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=$\frac{8x-1}{4x+4}$-$\frac{12x-4}{4-4x}$
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}
By Genesis
Đáp án:
Giải thích các bước giải:
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}`