Cho a+b+c=0 .Tính
A=[ab.(a-b)+bc.(b-c)+ac(c-a)].[\(\frac{1}{ab.\left(a-b\right)}+\frac{1}{bc.\left(b-c\right)}+\frac{1}{ca.\left(c-a\right)}\)]
Những câu hỏi liên quan
Cho a,b,c và a+b+c=1.Tính
P=\(\sqrt{\frac{\left(a+bc\right)\left(b+ca\right)}{c+ab}}+\sqrt{\frac{\left(b+ac\right)\left(c+ab\right)}{a+bc}}\sqrt{\frac{\left(c+ab\right)\left(a+bc\right)}{b+ca}}\)
\(\text{a+b+c = 1}\Rightarrow a=1-b-c\Rightarrow a+bc=1-b-c+bc=\left(b-1\right)\left(c-1\right)\)
tương tự \(b+ca=\left(a-1\right)\left(c-1\right);c+ab=\left(a-1\right)\left(b-1\right)\)
đặt a-1=x ; b-1=y ; c-1=z , ta có
\(P=\sqrt{\frac{yzzx}{xy}}+\sqrt{\frac{xzxy}{yz}}+\sqrt{\frac{xyyz}{xz}}=\sqrt{z^2}+\sqrt{x^2}+\sqrt{y^2}=x+y+z=1\)
Đúng 0
Bình luận (0)
Cho a,b,c >0 thỏa mãn a+b+c=1 .Tính P=\(\sqrt{\frac{\left(a+bc\right)\left(b+ca\right)}{c+ab}}+\sqrt{\frac{\left(b+ca\right)\left(c+ab\right)}{a+bc}}+\sqrt{\frac{\left(c+ab\right)\left(a+bc\right)}{b+ac}}\)
Ai giúp với
\(a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\)
Tương tự: \(b+ca=\left(a+b\right)\left(b+c\right)\) ; \(c+ab=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow P=a+b+b+c+c+a=2\left(a+b+c\right)=2\)
cho a,b,c>0 thỏa mãn a+b+c=1.
tính P=\(\sqrt{\frac{\left(a+bc\right)\left(b+ca\right)}{c+ab}}+\sqrt{\frac{\left(c+ab\right)\left(b+ca\right)}{a+bc}}+\sqrt{\frac{\left(a+bc\right)\left(c+ab\right)}{b+ca}}\)
Ta có : \(\left\{{}\begin{matrix}a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\\b+ca=b\left(a+b+c\right)+ca=\left(b+c\right)\left(a+b\right)\\c+ab=c\left(a+b+c\right)+ab=\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)
Từ đó ta có :
\(P=\Sigma\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(a+b\right)}{\left(a+c\right)\left(b+c\right)}}\)
\(P=\Sigma\sqrt{\left(a+b\right)^2}\)
\(P=\Sigma\left(a+b\right)\)
\(P=2\left(a+b+c\right)\)
\(P=2\)
Đúng 0
Bình luận (2)
Afrac{a^2+bc}{b+ac}+frac{b^2+ca}{c+ab}+frac{c^2+ab}{a+bc}frac{3left(a^2+bcright)}{left(a+b+cright)b+3ac}+frac{3left(b^2+caright)}{left(a+b+cright)c+3ab}+frac{3left(c^2+abright)}{left(a+b+cright)a+3bc}gefrac{3left(a^2+bcright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(b^2+caright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(c^2+abright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}3
Đọc tiếp
\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
Cho a>0 b>0 c>0 thỏa mãn a+b+c=1 tính gt bt
\(P=\sqrt{\frac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}+\sqrt{\frac{\left(c+ab\right)\left(b+ac\right)}{a+bc}}+\sqrt{\frac{\left(c+ab\right)\left(a+bc\right)}{b+ac}}\)
\(\sqrt{\frac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\frac{\left(a^2+ab+ac+bc\right)\left(b^2+bc+ba+ac\right)}{c^2+ca+cb+ab}}=\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+a\right)\left(b+c\right)}{\left(c+a\right)\left(c+b\right)}}=a+b\left(a,b,c>0;a+b+c=1\right)\)
Bạn làm tương tự nha
\(\Rightarrow P=a+b+c+a+b+c=2\left(a+b+c\right)=2\)
Đúng 0
Bình luận (0)
Cho a; b; c > 0 sao cho a+b+c=3. Chứng minh rằng
\(\frac{a}{b^2\left(ca+1\right)}+\frac{b}{c^2\left(ab+1\right)}+\frac{c}{a^2\left(bc+1\right)}\ge\frac{9}{\left(1+abc\right)\left(ab+bc+ca\right)}\)
30 tháng 3 2021 lúc 23:00
Cho a,b,c là các số thực dương thỏa mãn a+b+c=3abc. Chứng minh rằng :
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\left[\frac{a^4}{\left(ab+1\right)\left(ac+1\right)}+\frac{b^4}{\left(bc+1\right)\left(ab+1\right)}+\frac{c^4}{\left(ca+1\right)\left(bc+1\right)}\right]\ge\frac{27}{4}\)
(
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
hhhhhhhhhhhhhhhhh
hhhhhhhhhhhhhhhhhh
hhhhhhhhhhhhhhh
hhhhhhhhhhhhh
tính
\(B=\frac{\sqrt{\left(a+bc\right)\left(b+ca\right)}}{\sqrt{c+ab}}+\frac{\sqrt{\left(b+ca\right)\left(c+ab\right)}}{\sqrt{a+bc}}+\frac{\sqrt{\left(c+ab\right)\left(a+bc\right)}}{\sqrt{b+ac}}\)
(với a, b, c là số thực và a+b+c=1)
Do a + b + c = 1 nên \(\frac{\sqrt{\left(a+bc\right)\left(b+ca\right)}}{\sqrt{c+ab}}=\frac{\sqrt{\left[a\left(a+b+c\right)+bc\right]\left[b\left(a+b+c\right)+ca\right]}}{\sqrt{c\left(a+b+c\right)+ab}}\)
\(=\frac{\sqrt{\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)}}{\sqrt{ac+bc+c^2+ab}}=\frac{\sqrt{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\left(a+b\right)^2}=a+b\) (1)
Tương tự \(\hept{\begin{cases}\frac{\sqrt{\left(b+ca\right)\left(c+ab\right)}}{\sqrt{a+bc}}=b+c\text{ }\left(2\right)\\\frac{\sqrt{\left(c+ab\right)\left(a+bc\right)}}{\sqrt{b+ac}}=a+c\text{ }\left(3\right)\end{cases}}\)
Cộng vế với vế của (1)(2)(3) lại ta được :
\(\frac{\sqrt{\left(a+bc\right)\left(b+ca\right)}}{\sqrt{c+ab}}+\frac{\sqrt{\left(b+ca\right)\left(c+ab\right)}}{\sqrt{a+bc}}+\frac{\sqrt{\left(c+ab\right)\left(a+bc\right)}}{\sqrt{b+ac}}=2\left(a+b+c\right)=2\)
Đúng 0
Bình luận (0)
Cho các số dương a, b, c thỏa mãn ab+bc+ca=1.
CMR: \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge3+\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{a^2}}+\sqrt{\frac{\left(b+c\right)\left(b+a\right)}{b^2}}+\sqrt{\frac{\left(c+a\right)\left(c+b\right)}{c^2}}\)