Giải các pt sau :
`a). {5x+1}/5-{2x-1}/{2x+2}=2+{x^2+4x+1}/{x+1}`
`b). 1/{x+2}+1/{x^2-2x}=8/{x^3-4x}`
`c). {x+5}/{x^2-5x}+{5-x}/{2x^2+10x}={x-5}/{2x^2-50}`
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a){5x+1}/{5}-{2x-1}/{2x+2}=2+{x^2+4x+1}/{x+1} Điều kiện : x\ne-1<=>{5x+1}/{5}-{2x-1}/{2(x+1)}={2(x+1)+x^2+4x+1}/{x+1}<=>{2(x+1)(5x+1)}/{10(x+1)}-{5(2x-1)}/{10(x+1)}={10(x^2+6x+3)}/{10(x+1)}=>2(x+1)(5x+1)-5(2x-1)=10(x^2+6x+3)<=>10x^2+12x+2-10x+5=10x^2+60x+30<=>10x^2+2x+7-(10x^2+60x+30)=0<=>-58x=23<=>x=-23/58 (thỏa)Vậy S={-23/58}b){1}/{x+2}+{1}/{x^2-2x}={8}/{x^3-4x} Điều kiện : x\ne0;x\ne+-2<=>{1}/{x+2}+{1}/{x(x-2)}={8}/{x(x+2)(x-2)}<=>{x(x-2)}/{x(x-2)(x+2)}+{x+2}/{x(x-2)(x+2)}={8}/{x(x+2)(x-2)}<=>x(x-2)+x+2=8<=>x^2-x+2-8=0<=>x^2-x-6=0<=>x^2+2x-3x-6=0<=>x(x+2)-3(x+2)=0<=>(x-3)(x+2)=0<=>\(\left[ \begin{array}{l}x-3=0\\x+2=0\end{array} \right.\)<=>\(\left[ \begin{array}{l}x=3(tm)\\x=-2(ktm)\end{array} \right.\)Vậy S={3}c){x+5}/{x^2-5x}+{5-x}/{2x^2+10x}={x-5}/{2x^2-50} Điều kiện : x\ne0;x\ne+-5<=>{x+5}/{x(x-5)}+{5-x}/{2x(x+5)}={x-5}/{2(x-5)(x+5)}<=>{2(x+5)^2}/{2x(x-5)(x+5)}-{(x-5)^2}/{2x(x+5)(x-5)}={x(x-5)}/{2x(x-5)(x+5)}=>2(x+5)^2-(x-5)^2=x(x-5)<=>2x^2+20x+50-x^2+10x-25=x^2-5x<=>x^2+30x+25-(x^2-5x)=0<=>35x=-25<=>x=-5/7 (thỏa )Vậy S={-5/7}
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a)(5x+1)/5 – (2x-1)/(2x+2) = 2 + (x^2 + 4x + 1)/(x+1) ĐKXĐ:x≠-1<=> (5x+1)/5 – (2x-1)/[2(x+1)] = 2 + (x^2 + 4x + 1)/(x+1)<=> 2(5x+1)(x+1) – 5(2x-1) = 20(x+1) + 10(x^2 + 4x + 1)<=> 2(5x^2 + 5x + x + 1) – 10x + 5 = 20x + 20 + 10x^2 + 40x + 10<=> 10x^2 + 12x + 2 – 10x + 5 – 20x – 20 – 10x^2 – 40x – 10 =0<=> -58x -23 = 0<=> -58x = 23<=> x=-23/58 ™Vậy S={-23/58}b)1/(x+2) + 1/(x^2 – 2x) = 8/(x^3 – 4x) ĐKXĐ:x≠0;±2<=> 1/(x+2) + 1/[x(x-2)] = 8/[x(x-2)(x+2)]<=> x(x-2) + x+2 = 8<=> x^2 – 2x + x + 2 – 8=0<=> x^2 – x – 6 =0<=> x^2 – 3x + 2x – 6=0<=> x(x-3) + 2(x-3) =0<=> (x-3)(x+2)=0<=>\(\left[ \begin{array}{l}x-3=0\\x+2=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=3(tm)\\x=-2(ktm) \end{array} \right.\)Vậy S={3}c)(x+5)/(x^2 – 5x) + (5-x)/(2x^2 + 10x) = (x-5)/(2x^2 + 5x) ĐKXĐ:x≠±5<=> (x+5)/[x(x-5)] + (5-x)/[2x(x+5)] = (x-5)/[2(x-5)(x+5)]<=> 2(x+5)^2 – (x-5)^2 = x(x-5)<=> 2(x^2 + 10x + 25) – x^2 + 10x = x^2 – 5x<=> 2x^2 + 20x + 50 – x^2 + 10x – 25 – x^2 + 5x =0<=> 35x + 25 =0<=> 35x = -25<=> x = -5/7 ™Vậy S={-5/7}$\color{lightblue}{eriet}$