Toán `36x^2-10x+1/(18x)+2015` Tìm min : làm chơi thôi hơi dễ á 20/08/2021 By Nevaeh `36x^2-10x+1/(18x)+2015` Tìm min : làm chơi thôi hơi dễ á
Đáp án: Giải thích các bước giải: x=-13595*4^(1/3)/(3^(11.1(6))*(1.098788580951163*10^7-204017*căn bậc hai(3))^(1/3))+(1.098788580951163*10^7-204017*căn bậc hai(3))^(1/3)/(3^(19/6)*4^(1/3))+0.0(925); x = -((27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(2/3)*(căn bậc hai(3)*i+1)+13595*2^(4/3)*3^(5/6)*i-5*2^(2/3)*3^(1/6)*(27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(1/3)-13595*2^(4/3)*3^(1/3))/(2^(5/3)*3^(19/6)*(27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(1/3));x = ((27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(2/3)*(căn bậc hai(3)*i-1)+13595*2^(4/3)*3^(5/6)*i+5*2^(2/3)*3^(1/6)*(27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(1/3)+13595*2^(4/3)*3^(1/3))/(2^(5/3)*3^(19/6)*(27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(1/3)); Trả lời
Đáp án: Ta có : 36$x^{2}$ – 10$x^{}$ + $\frac{1}{18x^{}}$ + 2015=> ( 2$x^{}$ + $\frac{1}{18x^{}}$ ) + ( 36$x^{2}$ – 12$x^{}$ + 1 ) + 2015 – 1=> ( 2$x^{}$ + $\frac{1}{18x^{}}$ ) + $( x – \frac{1}{6^{}} )^{2}$ + 2014 Áp dụng bất đẳng thức Cauchy ( Cô si ) cho 2 số 2$x^{}$ , $\frac{1}{18x^{}}$ > 0 : 2$x^{}$ + $\frac{1}{18x^{}}$ $\geq$ 2 .$\sqrt[]{2x^{} .\frac{1}{18x^{}}}$ => 18$x^{}$ + $\frac{1}{18x^{}}$ $\geq$ $\frac{2}{3^{}}$Mặt khác : $( x – \frac{1}{6^{}} )^{2}$ $\geq$ 0=> ( 2$x^{}$ + $\frac{1}{18x^{}}$ ) + $( x – \frac{1}{6^{}} )^{2}$ + 2014 $\geq$ 2014 + $\frac{2}{3^{}}$=> ( 2$x^{}$ + $\frac{1}{18x^{}}$ ) + $( x – \frac{1}{6^{}} )^{2}$ + 2014 $\geq$ $\frac{6044}{3^{}}$=> Min = $\frac{6044}{3^{}}$Dấu ” = ” xảy ra ⇔ $\left \{ {{( x – \frac{1}{6^{}} )^{2}= 0} \atop {2x^{} = \frac{1}{18x^{}}}} \right.$ ⇔ $\left \{ {{ x – \frac{1}{6^{}} ^{}= 0} \atop {36x^{2} = 1}} \right.$ ⇔ $\left \{ {{ x = \frac{1}{6^{}} ^{}} \atop {x^{} = ± \frac{1}{6} }} \right.$ ⇔ $x^{}$ = $\frac{1}{6}$ Vậy Min = $\frac{6044}{3^{}}$ ⇔ $x^{}$ = $\frac{1}{6}$ Trả lời
Đáp án:
Giải thích các bước giải:
x=-13595*4^(1/3)/(3^(11.1(6))*(1.098788580951163*10^7-204017*căn bậc hai(3))^(1/3))+(1.098788580951163*10^7-204017*căn bậc hai(3))^(1/3)/(3^(19/6)*4^(1/3))+0.0(925); x = -((27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(2/3)*(căn bậc hai(3)*i+1)+13595*2^(4/3)*3^(5/6)*i-5*2^(2/3)*3^(1/6)*(27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(1/3)-13595*2^(4/3)*3^(1/3))/(2^(5/3)*3^(19/6)*(27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(1/3));x = ((27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(2/3)*(căn bậc hai(3)*i-1)+13595*2^(4/3)*3^(5/6)*i+5*2^(2/3)*3^(1/6)*(27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(1/3)+13595*2^(4/3)*3^(1/3))/(2^(5/3)*3^(19/6)*(27*căn bậc hai(1511)*căn bậc hai(109606493)-204017*căn bậc hai(3))^(1/3));
Đáp án:
Ta có :
36$x^{2}$ – 10$x^{}$ + $\frac{1}{18x^{}}$ + 2015
=> ( 2$x^{}$ + $\frac{1}{18x^{}}$ ) + ( 36$x^{2}$ – 12$x^{}$ + 1 ) + 2015 – 1
=> ( 2$x^{}$ + $\frac{1}{18x^{}}$ ) + $( x – \frac{1}{6^{}} )^{2}$ + 2014
Áp dụng bất đẳng thức Cauchy ( Cô si ) cho 2 số 2$x^{}$ , $\frac{1}{18x^{}}$ > 0 :
2$x^{}$ + $\frac{1}{18x^{}}$ $\geq$ 2 .$\sqrt[]{2x^{} .\frac{1}{18x^{}}}$
=> 18$x^{}$ + $\frac{1}{18x^{}}$ $\geq$ $\frac{2}{3^{}}$
Mặt khác : $( x – \frac{1}{6^{}} )^{2}$ $\geq$ 0
=> ( 2$x^{}$ + $\frac{1}{18x^{}}$ ) + $( x – \frac{1}{6^{}} )^{2}$ + 2014 $\geq$ 2014 + $\frac{2}{3^{}}$
=> ( 2$x^{}$ + $\frac{1}{18x^{}}$ ) + $( x – \frac{1}{6^{}} )^{2}$ + 2014 $\geq$ $\frac{6044}{3^{}}$
=> Min = $\frac{6044}{3^{}}$
Dấu ” = ” xảy ra ⇔ $\left \{ {{( x – \frac{1}{6^{}} )^{2}= 0} \atop {2x^{} = \frac{1}{18x^{}}}} \right.$
⇔ $\left \{ {{ x – \frac{1}{6^{}} ^{}= 0} \atop {36x^{2} = 1}} \right.$
⇔ $\left \{ {{ x = \frac{1}{6^{}} ^{}} \atop {x^{} = ± \frac{1}{6} }}
\right.$
⇔ $x^{}$ = $\frac{1}{6}$
Vậy Min = $\frac{6044}{3^{}}$ ⇔ $x^{}$ = $\frac{1}{6}$