`f)`
`(x+1/x)^2+2(x+1/x)-8=0`
ĐK : `x\ne0`
Đặt `x+1/x =t (t\ne0)`
`<=>t^2+2t-8=0`
`<=>t^2-2t+4t-8=0`
`<=>t(t-2)+4(t-2)=0`
`<=>(t+4).(t-2)=0`
`<=>`\(\left[ \begin{array}{l}t+4=0\\t-2=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}t=-4\\t=2\end{array}(tm) \right.\)
`+)x+1/x=-4`
`<=>x+1/x+4=0`
`<=>x^2+1+4x=0`
`<=>x^2+2.2x+2^2-3=0`
`<=>(x+2)^2=3=(+-\sqrt{3})^2`
`<=>`\(\left[ \begin{array}{l}x+2=\sqrt{3}\\x+2=-\sqrt{3}\end{array} \right.\)`<=>`\(\left[ \begin{array}{l}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{array} (tm)\right.\)
`+)x+1/x=2`
`<=>x+1/x-2=0`
`<=>x^2+1-2x=0`
`<=>(x-1)^2=0`
`<=>x-1=0`
`<=>x=1(tm)`
Vậy `S={1;-2-\sqrt{3};-2+\sqrt{3}}`
Bài `3`:
`a)`
`1/(x+1)-5/(x-2)=15/((x+1).(2-x))`
Đk : `x\ne-1;x\ne2`
`<=>(x-2)/((x+1).(x-2))-(5.(x+1))/((x+1).(x-2))=15/((x+1).(x-2))`
`<=>x-2-5(x+1)=15`
`<=>x-2-5x-5=15`
`<=>-4x=22`
`<=>x=-11/2(tmđk)`
Vậy `S={-11/2}`
`b)`
`(x-1)/(x+2)-(x)/(x-2)=(5x-2)/(4-x^2)`
ĐK :`x\ne+-2`
`<=>(x-1)/(x+2)-(x)/(x-2)=(2-5x)/(x^2-4)`
`<=>((x-1).(x-2))/((x+2).(x-2))-(x.(x+2))/((x-2).(x+2))=(2-5x)/((x-2).(x+2))`
`<=>(x-1).(x-2)-x.(x+2)=2-5x`
`<=>x^2-2x-x+2-x^2-2x=2-5x`
`<=>-5x+2=2-5x`
`<=>0x=0(lđ)`
Vậy `S={x|x\ne+-2}`
`c)`
`(x+5)/(x^2-5x)-(x-5)/(2x^2+10x)=(x+25)/(2x^2-50)`
ĐK: `x\ne0;x\ne+-5`
`<=>(x+5)/(x.(x-5))-(x-5)/(2x.(x+5))=(x+25)/(2.(x+5).(x-5))`
`<=>(2.(x+5)^2)/(2x.(x-5).(x+5))-((x-5)^2)/(2x.(x-5).(x+5))=(x.(x+25))/(2x.(x+5).(x-5))`
`<=>2.(x+5)^2-(x-5)^2=x(x+25)`
`<=>2(x^2+10x+25)-(x^2-10x+25)=x(x+25)`
`<=>2x^2+20x+50-x^2+10x-25=x^2+25x`
`<=>5x=-25`
`<=>x=-5(ktm)`
Vậy `S=∅`
`d)`
`(1)/(x-1)-(3x^2)/(x^3-1)=(2x)/(x^2+x+1)`
ĐK `x\ne1`
`<=>(x^2+x+1)/((x-1).(x^2+x+1))-(3x^2)/((x^2+x+1).(x-1))=(2x.(x-1))/((x^2+x+1).(x-1))`
`<=>x^2+x+1-3x^2=2x(x-1)`
`<=>x^2+x+1-3x^2=2x^2-2x`
`<=>-2x^2+x+1=2x^2-2x`
`<=>4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x.(x-1)+(x-1)=0`
`<=>(4x+1).(x-1)=0`
`<=>`\(\left[ \begin{array}{l}4x+1=0\\x-1=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=\dfrac{-1}{4}(tmđk)\\x=1(ktm)\end{array} \right.\)
Vậy `S={-1/4}`
`e)`
`(7)/(8x)+(5-x)/(4x^2-8x)=(x-1)/(2x(x-2))+1/(8x-16)`
Đk: `x\ne0;x\ne2`
`<=>(7)/(8x)+(5-x)/(4x.(x-2))=(x-1)/(2x(x-2))+1/(8.(x-2))`
`<=>(7.(x-2))/(8x.(x-2))+(2.(5-x))/(8x(x-2))=(4(x-1))/(8x(x-2))+x/(8x(x-2))`
`<=>7(x-2)+2(5-x)=4(x-1)+x`
`<=>7x-14+10-2x=4x-4+x`
`<=>5x-4=5x-4`
`<=>0x=0(lđ)`
Vậy `S={x|x\ne0;x\ne2}`
`f)`
`(2)/(x^2+3x+2)+(1)/(x^2+5x+6)=(1)/(x^2+4x+3)`
ĐK: `x\ne-1;x\ne-2;x\ne-3`
`<=>(2)/(x^2+2x+x+2)+(1)/(x^2+3x+2x+6)=(1)/(x^2+3x+x+3)`
`<=>(2)/(x.(x+2)+1(x+2))+(1)/(x.(x+3)+2(x+3))=(1)/(x(x+3)+1(x+3))`
`<=>(2)/((x+1).(x+2))+1/((x+2).(x+3))=1/((x+1).(x+3))`
`<=>(2(x+3))/((x+1).(x+2).(x+3))+(x+1)/((x+1).(x+2).(x+3))=(x+2)/((x+1).(x+2).(x+3))`
`<=>2(x+3)+x+1=x+2`
`<=>2x+6+x+1=x+2`
`<=>3x+7=x+2`
`<=>2x=-5`
`<=>x=-5/2(tmđk)`
Vậy `S={-5/2}`
