\(x^2+\frac{1}{x^2}+16y^2+\frac{1}{y^2}-10=0\)

<=>\(\left(x^2-2+\frac{1}{x^2}\right)+\left(16y^2-8+\frac{1}{y^2}\right)=0\)

<=>\(\left[x^2-2\cdot x\cdot\frac{1}{x}+\left(\frac{1}{x}\right)^2\right]+\left[\left(4y\right)^2-2\cdot4y\cdot\frac{1}{y}+\left(\frac{1}{y}\right)^2\right]=0\)

<=>\(\left(x-\frac{1}{x}\right)^2+\left(4y-\frac{1}{y}\right)^2=0\)

Mà \(\left(x-\frac{1}{x}\right)^2;\left(4y-\frac{1}{y}\right)^2>hoac=0\)

=>\(\hept{\begin{cases}\left(x-\frac{1}{x}\right)^2=0\\\left(4y-\frac{1}{y}\right)^2=0\end{cases}}\)

<=>\(\hept{\begin{cases}x-\frac{1}{x}=0\\4y-\frac{1}{y}=0\end{cases}}\)

đoạn này bạn tự giải tiếp

Vậy x=1 và y=1/2

Sorry

Ở trên mình KL thiếu

Còn có x= -1;y=-1/2

a) \(x^4-x^3+2x^2-x+1=0\)

\(\Leftrightarrow\left(x^4+x^2\right)-\left(x^3+x\right)+\left(x^2+1\right)=0\)

\(\Leftrightarrow x^2\left(x^2+1\right)-x\left(x^2+1\right)+\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x^2-x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\left(ktm\right)\\x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(ktm\right)\end{cases}}\)

Vậy phương trình vô nghiệm (ĐPCM)

b) \(x^4-2x^3+4x^2-3x+2=0\)

\(\Leftrightarrow\left(x^4-2x^3+x^2\right)+\left(x^2-2x+1\right)+\left(x^2-x+\frac{1}{4}\right)+\left(x^2+\frac{3}{4}\right)=0\)

\(\Leftrightarrow\left(x^2-x\right)^2+\left(x-1\right)^2+\left(x-\frac{1}{2}\right)^2+\left(x^2+\frac{3}{4}\right)=0\)

Có : \(\left(x^2-x\right)^2\ge0\)

        \(\left(x-1\right)^2\ge0\)

        \(\left(x-\frac{1}{2}\right)^2\ge0\)

          \(x^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Leftrightarrow\left(x^2-x\right)^2+\left(x-1\right)^2+\left(x-\frac{1}{2}\right)^2+\left(x^2+\frac{3}{4}\right)\ge\frac{3}{4}\)

Vậy phương trình vô nghiệm.(ĐPCM)

\(2x\left(x-2\right)+1=x-1\)

\(\Leftrightarrow2x^2-4x+1-x+1=0\)

\(\Leftrightarrow2x^2-5x+2=0\)

\(\Leftrightarrow2x^2-4x-x+2=0\)

\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\2x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}}\)

\(a+b=2\Rightarrow\left(a+b\right)^2=4\Rightarrow a^2+b^2+2ab=4\Rightarrow20+2ab=4\Rightarrow2ab=-16\Rightarrow ab=-8\)

\(\Rightarrow a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=2\left(20+8\right)=2.28=56\)

Ta có 

  \(a+b=2\)

\(\Leftrightarrow a^2+b^2+2ab=4\)

\(\Leftrightarrow2ab=4-\left(a^2+b^2\right)\)

\(\Leftrightarrow ab=-8\)

\(\Leftrightarrow\hept{\begin{cases}a^2b=-8a\\ab^2=-8b\end{cases}}\)

   Lại có 

    \(\left(a+b\right)\left(a^2+b^2\right)=a^3+b^3+a^2b+ab^2\)

                                            \(=a^3+b^3-8a-8b\)

                                            \(=a^3+b^3-8\left(a+b\right)\)

                                            \(=a^3+b^3-16\)

   Mà \(\left(a+b\right)\left(a^2+b^2\right)=2.20=40\)

  Nên \(a^3+b^3-16=40\)

           \(a^3+b^3=56\)

   Vậy \(a^3+b^3=56\)

Áp dụng BDT Svacxo ta có :

\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

Cách khác sử dụng Cosi : Dự đoán điểm rơi và ghép hợp lí !

Áp dụng bất đẳng thức cô - si với hai số dương:

\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}}=a\)

\(\frac{b^2}{c+a}+\frac{a+c}{4}\ge b\)

\(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)

\(\frac{a^2}{b+c}+\frac{b+c}{4}+\frac{b^2}{c+a}+\frac{a+c}{4}+\frac{c^2}{a+b}+\frac{a+b}{4}\ge a+b+c\)

=> => \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)

Dâu "=" xảy ra <=> a = b = c

\(\frac{x+3}{x+2}-\frac{x+4}{x+3}=\frac{x+5}{x+4}-\frac{x+6}{x+5}\)

\(\Leftrightarrow\frac{x+3}{x+2}-\frac{x+4}{x+3}=\frac{x+5}{x+4}-\frac{x+6}{x+5}\)

\(\Leftrightarrow\left(\frac{x+3}{x+2}-1\right)-\left(\frac{x+4}{x+3}-1\right)=\left(\frac{x+5}{x+4}-1\right)-\left(\frac{x+6}{x+5}-1\right)\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}=\frac{1}{x+4}-\frac{1}{x+5}\)

\(\Leftrightarrow\frac{1}{x+2}+\frac{1}{x+5}=\frac{1}{x+4}+\frac{1}{x+3}\)

\(\Leftrightarrow\frac{x+5+x+2}{\left(x+2\right)\left(x+5\right)}=\frac{x+4+x+3}{\left(x+4\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{2x+7}{x^2+7x+10}=\frac{2x+7}{x^2+7x+12}\)

\(\Leftrightarrow\orbr{\begin{cases}2x+7=0\\x^2+7x+10=x^2+7x+12\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{7}{2}\left(tm\right)\\0=2\left(ktm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{7}{2}\right\}\)

Thêm cho mik : \(ĐKXĐ:\hept{\begin{cases}x\ne-2;x\ne-3\\x\ne-4;x\ne-5\end{cases}}\)

Ta có : \(a^3+b^3+c^3=3abc\)

\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ca=0\left(2\right)\end{cases}}\)

Từ (1) \(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)

Khi đo s: \(P=\frac{abc}{\left(-a\right)\left(-b\right)\left(-c\right)}=-1\)

Từ (2) \(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow a=b=c\)

Khi đó : \(P=\frac{a^3}{2a\cdot2a\cdot2a}=\frac{1}{8}\)

Vậy : \(P=\frac{1}{8}\) hoặc \(P=-1\) với a,b,c thỏa mãn đề.