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Giúp mình giải bài 1 này với
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Helga
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Đáp án:
`a,(3x-7)(x+5)=(x+5)(3-2x)`
`<=>(3x-7)(x+5)-(x+5)(3-2x)=0`
`<=>(x+5)(3x-7-3+2x)=0`
`<=>(x+5)(5x-10)=0`
`<=>` \(\left[ \begin{array}{l}x+5=0\\5x-10=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.\)
Vậy `S={-5;2}`
`b,(-x+3)/2=(x-2)/3`
`<=>(3(-x+3))/6=(2(x-2))/6`
`<=>-3x+9=2x-4`
`<=>-5x=-13`
`<=>x=13/5`
Vậy `S={13/5}`
`c,(x-1)/(x-2)+5/(x+2)=12/(x^2-4)+1` ĐKXĐ: `x\ne+-2`
`<=>((x-1)(x+2)+5(x-2))/(x^2-4)=(12+x^2-4)/(x^2-4)`
`=>x^2+6x-12=x^2+8`
`<=>6x=20`
`<=>x=10/3(\text{tmđk})`
Vậy `S={10/3}`
Đáp án:
a) $\rm S=\{-5;2\}$
b) $\rm S=\{\dfrac{13}{5}\}$
c) $\rm S=\{\dfrac{10}{3}\}$
Giải thích các bước giải:
a)
`(3x-7)(x+5)=(x+5)(3-2x)`
`<=> (3x-7)(x+5)-(x+5)(3-2x)=0`
`<=> (x+5)[3x-7-(3-2x)]=0`
`<=> (x+5)(3x-7-3+2x)=0`
`<=> (x+5)(5x-10)=0`
`<=>` \(\left[ \begin{array}{l}x+5=0\\5x-10=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.\)
Vậy $\rm S=\{-5;2\}$
b)
`(-x+3)/2=(x-2)/3`
`<=> 3(-x+3)=2(x-2)`
`<=> -3x+9=2x-4`
`<=> -3x-2x=-4-9`
`<=> -5x=-13`
`<=> x=13/5`
Vậy $\rm S=\{\dfrac{13}{5}\}$
c)
`(x-1)/(x-2)+5/(x+2)=12/(x^2-4)+1`
$\rm ĐKXĐ : x \ne \pm 2$
`<=> ((x-1)(x+2)+5(x-2))/(x^2-4)=(12+x^2-4)/(x^2-4)`
`=> (x-1)(x+2)+5(x-2)=12+x^2-4`
`<=> x^2+6x-12=x^2+8`
`<=> 6x=20`
`<=> x=10/3 ™`
Vậy $\rm S=\{\dfrac{10}{3}\}$