Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng : x + y + z = 0 suy ra x3 + y3 + z3 = 3xyz
1/a + 1/2b + 1/3c = 0 = >... rồi biến đổi nhé
A=\(\frac{a^2}{bc}\)+\(\frac{b^2}{ac}\)+\(\frac{c^2}{ab}\)=\(\frac{a^3}{abc}\)+\(\frac{b^3}{abc}\)+\(\frac{c^3}{abc}\)=\(\frac{a^3+b^3+c^3}{abc}\)
Mà a^3+b^3+c^3=3abc ( Tự chứng minh )
\(\Rightarrow\)A= \(\frac{3abc}{abc}\)= 3
có ở trong câu hỏi tương tự nhé
\(S=13\left(\frac{a}{18}+\frac{c}{24}\right)+13\left(\frac{b}{24}+\frac{c}{48}\right)+\left(\frac{a}{9}+\frac{b}{6}+\frac{2}{ab}\right)+\left(\frac{a}{18}+\frac{c}{24}+\frac{2}{ac}\right)+\left(\frac{b}{8}+\frac{c}{16}+\frac{2}{bc}\right)+\left(\frac{a}{9}+\frac{b}{6}+\frac{c}{12}+\frac{8}{abc}\right)\)Cô si các ngoặc là được nhé
1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)
\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)
Bài 2:
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=3\Leftrightarrow\frac{1}{a+1}=1-\frac{1}{b+1}+1-\frac{1}{c+1}+1-\frac{1}{d+1}\)
\(\Leftrightarrow\frac{1}{a+1}=\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\ge3\sqrt[3]{\frac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}>0\)
Tương tự:
\(\frac{1}{b+1}\ge3\sqrt[3]{\frac{cda}{\left(c+1\right)\left(d+1\right)\left(a+1\right)}}>0\);\(\frac{1}{c+1}\ge3\sqrt[3]{\frac{dab}{\left(d+1\right)\left(a+1\right)\left(b+1\right)}}>0\);
\(\frac{1}{d+1}\ge3\sqrt[3]{\frac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}>0\)
\(\Rightarrow\frac{1}{a+1}.\frac{1}{b+1}.\frac{1}{c+1}.\frac{1}{d+1}\ge3^4\sqrt[3]{\frac{\left(abcd\right)^3}{\left[\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)\right]^3}}\)
\(\Leftrightarrow\frac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\ge81\frac{abcd}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\)
\(\Leftrightarrow abcd\le\frac{1}{81}\)
Dấu "="xảy ra khi \(a=b=c=d?\). Không chắc lắm.
Sửa một chút:
Bài 2: Thay dấu "=" bởi lớn hơn hoặc bằng, không có gì cả (nãy nhìn nhầm)
\(\Leftrightarrow\frac{1}{a+1}\ge1-\frac{1}{b+1}+1-\frac{1}{c+1}+1-\frac{1}{d+1}=\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\)
\(\ge3\sqrt[3]{\frac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}>0\left(AM-GM\right)\)
Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Tính \(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}\)
P= abc(\(\frac{1}{^{a^3}}\)+\(\frac{1}{b^3}\)+\(\frac{1}{c^3}\)) = abc[(\(\frac{1}{a}\)+\(\frac{1}{b}\))3+\(\frac{1}{c^3}\)-\(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)]=abc[(\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\))(....)- \(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)]
=abc.(- \(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)) =-3(\(\frac{c}{a}\)+\(\frac{c}{b}\)) = -3c(\(\frac{1}{a}\)+\(\frac{1}{b}\)) = -3c.\(\frac{-1}{c}\)=3
P = 3
Đầu tiên,bạn cần chứng minh x + y + z = 0 thì x3 + y3 + z3 = 3xyz ( Bạn ko biết c/m thì hỏi nhé)
Thay\(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}=\frac{3}{abc}\)
\(\Rightarrow M=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}=abc\left(\frac{1}{c^3}+\frac{1}{a^3}+\frac{1}{b^3}\right)=abc.\frac{3}{abc}=3\)
\(A=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\left(a^3+b^3+c^3\right)\frac{1}{abc}\)
Cm với a+b+c=0 thì \(a^3+b^3+c^3=3abc\)(1) .Từ đó tính dc A, muốn cm(1) bạn xét hiệu nhé
\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)(luôn đúng vì a+b+c=0)
\(gt\Rightarrow\frac{ab+bc+ca}{abc}=0\) \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-czx\right)+3xyz\)
+ \(A=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\left[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{bc}-\frac{1}{ca}\right)+\frac{3}{abc}\right]\)
\(=abc\cdot\frac{3}{abc}=3\)
Ta có:
ab + ac + bc = 0
\(\Rightarrow\) \(\frac{ab+ac+bc}{abc}=0\)
\(\Rightarrow\) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Đặt \(\frac{1}{a}=x;\) \(\frac{1}{b}=y;\) \(\frac{1}{c}=z\)
Mà x + y + z = 0
=> x3 + y3 + z3 = 3xyz (Tự chứng minh nhé bạn, nếu không chứng minh được thì bình luận nhé!)
\(\Rightarrow\) \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Ta có:
\(A=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}\)
\(A=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)
\(A=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(A=abc.\frac{3}{abc}\)
\(A=3\)