\(x^5+1+1+1+1\) \(\ge5\sqrt[5]{x^5\cdot1\cdot1\cdot1\cdot1}=5x\)

\(y^5+1+1+1+1\ge5\sqrt[5]{y^5\cdot1\cdot1\cdot1\cdot1}=5y\)

 \(\Rightarrow x^5+y^5+8\ge5x+5y=10\)

  \(\Rightarrow x^5+y^5\ge2\)

sao mình ko hiểu bạn trả lời sao hết vậy

CŨNG LÀM NÂNG CAO P/ T SAO

a) 2a(x+y)+y+x = 2a(x+y)+(x+y)

                       = (2a+1)(x+y)

b) 2(x-y)^2-x+y = \(2\left(x-y\right)^2-\left(x-y\right)=\left(2x-2y-1\right)\left(x-y\right)\)

ta có 

\(\frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}\)

\(3+\frac{bc\left(b+c\right)+ac\left(b+c\right)+ab\left(a+b\right)}{abc}=0\) 

\(\frac{b^2c+bc^2}{abc}>0\)

tương tự các phân thức còn lại  suy ra a=b=c

\(3x\left(2x+3\right)-4x^2=-9\)

\(3x\left(2x+3\right)-4x^2+9=0\)

\(3x\left(2x+3\right)-\left[\left(2x\right)^2-3^2\right]=0\)

\(3x\left(2x+3\right)-\left(2x-3\right)\left(2x+3\right)=0\)

\(\left(2x+3\right)\left(3x-2x+3\right)=0\)

\(\left(2x+3\right)\left(x+3\right)=0\)

\(\orbr{\begin{cases}2x+3=0\\x+3=0\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{-3}{2}\\x=-3\end{cases}}\)

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

mả \(a^2+b^2+c^2=2\Rightarrow2\left(ab+bc+ca\right)=-2\) 

\(\Leftrightarrow ab+bc+ca=-1\Leftrightarrow\left(ab+bc+ca\right)^2=1\)

\(\Leftrightarrow a^2b^2+c^2b^2+c^2a^2+2abc\left(a+b+c\right)=1\)

mả \(a+b+c=0\Rightarrow a^2b^2+c^2a^2+b^2c^2=1\)

mặt khác \(\left(a^2+b^2+c^2\right)^2=4\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+c^2b^2+a^2c^2\right)=4\)

\(\Rightarrow a^4+b^4+c^4=2\)

\(a^4+b^4+c^4=2\)kq của mk đó