Đáp án:
Giải thích các bước giải:
`c)`
`ĐK : x ne +-1`
`<=> ((x-1)-(x+1))/(x^2-1)=(2x)((x-1)(x^2+x+1))`
`<=> (-2)/(x^2-1)=(2x)((x-1)(x^2+x+1))`
`<=> ((-2)(x^2+x+1))/((x+1)(x-1)(x^2+x+1))=(2x(x+1))/((x+1)(x-1)(x^2+x+1))`
`=> (-2)(x^2+x+1)=2x(x+1)`
`=> -2x^2-2x-2=2x^2+2x`
`<=> -4x^2-4x-2=0`
`<=> 4x^2+4x+2=0`
`<=> (2x+1)^2+1=0`
`<=> (2x+1)^2=-1`
`=>` Pt vô nghiệm
`d)`
`ĐK : x ne -2`
`<=> ((x^3+8)+(x^2-2x+4))/(x^3+8)=(12)/(x^3+8)`
`=> x^3+8+x^2-2x+4=12`
`<=> x^3+x^2-2x=0`
`<=> x^3-x^2+2x^2-2x=0`
`<=> x^2(x-1)+2x(x-1)=0`
`<=> x(x+2)(x-1)=0`
`<=> x=0` hoặc `x=-2` hoặc `x=1`
`=> S={-2;0;1}`
`e)`
`ĐK : x ne 0`
`<=> (x(x-1)+3(x+3))/(3x)=(6x)/(3x)`
`=> x(x-1)+3(x+3)=6x`
`<=> x^2-x+3x+9=6x`
`<=> x^2-4x+9=0`
`<=> (x-2)^2+5=0`
`<=> (x-2)^2=-5`
`=>` Pt vô nghiệm
`f)`
`<=> (x-5)(x+5)=(2x-1)(x+5)`
`<=> (x-5)(x+5)-(2x-1)(x+5)=0`
`<=> (x+5)(x-5-2x+1)=0`
`<=> (x+5)(-4-x)=0`
`<=> x=-5` hoặc `x=-4`
`=> S={-4;-5}`
`g)`
`ĐK : x ne -2;0`
`<=> (x(x-2)-(x^2+2))/(x(x+2))=(3(x+2))/(x(x+2))`
`=> x(x-2)-(x^2+2)=3(x+2)`
`<=> x^2-2x-x^2-2=3x+6`
`<=> -5x-8=0`
`<=> x=-8/5`
`=> S={-8/5}`
