x/2^2x/2^3x/2^4=x/3^2+x/3^3+x/3^4 Tính x PLEASE HELP ME 13/11/2021 Bởi Camila x/2^2x/2^3x/2^4=x/3^2+x/3^3+x/3^4 Tính x PLEASE HELP ME
`\qquad x/{2^2}+x/{2^3}+x/{2^4}=x/{3^2}+x/{3^3}+x/{3^4}` `<=>x/4+x/8+x/{16}=x/9+x/{27}+x/{81}` `<=>(x/4+x/8+x/{16}-x/9-x/{27}-x/{81})=0` `<=>x(1/ 4 + 1/ 8 + 1/{16}- 1/9 – 1/{27}- 1/{81})=0` $(1)$ Vì `1/ 4>1/9; 1/ 8>1/{27};1/{16}>1/{81}` `=>1/ 4-1/9>0; 1/ 8-1/{27}>0;1/{16}-1/{81}>0` `=>1/ 4 + 1/ 8 + 1/{16}- 1/9 – 1/{27}- 1/{81}=(1/ 4-1/9)+(1/ 8-1/{27})+(1/16-1/81)>0` Từ `(1)=>x=0` Vậy `x=0` Bình luận
`x/2^2*x/2^3*x/2^4=x/3^2+x/3^3+x/3^4` `<=>x^3/2^9=x/9+x/27+x/81` `<=>x^3/2^9=(9x)/81+(3x)/81+x/81` `<=>x^3/512=(13x)/81` `=>81x^3=13x*512` `<=>81x^3-13x*512=0` `<=>x(81x^2-6656)=0` `<=>` \(\left[ \begin{array}{l}81x^2=6656\\x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}81x^2=6656\\x=0\end{array} \right.\) ` Vậy `S={ 0; +-\frac{16\sqrt[26]}{9}}` Bình luận
`\qquad x/{2^2}+x/{2^3}+x/{2^4}=x/{3^2}+x/{3^3}+x/{3^4}`
`<=>x/4+x/8+x/{16}=x/9+x/{27}+x/{81}`
`<=>(x/4+x/8+x/{16}-x/9-x/{27}-x/{81})=0`
`<=>x(1/ 4 + 1/ 8 + 1/{16}- 1/9 – 1/{27}- 1/{81})=0` $(1)$
Vì `1/ 4>1/9; 1/ 8>1/{27};1/{16}>1/{81}`
`=>1/ 4-1/9>0; 1/ 8-1/{27}>0;1/{16}-1/{81}>0`
`=>1/ 4 + 1/ 8 + 1/{16}- 1/9 – 1/{27}- 1/{81}=(1/ 4-1/9)+(1/ 8-1/{27})+(1/16-1/81)>0`
Từ `(1)=>x=0`
Vậy `x=0`
`x/2^2*x/2^3*x/2^4=x/3^2+x/3^3+x/3^4`
`<=>x^3/2^9=x/9+x/27+x/81`
`<=>x^3/2^9=(9x)/81+(3x)/81+x/81`
`<=>x^3/512=(13x)/81`
`=>81x^3=13x*512`
`<=>81x^3-13x*512=0`
`<=>x(81x^2-6656)=0`
`<=>` \(\left[ \begin{array}{l}81x^2=6656\\x=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}81x^2=6656\\x=0\end{array} \right.\) `
Vậy `S={ 0; +-\frac{16\sqrt[26]}{9}}`