}{8}\end{array} \right.\) 

   Vậy `S={-1/2;3/8}`

   `e)`

   `(x+3)/(x-3)-(x-3)/(x+3)=9/(x^2-9)` (đk: `x\ne+-3`)

   `<=>((x+3)^2-(x-3)^2)/(x^2-9)=9/(x^2-9)`

   `=>x^2+6x+9-x^2+6x-9=9`

   `<=>12x=9`

   `<=>x=3/4` (tmđk)

   Vậy `S={3/4}`

   Trả lời