Giải pt $\frac{1}{x(x+1)}$ +$\frac{1}{(x+1)(x+2)}$ +$\frac{1}{(x+2)(x+3)}$ =$\frac{3}{10}$ 20/09/2021 Bởi Bella Giải pt $\frac{1}{x(x+1)}$ +$\frac{1}{(x+1)(x+2)}$ +$\frac{1}{(x+2)(x+3)}$ =$\frac{3}{10}$
Đáp án: Giải thích các bước giải: ta có: $\frac{1}{x(x+1)}$+$\frac{1}{(x+1)(x+2)}$+$\frac{1}{(x+2)(x+3)}$=$\frac{3}{10}$ ⇒$\frac{(x+2)(x+3)}{x(x+1)(x+3)(x+2)}$+$\frac{(x+3)x}{x(x+3)(x+1)(x+2)}$+$\frac{x(x+1)}{(x(x+1)(x+2)(x+3)}$=$\frac{3}{10}$ ⇒$\frac{(x+2)(x+3)+x(x+3)+(x(x+1)}{x(x+1)(x+2)(x+3)}$=$\frac{3}{10}$ ⇒$\frac{x^{2}+5x+6+x^2+3x+x^2+x}{(x+1)(x+2)(x+3)x}$=$\frac{3}{10}$ ⇒$\frac{3x^2+9x+6}{x(x+1)(x+2)(x+3)}$=$\frac{3}{10}$ ⇒30$x^{2}$+90x+60=3x(x+1)(x+2)(x+3) ⇒30$x^{2}$+90x+60=3$x^{4}$+18$x^{3}$+33$x^{2}$ +18x ⇒3$x^{4}$+18$x^{3}$+33$x^{2}$ +18x-30$x^{2}$-90x-60=0 ⇒3$x^{4}$+18$x^{3}$+3$x^{2}$-72x-60=0 ⇒$x^{4}$+6$x^{3}$+$x^{2}$-24x-20=0 ⇒(x-2)(x+2)(x+5)(x+1)=0 ⇒x=2;=-2;=-5;=-1 mà đkxđ là x khác 0;-1;-2;-3 nên’ x=2;-5 Bình luận
`((x+2)(x+3))/(x(x+1)(x+2)(x+3))“+(10x(x+3))/(x(x+1)(x+2)(x+3))“+(10x(x+1))/(x(x+1)(x+2)(x+3))`=`(3x(x+1)(x+2)(x+3))/(x(x+1)(x+2)(x+3))` `=>10(x+2)(x+3)+10x(x+3)+10x(x+1)=3x(x+1)(x+2)(x+3)` `=>10x^2+30x+20x+60+10x^2+30x+10x^2+10x=3x^4+18x^3+33x^2+18x` `=>-3x^4-18x^3-3x^2+72x+60=0` `=>-3x^4-15x^3-3x^3-15x^2+12x^2+60=0` `=>-2x^3(x+5)-3x^2(x+5)+12x(x+5)=0` `=>(-2x^3-3x^2+12x)(x+5)=0` `……` `=>(x-2)(x+5)=0` `=>`\(\left[ \begin{array}{l}x+2=0\\x-5=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=-2\\x=5\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=-2\\x=5\end{array} \right.\) Bình luận
Đáp án:
Giải thích các bước giải:
ta có:
$\frac{1}{x(x+1)}$+$\frac{1}{(x+1)(x+2)}$+$\frac{1}{(x+2)(x+3)}$=$\frac{3}{10}$
⇒$\frac{(x+2)(x+3)}{x(x+1)(x+3)(x+2)}$+$\frac{(x+3)x}{x(x+3)(x+1)(x+2)}$+$\frac{x(x+1)}{(x(x+1)(x+2)(x+3)}$=$\frac{3}{10}$
⇒$\frac{(x+2)(x+3)+x(x+3)+(x(x+1)}{x(x+1)(x+2)(x+3)}$=$\frac{3}{10}$
⇒$\frac{x^{2}+5x+6+x^2+3x+x^2+x}{(x+1)(x+2)(x+3)x}$=$\frac{3}{10}$
⇒$\frac{3x^2+9x+6}{x(x+1)(x+2)(x+3)}$=$\frac{3}{10}$
⇒30$x^{2}$+90x+60=3x(x+1)(x+2)(x+3)
⇒30$x^{2}$+90x+60=3$x^{4}$+18$x^{3}$+33$x^{2}$ +18x
⇒3$x^{4}$+18$x^{3}$+33$x^{2}$ +18x-30$x^{2}$-90x-60=0
⇒3$x^{4}$+18$x^{3}$+3$x^{2}$-72x-60=0
⇒$x^{4}$+6$x^{3}$+$x^{2}$-24x-20=0
⇒(x-2)(x+2)(x+5)(x+1)=0
⇒x=2;=-2;=-5;=-1
mà đkxđ là x khác 0;-1;-2;-3
nên’
x=2;-5
`((x+2)(x+3))/(x(x+1)(x+2)(x+3))“+(10x(x+3))/(x(x+1)(x+2)(x+3))“+(10x(x+1))/(x(x+1)(x+2)(x+3))`=`(3x(x+1)(x+2)(x+3))/(x(x+1)(x+2)(x+3))`
`=>10(x+2)(x+3)+10x(x+3)+10x(x+1)=3x(x+1)(x+2)(x+3)`
`=>10x^2+30x+20x+60+10x^2+30x+10x^2+10x=3x^4+18x^3+33x^2+18x`
`=>-3x^4-18x^3-3x^2+72x+60=0`
`=>-3x^4-15x^3-3x^3-15x^2+12x^2+60=0`
`=>-2x^3(x+5)-3x^2(x+5)+12x(x+5)=0`
`=>(-2x^3-3x^2+12x)(x+5)=0`
`……`
`=>(x-2)(x+5)=0`
`=>`\(\left[ \begin{array}{l}x+2=0\\x-5=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=-2\\x=5\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=-2\\x=5\end{array} \right.\)