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
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
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}`
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}`