Cho a,b.c>0 và \(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\).Tính M=\(\dfrac{1+a}{\sqrt{a}+\sqrt{b}}+\dfrac{1+b}{\sqrt{b}+\sqrt{c}}+\dfrac{1+c}{\sqrt{c}+\sqrt{a}}\)
Những câu hỏi liên quan
Cho 3 số thực dương a,b.c thỏa mãn abc=1 cmr:\(\dfrac{b+c}{\sqrt{a}}+\dfrac{c+a}{\sqrt{b}}+\dfrac{a+b}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Cho a,b,c>0;abc=4
Tính M=\(\sqrt{\dfrac{\sqrt{a}}{\sqrt{ab}+\sqrt{a}+2}}+\sqrt{\dfrac{\sqrt{b}}{\sqrt{bc}+\sqrt{b}+1}}+\sqrt{\dfrac{\sqrt{a}}{\sqrt{ac}+\sqrt{c}+1}}\)
Cho a,b,c>0 và a+b+c=căn a +căn b +căn c=2.Tính A=
\(\left(\dfrac{\sqrt{a}}{1+a}+\dfrac{\sqrt{b}}{1+b}+\dfrac{\sqrt{c}}{1+c}\right)\left(\sqrt{1+a}\right)\left(\sqrt{1+b}\right)\left(\sqrt{1+c}\right)\)
Lời giải:
\(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\)
\(\Rightarrow (\sqrt{a}+\sqrt{b}+\sqrt{c})^2=4\)
\(\Leftrightarrow a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=4\)
\(\Leftrightarrow \sqrt{ab}+\sqrt{bc}+\sqrt{ac}=\frac{4-(a+b+c)}{2}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})\)
Tương tự:
$b+1=(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})$
$c+1=(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})$
Khi đó:
\(A=\left[\frac{\sqrt{a}}{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})}+\frac{\sqrt{b}}{(\sqrt{b}+\sqrt{a})(\sqrt{b}+\sqrt{c})}+\frac{\sqrt{c}}{(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})}\right]\sqrt{(a+1)(b+1)(c+1)}\)
\(\frac{\sqrt{a}(\sqrt{b}+\sqrt{c})+\sqrt{b}(\sqrt{c}+\sqrt{a})+\sqrt{c}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.\sqrt{(\sqrt{a}+\sqrt{b})^2(\sqrt{b}+\sqrt{c})^2(\sqrt{c}+\sqrt{a})^2}\)
\(=\frac{2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})\)
\(=2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=2\)
Đúng 2
Bình luận (0)
Cho a,b,c >0 và abc=1
\(\dfrac{\sqrt{a}}{2+b\sqrt{a}}+\dfrac{\sqrt{b}}{2+c\sqrt{b}}+\dfrac{\sqrt{c}}{2+a\sqrt{c}}\ge1\)
Cho a,b,c>0 t/m abc=1
\(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
Đặt \(\left(a;b;c\right)=\left(x^4;y^4;z^4\right)\Rightarrow xyz=1\)
\(VT=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)
\(VT=\dfrac{1}{x^2+y^2+y^2+1+2}+\dfrac{1}{y^2+z^2+z^2+1+2}+\dfrac{1}{z^2+x^2+x^2+1+2}\)
\(VT\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Đúng 2
Bình luận (0)
cho M= \(\dfrac{\sqrt{a}+\sqrt{b}-1}{a+a\sqrt{b}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\dfrac{\sqrt{b}}{a\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
a) tìm điều kiện a và b để M xác định
b) c/m M>0
Bài: Cho M=\(\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}\) + \(\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\) . ( \(\dfrac{b}{a-\sqrt{ab}}\) + \(\dfrac{\sqrt{b}}{a+\sqrt{ab}}\) )
a) Tìm đk của a và b để M xác định
b) C/m M > 0
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a< >b\end{matrix}\right.\)
b: \(M=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{b}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{b\left(a+\sqrt{ab}\right)+\sqrt{b}\left(a-\sqrt{ab}\right)}{a^2-ab}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(a-b\right)}\cdot\dfrac{ab+b\sqrt{ab}+a\sqrt{b}-b\sqrt{a}}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(\sqrt{a}-\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}\left(\sqrt{ab}+b+\sqrt{a}-\sqrt{b}\right)}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{a\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2}\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}+\sqrt{b}-1\right)+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+2\sqrt{ab}-2\sqrt{a}+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}-\sqrt{a}+b-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}+b-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\left(2\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{a}+\sqrt{b}-1}{2a}\)
Giả sử như a=0,1 và b=0,11 thì M<0 nha bạn
=>Đề này sai rồia: ĐKXĐ:
b: \(M=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{b}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{b\left(a+\sqrt{ab}\right)+\sqrt{b}\left(a-\sqrt{ab}\right)}{a^2-ab}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(a-b\right)}\cdot\dfrac{ab+b\sqrt{ab}+a\sqrt{b}-b\sqrt{a}}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(\sqrt{a}-\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}\left(\sqrt{ab}+b+\sqrt{a}-\sqrt{b}\right)}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{a\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2}\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}+\sqrt{b}-1\right)+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+2\sqrt{ab}-2\sqrt{a}+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}-\sqrt{a}+b-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}+b-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\left(2\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{a}+\sqrt{b}-1}{2a}\)
Giả sử như a=0,1 và b=0,11 thì M<0 nha bạn
=>Đề này sai rồi
Đúng 1
Bình luận (0)
1, Cho a, b, c là 3 số dương. CMR:
a, \(\dfrac{a}{\sqrt{a+b}\sqrt{a+c}}+\dfrac{b}{\sqrt{a+b}\sqrt{b+c}}+\dfrac{c}{\sqrt{a+c}\sqrt{b+c}}\le\dfrac{3}{2}\)
b, \(\dfrac{a}{\sqrt{a+b}\sqrt{b+c}}+\dfrac{b}{\sqrt{a+c}\sqrt{b+c}}+\dfrac{c}{\sqrt{a+c}\sqrt{b+a}}\ge\dfrac{3}{2}\)
Cho a, b, c > 0 thoả mãn: \(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\). Chứng minh rằng: \(\dfrac{\sqrt{a}}{a+1}+\dfrac{\sqrt{b}}{b+1}+\dfrac{\sqrt{c}}{c+1}=\dfrac{2}{\sqrt{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=4\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
Tương tự: \(b+1=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(c+1=\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(VT=\sum\dfrac{\sqrt{a}}{a+1}=\sum\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(=\dfrac{2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(VP=\dfrac{2}{\sqrt{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}=\dfrac{2}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2}}\)
\(=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(\Rightarrow VT=VP\) (đpcm)
Đúng 2
Bình luận (0)