$\frac{1}{x^2+3x+2}$+$\frac{1}{x^2+5x+6}$+$\frac{1}{x^2+7x+12}$=`1/3` giải chi tiết giùm 15/08/2021 Bởi Abigail $\frac{1}{x^2+3x+2}$+$\frac{1}{x^2+5x+6}$+$\frac{1}{x^2+7x+12}$=`1/3` giải chi tiết giùm
Đáp án: `S=\{\frac{3\sqrt5-5}{2};\frac{-3\sqrt5-5}{2}\}` Giải thích các bước giải: Ta có: `x^2+3x+2=x^2+x+2x+2=x(x+1)+2(x+1)=(x+1)(x+2)` `x^2+5x+6=x^2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)` `x^2+7x+12=x^2+3x+4x+12=x(x+3)+4(x+3)=(x+3)(x+4)` `\to ĐKXĐ: x\ne-1;x\ne-2;x\ne-3;x\ne-4` `1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)=1/3` `⇔1/((x+1)(x+2))+1/((x+2)(x+3))+1/((x+3)(x+4))=1/3` `⇔1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=1/3` `⇔1/(x+1)-1/(x+4)=1/3` `⇔(x+4-(x+1))/((x+1)(x+4))=1/3` `⇔(x+4-x-1)/((x+1)(x+4))=1/3` `⇔3/((x+1)(x+4))=1/3` `⇔9=(x+1)(x+4)` `⇔x^2+5x+4=9` `⇔x^2+5x-5=0` `⇔x^2+2.(5)/2x+25/4-45/4=0` `⇔(x+5/2)^2-45/4=0` `⇔(x+5/2)^2=45/4` `⇔(x+5/2)^2=((3\sqrt5)/2)^2` `⇔|x+5/2|=(3\sqrt5)/2` \(⇔\left[ \begin{array}{l}x=\dfrac{3\sqrt5-5}{2}(TM)\\x=\dfrac{-3\sqrt5-5}{2}(TM)\end{array} \right.\) Vậy `S=\{\frac{3\sqrt5-5}{2};\frac{-3\sqrt5-5}{2}\}` Bình luận
Đáp án + Giải thích các bước giải: `(1)/(x^{2}+3x+2)+(1)/(x^{2}+5x+6)+(1)/(x^{2}+7x+12)=(1)/(3)` `(ĐKXĐ:x\ne{-1;-2;-3;-4})` `<=>(1)/((x^{2}+x)+(2x+2))+(1)/((x^{2}+2x)+(3x+6))+(1)/((x^{2}+3x)+(4x+12))=(1)/(3)` `<=>(1)/(x(x+1)+2(x+1))+(1)/(x(x+2)+3(x+2))+(1)/(x(x+3)+4(x+3))=(1)/(3)` `<=>(1)/((x+1)(x+2))+(1)/((x+2)(x+3))+(1)/((x+3)(x+4))=(1)/(3)` `<=>(1)/(x+1)-(1)/(x+2)+(1)/(x+2)-(1)/(x+3)+(1)/(x+3)-(1)/(x+4)=(1)/(3)` `<=>(1)/(x+1)-(1)/(x+4)=(1)/(3)` `<=>(x+4-(x+1))/((x+1)(x+4))=(1)/(3)` `<=>(3)/((x+1)(x+4))=(1)/(3)` `<=>(x+1)(x+4)=9` `<=>x^{2}+x+4x+4=9` `<=>x^{2}+5x-5=0` `<=>(x^{2}+2.x.(5)/(2)+(25)/(4))-(45)/(4)=0` `<=>(x+(5)/(2))^{2}=(45)/(4)` `<=>x+(5)/(2)=±(3\sqrt{5})/(2)` `<=>x=(-5±3\sqrt{5})/(2)` Vậy phương trình có tập nghiệm là : `S={(-5±3\sqrt{5})/(2)}` Bình luận
Đáp án:
`S=\{\frac{3\sqrt5-5}{2};\frac{-3\sqrt5-5}{2}\}`
Giải thích các bước giải:
Ta có:
`x^2+3x+2=x^2+x+2x+2=x(x+1)+2(x+1)=(x+1)(x+2)`
`x^2+5x+6=x^2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)`
`x^2+7x+12=x^2+3x+4x+12=x(x+3)+4(x+3)=(x+3)(x+4)`
`\to ĐKXĐ: x\ne-1;x\ne-2;x\ne-3;x\ne-4`
`1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)=1/3`
`⇔1/((x+1)(x+2))+1/((x+2)(x+3))+1/((x+3)(x+4))=1/3`
`⇔1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=1/3`
`⇔1/(x+1)-1/(x+4)=1/3`
`⇔(x+4-(x+1))/((x+1)(x+4))=1/3`
`⇔(x+4-x-1)/((x+1)(x+4))=1/3`
`⇔3/((x+1)(x+4))=1/3`
`⇔9=(x+1)(x+4)`
`⇔x^2+5x+4=9`
`⇔x^2+5x-5=0`
`⇔x^2+2.(5)/2x+25/4-45/4=0`
`⇔(x+5/2)^2-45/4=0`
`⇔(x+5/2)^2=45/4`
`⇔(x+5/2)^2=((3\sqrt5)/2)^2`
`⇔|x+5/2|=(3\sqrt5)/2`
\(⇔\left[ \begin{array}{l}x=\dfrac{3\sqrt5-5}{2}(TM)\\x=\dfrac{-3\sqrt5-5}{2}(TM)\end{array} \right.\)
Vậy `S=\{\frac{3\sqrt5-5}{2};\frac{-3\sqrt5-5}{2}\}`
Đáp án + Giải thích các bước giải:
`(1)/(x^{2}+3x+2)+(1)/(x^{2}+5x+6)+(1)/(x^{2}+7x+12)=(1)/(3)` `(ĐKXĐ:x\ne{-1;-2;-3;-4})`
`<=>(1)/((x^{2}+x)+(2x+2))+(1)/((x^{2}+2x)+(3x+6))+(1)/((x^{2}+3x)+(4x+12))=(1)/(3)`
`<=>(1)/(x(x+1)+2(x+1))+(1)/(x(x+2)+3(x+2))+(1)/(x(x+3)+4(x+3))=(1)/(3)`
`<=>(1)/((x+1)(x+2))+(1)/((x+2)(x+3))+(1)/((x+3)(x+4))=(1)/(3)`
`<=>(1)/(x+1)-(1)/(x+2)+(1)/(x+2)-(1)/(x+3)+(1)/(x+3)-(1)/(x+4)=(1)/(3)`
`<=>(1)/(x+1)-(1)/(x+4)=(1)/(3)`
`<=>(x+4-(x+1))/((x+1)(x+4))=(1)/(3)`
`<=>(3)/((x+1)(x+4))=(1)/(3)`
`<=>(x+1)(x+4)=9`
`<=>x^{2}+x+4x+4=9`
`<=>x^{2}+5x-5=0`
`<=>(x^{2}+2.x.(5)/(2)+(25)/(4))-(45)/(4)=0`
`<=>(x+(5)/(2))^{2}=(45)/(4)`
`<=>x+(5)/(2)=±(3\sqrt{5})/(2)`
`<=>x=(-5±3\sqrt{5})/(2)`
Vậy phương trình có tập nghiệm là : `S={(-5±3\sqrt{5})/(2)}`