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bđt \(\Leftrightarrow\dfrac{\sqrt{bc}}{\sqrt{a}}+\dfrac{\sqrt{ca}}{\sqrt{b}}+\dfrac{\sqrt{ab}}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\)
Ta có: \(\left(\dfrac{\sqrt{bc}}{\sqrt{a}}+\dfrac{\sqrt{ab}}{\sqrt{c}}\right)+\left(\dfrac{\sqrt{ca}}{\sqrt{b}}+\dfrac{\sqrt{bc}}{\sqrt{a}}\right)+\left(\dfrac{\sqrt{ab}}{\sqrt{c}}+\dfrac{\sqrt{ca}}{\sqrt{b}}\right)\ge2\sqrt{b}+2\sqrt{c}+2\sqrt{a}\)
\(\Leftrightarrow2\left(\dfrac{\sqrt{bc}}{\sqrt{a}}+\dfrac{\sqrt{ca}}{\sqrt{b}}+\dfrac{\sqrt{ab}}{\sqrt{c}}\right)\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\Leftrightarrow\dfrac{\sqrt{bc}}{\sqrt{a}}+\dfrac{\sqrt{ca}}{\sqrt{b}}+\dfrac{\sqrt{ab}}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\left(đpcm\right)\)
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
\(A=\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a}+3}\right)=\dfrac{7\sqrt{a}}{a-9}-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{7\sqrt{a}}{a-9}-\dfrac{a+3\sqrt{a}-a+3\sqrt{a}+\sqrt{a}-3}{a-9}=\dfrac{3}{a-9}\)\(B=\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\right)=\dfrac{\sqrt{a}-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-3\right)}:\dfrac{a-9-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{3}{\sqrt{a}\left(\sqrt{a}-3\right)}.\dfrac{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}{-5}=\dfrac{3\sqrt{a}-6}{-5\sqrt{a}}\)
\(C=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\sqrt{a}-a}\right).\left(\dfrac{1}{a}-2\right)=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\left(\sqrt{a}-1\right)}\right).\dfrac{1-2a}{a}=\dfrac{a\sqrt{a}-a}{\sqrt{a}-1}.\dfrac{1-2a}{a}=\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}-1}.\dfrac{1-2a}{a}=1-2a\)\(D=\dfrac{a\sqrt{a}+1}{a-1}-\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a\sqrt{a}+1-\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1}=\dfrac{a\sqrt{a}+1-a\sqrt{a}+a+\sqrt{a}-1}{a-1}=\dfrac{a+\sqrt{a}}{a-1}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-1}\)
1. Đặt $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=T$
$\frac{a}{b+c}> \frac{a}{a+b+c}$
$\frac{b}{c+a}> \frac{b}{c+a+b}$
$\frac{c}{a+b}> \frac{c}{a+b+c}$
$\Rightarrow T> \frac{a+b+c}{a+b+c}=1$ (đpcm)
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Xét hiệu:
$\frac{a}{b+c}-\frac{2a}{a+b+c}=\frac{-a(b+c-a)}{(b+c)(a+b+c)}<0$ theo BĐT tam giác
$\Rightarrow \frac{a}{b+c}< \frac{2a}{a+b+c}$
Tương tư: $\frac{b}{c+a}< \frac{2b}{c+a+b}$
$\frac{c}{a+b}< \frac{2c}{a+b+c}$
Cộng theo vế:
$T< \frac{2(a+b+c)}{a+b+c}=2$
$\frac{b}{a+c}
2.
Áp dụng BĐT AM-GM:
\(\frac{b+c}{a}.1\leq \frac{1}{4}(\frac{b+c}{a}+1)^2=\frac{(b+c+a)^2}{4a^2}\)
\(\Rightarrow \sqrt{\frac{a}{b+c}}\geq \frac{2a}{a+b+c}\)
Tương tự với các phân thức còn lại và cộng theo vế:
$\Rightarrow T\geq \frac{2(a+b+c)}{a+b+c}=2$
Dấu "=" xảy ra khi $b+c=a; c+a=b; a+b=c\Rightarrow a=b=c=0$ (vô lý)
Vậy dấu "=" không xảy ra, tức là $T>2>1$ (đpcm)
từ giả thiết ,ta có:\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=4\)\(\Leftrightarrow a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)=4\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=1\)---> thay 1= vào ...