Giải hệ phương trình mũ

L

levanvu12a1

Đặt a=2x,b=3x,c=4x a=2^x,b=3^x,c=4^x a,b,c>0
Pt trở thành ab+c+ba+c+ca+b=32\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{3}{2}
Đặt VT=P giờ ta c/m P\geq 32\frac{3}{2}
\Rightarrow P+3=a+b+cb+c+a+b+cb+c+a+b+ca+c P+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}
\Leftrightarrow 2P+6=((a+b)+(b+c)+(a+c))(1a+b+1b+c+1a+c) 2P+6 = ((a+b)+(b+c)+(a+c))(\frac{1}{a+b}+\frac{1}{b+c}+ \frac{1}{a+c}) \geq 9
\Rightarrow P \geq 32 \frac{3}{2}
Xảy ra dấu"=" \Leftrightarrow a=b=c \Leftrightarrow x=0
 
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N

nguyentanlap1994

Đặt a=2x,b=3x,c=4x a=2^x,b=3^x,c=4^x a,b,c>0
Pt trở thành ab+c+ba+c+ca+b=32\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{3}{2}
Đặt VT=P giờ ta c/m P\geq 32\frac{3}{2}
\Rightarrow P+3=a+b+cb+c+a+b+cb+c+a+b+ca+c P+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}
\Leftrightarrow 2P+6=((a+b)+(b+c)+(a+c))(1a+b+1b+c+1a+c) 2P+6 = ((a+b)+(b+c)+(a+c))(\frac{1}{a+b}+\frac{1}{b+c}+ \frac{1}{a+c}) \geq 9
\Rightarrow P \geq 32 \frac{3}{2}
Xảy ra dấu"=" \Leftrightarrow a=b=c \Leftrightarrow x=0

Cám ơn bạn nhiều còn bài 1 thì sao bạn ?????????????
 
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