9+1465+1

Bài 4: (ý c chuyên toán Hình vào làm nốt nhé @@)

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

\(\Rightarrow2x.\left(x+2\right)=0\)

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

\(\left(2x+1\right)^2-\left(x-3\right)^2=-15\)

\(\Rightarrow\left(2x+1-x+3\right).\left(2x+1+x-3\right)=-15\)

\(\Rightarrow\left(x+4\right).\left(3x-2\right)=-15\)

\(\Rightarrow3x^2-2x+12x-8+15=0\)

\(\Rightarrow3x^2+\left(-3x+12x\right)+\left(-8+15\right)=0\)

\(\Rightarrow3x^2+10x+7=0\)

\(\Rightarrow3x^2+3x+7x+7=0\)

\(\Rightarrow(3x^2+3x)+\left(7x+7\right)=0\)

\(\Rightarrow3x.\left(x+1\right)+7.\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x+7=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=-7\\x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-7}{3}\\x=-1\end{cases}}\)

\(\Rightarrow\left(3x+7\right).\left(x+1\right)=0\)

Bạn sửa hai dòng cuối phần b) cho mình thành này nhé. (Tại nó bị nhảy dòng)

\(\Rightarrow\left(3x+7\right).\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x+7=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=-7\\x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-7}{3}\\x=-1\end{cases}}\)

\(x+y+z=0\)

\(\Rightarrow\left(x+y+z\right)^2=0\)

\(\Rightarrow x^2+y^2+z^2+2.\left(xy+yz+xz\right)=0\)

\(\Rightarrow1+2.\left(xy+yz+xz\right)=0\)

\(\Rightarrow xy+yz+xz=\frac{-1}{2}\)

\(\Rightarrow\left(xy+yz+xz\right)^2=\frac{1}{4}\)

\(\Rightarrow x^2y^2+y^2z^2+x^2z^2+2.\left(xy^2z+xyz^2+x^2yz\right)=\frac{1}{4}\)

\(\Rightarrow x^2y^2+y^2z^2+x^2z^2=\frac{1}{4}\)

\(x^2+y^2+z^2=1\)

\(\Rightarrow\left(x^2+y^2+z^2\right)^2=1\)

\(\Rightarrow x^4+y^4+z^4+2.\left(x^2y^2+y^2z^2+x^2z^2\right)=1\)

\(\Rightarrow x^4+y^4+z^4+2.\frac{1}{4}=1\)

\(\Rightarrow x^4+y^4+z^4=\frac{1}{2}\)

\(\Rightarrow S=\frac{1}{2}\)