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T

tuvuthanhthuy

tra loi nek

$3a^2+3b^2$=10ab=> $a^2+b^2$=$\dfrac{10ab}{3}$
xét $M^2$=$( \dfrac{a-b}{a+b} )^2$
= $\dfrac{a^2-2ab+b^2}{a^2+2ab+b^2}$

=$\dfrac{\dfrac{10}{3}ab-2ab}{\dfrac{10}{3}ab+2ab}$

=$\dfrac{\dfrac{4}{3}}{\dfrac{16}{3}}$=$\dfrac{1}{4}$
Vì a>b>0\RightarrowM>0\RightarrowM=$\dfrac{1}{2}$
Chú ý gõ Latex
 
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M

motminhdidem

Cho a>0,b>0 thoả [TEX]3a^2+3b^2=10ab.[/TEX]
tính:
[TEX]M=\frac{a-b}{a+b}.[/TEX] ..hết...

Ta có:

$M=\frac{a-b}{a+b}$

\Leftrightarrow $M^2=\frac{(a-b)^2}{(a+b)^2}=\frac{3(a^2-2ab+b^2)}{3(a^2-2ab+b^2)}=\frac{3a^2+3b^2-6ab}{3a^2+3b^2+6ab}=\frac{3a^2+3b^2-10ab+4ab}{3a^2+3b^2-10ab+16ab}=\frac{4ab}{16ab}=\frac{1}{4}$

\Rightarrow $M=\frac{1}{2}$ ( vì $a,b>0$)
 
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