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B1) Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{xy+yz+zx}{xyz}=0\)
\(\Rightarrow xy+yz+zx=0\)
Ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(=x^2+y^2+z^2+2.0\)
\(=x^2+y^2+z^2\left(đpcm\right)\)
B2) \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\forall a;b\\\left(b-c\right)^2\ge0\forall b;c\\\left(c-a\right)^2\ge0\forall c;a\end{cases}\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow a=b=c\left(đpcm\right)}\)
\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right).2=\left(ab+bc+ca\right).2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Ta có: \(\hept{\begin{cases}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(c-a\right)^2\ge0\forall a,c\end{cases}}\)\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
Mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\)
Vậy \(a^2+b^2+c^2=ab+bc+ca\)thì \(a=b=c\)
theo bài ra ta có : \(\left(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}\right)^2=1^2=1\)
Ta thấy
\(\left(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-2.\frac{1}{xy}-2.\frac{1}{xz}+2.\frac{1}{yz}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-2\left(\frac{1}{xy}+\frac{1}{xz}-\frac{1}{yz}\right)\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-2\left(\frac{z+y-x}{xyz}\right)\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-2\left(\frac{0}{xyz}\right)=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\) vì x = y+z nê y+z-x = 0
Vậy \(\left(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1ĐPCM\)
Bài 1 : (x + 5)3 - x3 - 125
= (x + 5 - x)[(x + 5)2 + x(x + 5) + x2] - 125
= 5(x2 + 10x + 25 + x2 + 5x + x2)
= 5(3x2 + 15x + 25) - 125
= 5(3x2 + 15x + 25 - 25)
= 5(3x2 + 15x)
Xét \(\frac{x}{y^3-1}+\frac{y}{x^3-1}=\frac{1-y}{y^3-1}+\frac{1-x}{x^3-1}=-\frac{1}{x^2+x+1}-\frac{1}{y^2+y+1}\)
\(=-\frac{x^2+y^2+x+y+2}{\left(x^2+x+1\right)\left(y^2+y+1\right)}=-\frac{x^2+y^2+3}{x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+x+y+1}\)
\(=-\frac{\left(x+y\right)^2-2xy+3}{x^2y^2+x^2+y^2+2xy+2}=-\frac{4-2xy}{x^2y^2+3}=\frac{2\left(xy-2\right)}{x^2y^2+3}\)
từ đó ta có đpcm
Điều kiện: \(x\ne1\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x-2}{x^2+x+1}\)
Vì \(x\ne1\)nên:
\(\Leftrightarrow\frac{x-2}{x^2+x+1}=\frac{x-2}{x^2+x+1}\)Đúng với mọi \(x\ne1\).ĐPCM
cách lm dễ nhất là bn nhân chéo.....roi phân tích đa thức thành nhân tử là đc