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Ta có: \(x^2+y^2=1\Leftrightarrow\left(x^2+y^2\right)^2=1\) (1)
Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) ta được:
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right)ab\)
\(\Leftrightarrow x^4ab+x^4b^2+y^4a^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)
\(\Leftrightarrow x^4b^2+y^4a^2=2x^2y^2ab\)
\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)
\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\)
\(\Leftrightarrow x^2b-y^2a=0\)
\(\Leftrightarrow x^2b=y^2a\)
\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\left(\frac{x^2}{a}\right)^{1009}=\left(\frac{y^2}{b}\right)^{1009}=\left(\frac{1}{a+b}\right)^{1009}\)
\(\Rightarrow\frac{x^{2018}}{a^{1009}}=\frac{y^{2018}}{b^{1009}}=\frac{1}{\left(a+b\right)^{1009}}\)
\(\Rightarrow\frac{x^{2018}}{a^{1009}}+\frac{y^{2018}}{b^{1009}}=\frac{1}{\left(a+b\right)^{1009}}+\frac{1}{\left(a+b\right)^{1009}}=\frac{2}{\left(a+b\right)^{1009}}\left(đpcm\right)\)
Ta có: \(a^3+b^3+c^3-a^2+b^2+c^2=0\)
\(\Leftrightarrow a^2\left(a-1\right)+b^2\left(b-1\right)+c^2\left(c-1\right)=0\)
Mà \(a^2+b^2+c^2=1\)
\(\Rightarrow\hept{\begin{cases}a\le1\\b\le1\\c\le1\end{cases}}\Rightarrow\hept{\begin{cases}1-a0\\1-b\ge0\\1-c\ge0\end{cases}}\)
\(\Rightarrow a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)\ge0\)
Dấu "=" xảy ra khi: \(a^2\left(1-a\right)=b^2\left(1-b\right)=c^2\left(1-c\right)\)
Kết hợp với giả thiết
=> a,b,c hoán vị 1;0;0
=> S= 1
\(2x^2+y^2+z^2-2xy-2x+1=0\)
\(\Rightarrow\left(x^2+y^2-2xy\right)+\left(x^2-2x+1\right)+z^2=0\)
\(\Rightarrow\left(x-y\right)^2+\left(x-1\right)^2+z^2=0\)
\(\Leftrightarrow x=y=1;=0\)
\(A=x^{2018}+y^{2019}+z^{2020}=1+1+0=2\)
2)
\(a+b+c=6\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=36\)
\(\Leftrightarrow12+2\left(ab+bc+ac\right)=36\Leftrightarrow ab+bc+ac=12\)
Kết hợp với \(a^2+b^2+c^2=12\Leftrightarrow a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
Kết hợp với \(a+b+c=6\Leftrightarrow a=b=c=2\)
\(P=\left(a-3\right)^{2019}+\left(b-3\right)^{2019}+\left(c-3\right)^{2019}=\left(-1\right)^{2019}+\left(-1\right)^{2019}+\left(-1\right)^{2019}=-3\)
Cộng vế với vế, ta có:
\(a^2-20b+81+b^2+18c+9+c^2+6a+100=0\)
\(\Rightarrow\left(a^2+6a+9\right)+\left(b^2-20b+100\right)+\left(c^2+18c+81\right)=0\)
\(\Rightarrow\left(a^2+2.a.3+3^2\right)+\left(b^2-2.b.10+10^2\right)+\left(c^2+2.9.c+9^2\right)=0\)
\(\Rightarrow\left(a+3\right)^2+\left(b-10\right)^2+\left(c+9\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}a+3=0\\b-10=0\\c+9=0\end{cases}\Rightarrow}\hept{\begin{cases}a=-3\\b=10\\c=-9\end{cases}}\)
Khi đó: \(M=\left(a+2\right)^{2017}+\left(b-9\right)^{2018}+\left(c+9\right)^{2018}\)
\(=\left(-3+2\right)^{2017}+\left(10-9\right)^{2018}+\left(-9+9\right)^{2018}\)
\(=-1+1+0=0\)
tau méc thầy hùng