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

## Answers ( )

1. Đáp án:

Giải thích các bước giải:

2. 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}