cho a,b,c thỏa mãn \(\hept{\begin{cases}a+b+c=0\\a^2+b^2+c^2=2009\end{cases}}\)tinh A= \(a^4\)+\(b^4\)+\(c^4\)
Cho ba số a, b, c thỏa mãn \(\hept{\begin{cases}a+b+c=0\\\\a^2+b^2+c^2=2009\end{cases}}\) tính \(A=a^4+b^4+c^4\)
\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\Rightarrow ab+bc+ac=-\frac{2009}{2}\)
\(\left(ab+bc+ac\right)^2=a^2b^2+a^2c^2+b^2c^2+2abc\left(a+c+b\right)=a^2b^2+a^2c^2+b^2c^2\)\(\Rightarrow a^2b^2+a^2c^2+b^2c^2=\frac{2009^2}{4}\)
\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Rightarrow2009^2=a^4+b^4+c^4+\frac{2009^2}{4}\cdot2\)
\(\Rightarrow a^4+b^4+c^4=\frac{2009^2}{2}\)
Ta có \(a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)=-2\left(ab+bc+ca\right)\)
\(a^2b^2+b^2c^2+c^2a^2=\left(ab+bc+ca\right)^2-2abc\left(a+b+c\right)=\left(\frac{a^2+b^2+c^2}{2}\right)^2=\frac{2009^2}{4}\)
\(A=a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)=\frac{2009^2}{2}\)
Cho a,b,c thỏa mãn
\(\orbr{\hept{\begin{cases}a+b+c=0\\a^2+b^2+c^2=2009\end{cases}}}\)
Tính a4 + b4 + c4
\(\text{Chắc bn ghi thiếu đề :}\)
\(\hept{\begin{cases}a+b+c=0\\a^2+b^2+c^2=1\end{cases}}\)
\(Tính\)\(a^4+b^4+c^4\)
\(Giải:\)\(\text{Đặt}\)\(M=a^4+b^4+c^4\)
\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\)
\(1=M=\left(2a^2b^2+2b^2c^2+2c^2a^2\right)\)
\(M=1-\left(2a^2b^2+2b^2c^2+2c^2a^2\right)=1-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(0=1+2ab+2ac+2bc\)
\(2\left(ab+ac+bc\right)=-1\Rightarrow ab+ac+bc=-\frac{1}{2}\)
\(\left(ab+ac+bc\right)^2=a^2b^2+a^2c^2+b^2c^2+2\left(a^2bc+ab^2c+abc^2\right)\)
\(\frac{1}{4}=^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)\)
\(\Rightarrow^2b^2+a^2c^2+b^2c^2=\frac{1}{4}.0\left(vì\right)a+b+c=0\)
\(M=1-2.\frac{1}{4}=\frac{1}{2}\)
\(a+b+c=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0.\)
\(\Leftrightarrow ab+bc+ca=-\frac{2009}{2}.\)
\(\Rightarrow\left(ab+bc+ca\right)^2=\frac{2009^2}{4}.\)
\(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{2009^2}{4}.\)
\(a^2b^2+b^2c^2+c^2a^2=\frac{2009^2}{4}.\)
Ta có \(\left(a^2+b^2+c^2\right)^2=2009^2\)
\(a^4+b^4+c^4=2009^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=2009^2-2.\frac{2009^2}{4}=\frac{2009^2}{2}.\)
cho a b c thỏa mãn \(\hept{\begin{cases}a+b+c=0\\a^2+b^2+c^2=2009\end{cases}}\)
Tính A=\(a^4+b^4+c^4\)
Ta có: \(a+b+c=0\)
\(\Leftrightarrow a+b=-c\)
\(\Leftrightarrow a^2+2ab+b^2=c^2\)
\(\Leftrightarrow a^2+b^2-c^2=-2ab\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2-a^2c^2-b^2c^2\right)=4a^2b^2\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+a^2c^2+b^2c^2\right)\)
Ta lại có: \(a^2+b^2+c^2=2009\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=2009^2\)
\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=2009^2\)
\(\Leftrightarrow a^4+b^4+c^4=\frac{2009^2}{2}\)
cho 3 số a,b,c thỏa mãn: \(\hept{\begin{cases}a+b+c=-2\\\\a^2+b^2+c^2=2\end{cases}}\)cmr \(-\frac{4}{3}\)\(\le\)a,b,c\(\le\)0
Tìm a, b, c thỏa mãn:
\(\hept{\begin{cases}a^4-2b=\frac{-1}{2}\\b^4-2c=\frac{-1}{2}\\c^4-2a=\frac{-1}{2}\end{cases}}\)
Cho a,b,b thỏa mãn \(\hept{\begin{cases}a^2+b^2+c^2=2\\ab+bc+ca=1\end{cases}}\)CMR \(\frac{-4}{3}< a,b,c< \frac{4}{3}\)
Tìm a,b,c,d >0 thỏa mãn:
\(\hept{\begin{cases}a+b+c+d=4\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=4\end{cases}}\)
Ta có:
\(\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\ge\left(a+b+c+d\right).\frac{16}{\left(a+b+c+d\right)}=16\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge4\)
Dấu = xảy ra khi \(a=b=c=d=1\)
Cho 3 số a;b;c thỏa mãn
\(\hept{\begin{cases}a+b+c=-2\\a^2+b^2+c^2=2\end{cases}}\)
\(CMR:\frac{-4}{3}\le a;b;c\le0\)
Áp dụng bđt bu nhi a ta có
\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\Rightarrow\left(-2-c\right)^2\le2\left(2-c^2\right)\)
=> \(c^2+4c+4\le4-2c^2\)
=> \(3c^2+4c\le0\Rightarrow c\left(3c+4\right)\le0\Rightarrow-\frac{4}{3}\le c\le0\)
Cho a;b;c;d > 0 thỏa mãn đồng thời các đk \(\hept{\begin{cases}a^2+b^2=1\\\frac{a^4}{c}+\frac{b^4}{d}=\frac{1}{c+d}\end{cases}}\). CMR: \(\frac{a^2}{c}+\frac{d}{b^2}\ge2\)?
(P/s: Đang cần gấp nhé !)
\(\frac{d}{b^2}\) hay \(\frac{b^2}{d}\)hả bạn?
Ta có: \(\frac{a^4}{c}+\frac{b^4}{d}\ge\frac{\left(a^2+b^2\right)^2}{c+d}=\frac{1}{c+d}\)
Dấu = xảy ra khi \(\frac{a^2}{c}=\frac{b^2}{d}\)
Do đó: \(VT=\frac{a^2}{c}+\frac{b}{d^2}=\frac{d^2}{b}+\frac{b}{d^2}\ge2\sqrt{\frac{d^2}{b}.\frac{b}{d^2}}=2\)