\(a^3+b^3+c^3=3abc\) với\(a,b,c\ne0\)và \(a+b+c\ne0\)
tính \(P=\left(2006+\frac{a}{b}\right)\left(2006+\frac{b}{c}\right)\left(2006+\frac{c}{a}\right)\)
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh:
a) \(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
b)\(\frac{4a^4+5b^4}{4c^4+5d^4}=\frac{a^2b^2}{c^2d^2}\)
c)\(\left(\frac{a-b}{c-d}\right)^{2005}=\frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}\)
d)\(\frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}=\frac{\left(a+b\right)^{2005}}{\left(c+d\right)^{2005}}\)
e)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
f)\(\frac{\left(20a^{2007}-11c^{2007}\right)^{2006}}{\left(20a^{2006}+11c^{2006}\right)^{2007}}=\frac{\left(20b^{2007}-11d^{2007}\right)^{2006}}{\left(20b^{2006}+11d^{2006}\right)^{2007}}\)
ừ, bạn bik làm thì giúp mình nha ^^
Ta có:a+b+c\(\ne\)0
a,b,c\(\ne\)0
\(a^3,b^3,c^3=3abc\)
Hãy giải phương trình:
P=\(P=\left(2006+\frac{a}{b}\right)\left(2006+\frac{b}{c}\right)\left(2006+\frac{c}{a}\right)\)
Cho \(a^3+b^3+c^3=3abc\) với \(a+b+c\ne0;a,b,c\ne0\)
Tình giá trị biểu thức \(P=\left(2017+\frac{a}{b}\right)\left(2017+\frac{b}{c}\right)\left(2017+\frac{c}{a}\right)\)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\) (vì \(a+b+c\ne0\))
<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\) (nhân cả hai về với hai)
<=> \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
<=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
<=> a - b = b - c = c - a = 0 (vì 3 cái đấy đều lớn hơn hoặc bằng 0)
<=> a = b = c
Nên : P = \(\left(2017+\frac{a}{b}\right)\left(2017+\frac{b}{c}\right)\left(2017+\frac{c}{a}\right)=\left(2017+\frac{a}{a}\right)\left(2017+\frac{a}{a}\right)+\left(2017+\frac{a}{a}\right)\)
\(=\left(2017+1\right)\left(2017+1\right)\left(2017+1\right)=2018.2018.2018=2018^3\)
Bài 1 : cho a , b ,c là ba số khác 0 thỏa mãn điều kiện a^3 + b^3 +c^3 = 3abc . và a + b + c = 0 . Tính gtbt :
\(M=\left(1+\frac{a}{b}\right)\left(a+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Bài 2 :
Cho 3 số x , y , z \(\ge0\) và x2006 + y2006 + z2006 = 3 . Tìm GTLN của A = x^2 + y ^2 + x ^2
Ai làm được mk link cho nha
Sửa đề \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Ta có: \(a^3+b^3+c^3=3ab\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
TH1: a+b+c=0
=> \(\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}}\)
Thay vào M ta được M=\(\left(1-\frac{b+c}{b}\right)\left(1-\frac{a+c}{c}\right)\left(1-\frac{a+b}{a}\right)\)
\(\Rightarrow M=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)
TH2: \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow M=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Bài 1 : cho a , b ,c là ba số khác 0 thỏa mãn điều kiện a^3 + b^3 +c^3 = 3abc . và a + b + c = 0 . Tính gtbt :
\(M=\left(1+\frac{a}{b}\right)\left(a+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Bài 2 :
Cho 3 số x , y , z \(\ge0\) và x2006 + y2006 + z2006 = 3 . Tìm GTLN của A = x^2 + y ^2 + x ^2
Ai làm được mk link cho nha
Cho \(a^3+b^3+c^3=3abc\)và \(abc\ne0;a+b+c=0\)
CMR \(\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{c}+\frac{1}{a}\right)=0\)
Let \(a,b,c\ge0\) such that \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ne0\) . Prove that:
\(a^3+b^3+c^3+3abc-ab\left(a+b\right)-bc\left(b+c\right)-ca\left(c+a\right)\ge abc\left(\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}-3\right)\)
BĐT sau đây vẫn đúng: \(\Sigma a\left(a-c\right)\left(a-b\right)\ge abc\left(\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}-3\right)+\frac{16\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a+b+c\right)^3}\)
\(Cho\)\(abc\ne0\)\(và\)\(a^3+b^3+c^3=3abc\)
\(Tính\)\(A=\left(1+\frac{a}{b}\right).\left(1+\frac{b}{c}\right).\left(1+\frac{c}{a}\right)\)
https://vn.answers.yahoo.com/question/index?qid=20110907041853AA9iaBQ
giúp với mọi người
câu này mik cũng cần
pls nhé phfi
Câu 1: CMR : Nếu \(a^3+b^3+c^3=3abc\) thì \(a+b+c=0\) hoặc \(a=b=c\)
Câu 2: Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) . Tính \(\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}\)
Câu 3 : Cho \(a^3+b^3+c^3=3abc\left(a.b.c\ne0\right)\). Tính\(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Câu 1:
Chứng minh a3+b3+c3=3abc thì a+b+c=0\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(\Rightarrow\left[\left(a+b\right)^3+c^3\right]-3abc\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow0=0\) Đúng (Đpcm)
Chứng minh a3+b3+c3=3abc thì a=b=cÁp dụng Bđt Cô si 3 số ta có:
\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)
Dấu = khi a=b=c (Đpcm)
Câu 2
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\cdot\frac{1}{abc}\)
Ta có:
\(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\cdot3\cdot\frac{1}{abc}=3\)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{array}\right.\)
Xét \(a+b+c=0\)\(\Rightarrow\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}\)\(\Rightarrow A=\left(1-\frac{b+c}{b}\right)\left(1-\frac{a+c}{c}\right)\left(1-\frac{a+b}{a}\right)\)
\(=\left(1-1-\frac{c}{b}\right)\left(1-1-\frac{a}{c}\right)\left(1-1-\frac{b}{a}\right)\)
\(=\left(-\frac{c}{b}\right)\left(-\frac{a}{c}\right)\left(-\frac{b}{a}\right)=-1\)
Xét \(a^2+b^2+c^2-ab-bc-ca=0\)\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a-b=b-c=c-a=0\Leftrightarrow a=b=c\)
\(\Leftrightarrow A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)