Có:

\(x+y+z=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x=y+z\\-y=x+z\\-z=x+y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=\left(y+z\right)^2\\y^2=\left(x+z\right)^2\\z^2=\left(x+y\right)^2\end{matrix}\right.\)

\(\Rightarrow ax^2+by^2+cz^2\)

\(=a\left(y+z\right)^2+b\left(x+z\right)^2+c\left(x+y\right)^2\)

\(=x^2\left(b+c\right)+y^2\left(a+c\right)+z^2\left(a+b\right)+2\left(ayz+bxz+cxy\right)\)

Mà \(a+b+c=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+c=-a\\a+c=-b\\a+b=-c\end{matrix}\right.\)

Đồng thời có: \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)

\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)

\(\Leftrightarrow ayz+bxz+cxy=0\)

Từ đây ta có:)

\(ax^2+by^2+cz^2=-ax^2-by^2-cz^2\)

\(\Rightarrow2\left(ax^2+by^2+cz^2\right)=0\)

\(\Rightarrow ax^2+by^2+cz^2=0\left(đpcm\right)\)

20520l

tick se giai tiep

đây là một phép tính dài ;máy tính cũng ko tính đc chính vì vậy bạn nào hỏi câu này thì tự tính ra

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

\(\Rightarrow4x^2+4y^2+4z^2+4t^2=4xy+4xz+4xt\)

\(\Rightarrow4x^2+4y^2+4z^2+4t^2-4xy-4xz-4xt=0\)

\(\Rightarrow x^2-4xy+4y^2+x^2-4xz+4z^2+x^2-4xt+4t^2+x^2=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x-2y=0\\x-2z=0\\x-2t=0\\x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2y\\x=2z\\x=2t\\x=0\end{matrix}\right.\)

\(\Rightarrow x=y=z=t=0\)

Kết luận: \(x=y=z=t=0\)

\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Rightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-40=0\)

\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)

Đặt \(a=x^2+6x+5\)

\(\Rightarrow a\left(a+3\right)-40=0\)

\(\Rightarrow a^2+3a-40=0\)

\(\Rightarrow a^2+8a-5a-40=0\)

\(\Rightarrow a\left(a+8\right)-5\left(a+8\right)=0\)

\(\Rightarrow\left(a+8\right)\left(a-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+8=0\\a-5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}a=-8\\a=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+6x+5+8=0\\x^2+6x+5-5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+6x+13=0\\x^2+6x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+6x+13\ne0\\x\left(x+6\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Kết luận:\(S=\left\{-6;0\right\}\)

2)

\(a^3+b^3+c^3=3abc\)

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

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

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

\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-cb-ac\right)\)

\(\Rightarrow a+b+c=0\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)

\(\Rightarrow N=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(\Rightarrow N=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\)

\(\Rightarrow N=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\)

\(\Rightarrow N=-1\)

a)\(3x\left(x^2-2x\right)\)

\(=3x^3-6x^2\)

b) \(\left(27x^3-1\right):\left(9x^2+3x+1\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right):\left(9x^2+3x+1\right)\)

\(=3x-1\)

c) \(\dfrac{4y^3}{7x^2}.\dfrac{14x^3}{y}\)

\(=8xy^2\)

\(\)d)\(\dfrac{x^2-9}{2x+6}:\dfrac{3-x}{2}\)

\(=-\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}:\dfrac{x-3}{2}\)

\(=-\dfrac{\left(x+3\right)\left(x-3\right)}{2\left(x+3\right)}.\dfrac{2}{\left(x-3\right)}\)

\(=-1\)

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