² a)$\frac{5-2x}{3}$+$\frac{x}{2}$=$\frac{x-5}{4}$-2 b)$\frac{3}{x-1}$=$\frac{3x+2}{1-x²}$-$\frac{4}{x+1}$ c)$x^{2}$-25=(x+5)(3-2x) d)$\frac{6}{x²-1}

Question

² a)$\frac{5-2x}{3}$+$\frac{x}{2}$=$\frac{x-5}{4}$-2
b)$\frac{3}{x-1}$=$\frac{3x+2}{1-x²}$-$\frac{4}{x+1}$
c)$x^{2}$-25=(x+5)(3-2x)
d)$\frac{6}{x²-1}$+5=$\frac{8x-1}{4x+4}$-$\frac{12x-4}{4-4x}$

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Genesis 2 tuần 2021-10-10T10:59:33+00:00 2 Answers 2 views 0

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    0
    2021-10-10T11:01:03+00:00

    Đáp án:

     

    Giải thích các bước giải:

     

    0
    2021-10-10T11:01:16+00:00

    Đáp án:

    Giải thích các bước giải:

    `a)` `(5-2x)/3+x/2=(x-5)/4-2`

    `↔(4(5-2x)+6x)/12=(3(x-5)-2.12)/12`

    `→20-8x+6x=3x-15-24`

    `↔20-2x=3x-39`

    `↔-2x-3x=-39-20`

    `↔-5x=-59`

    `↔x=59/5`

    Vậy `S={59/5}`

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

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

    `→-3x-2-4x+4=3x+3`

    `↔-7x+2=3x+3`

    `↔-7x-3x=3-2`

    `↔-10x=1`

    `↔x=-1/10`

    Vậy `S={-1/10}`

    `c)x^2-25=(x+5)(3-2x)`

    `↔(x-5)(x+5)=(x+5)(3-2x)`

    `↔(x-5)(x+5)-(x+5)(3-2x)=0`

    `↔(x+5)(x-5-3+2x)=0`

    `↔(x+5)(3x-8)=0`

    `↔` \(\left[ \begin{array}{l}x+5=0\\3x-8=0\end{array} \right.\) 

    `↔` \(\left[ \begin{array}{l}x=-5\\x=\dfrac{8}{3}\end{array} \right.\) 

    Vậy `S={-5;8/3}`

    `d)` `6/(x^2-1)+5=(8x-1)/(4x+4)-(12x-4)/(4-4x)(ĐK:x\ne±1)`

    `↔6/((x-1)(x+1))+5=(8x-1)/(4(x+1))+(12x-4)/(4(x-1))`

    `↔(6.4+5.4(x^2-1))/(4(x-1)(x+1))=((8x-1)(x-1)+(12x-4)(x+1))/(4(x-1)(x+1))`

    `→24+20x^2-20=20x^2-x-3`

    `↔20x^2+4=20x^2-x-3`

    `↔20x^2-20x^2+x+4+3=0`

    `↔x+7=0`

    `↔x=-7`

    Vậy `S={-7}`

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