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Đặt \(\sqrt{\text{x}}-\sqrt{y}=a\); \(\sqrt{y}-\sqrt{z}=b\); \(\sqrt{z}-\sqrt{x}=c\)
\(\Rightarrow a+b+c=0\). Ta sẽ chứng minh : \(a^3+b^3+c^3=3abc\)
Ta có : \(a+b+c=0\Rightarrow a=-\left(b+c\right)\Rightarrow a^3=-\left(b+c\right)^3\)
\(\Rightarrow a^3=-\left[b^3+c^3+3bc\left(b+c\right)\right]\Rightarrow a^3+b^3+c^3=-3bc\left(-a\right)=3abc\)
Mặt khác, ta lại có : \(a^3+b^3+c^3=0\left(gt\right)\Rightarrow3abc=0\Rightarrow abc=0\)
\(\Rightarrow a=0\)hoặc \(b=0\)hoặc \(c=0\)
Tu do de dang giai tiep bai toan!
\(a^2+b^2=\left(a+b-c\right)^2=a^2+\left(b-c\right)^2+2a\left(b-c\right)=b^2+\left(a-c\right)^2+2b\left(a-c\right)\)
\(\Rightarrow\left\{{}\begin{matrix}b^2=\left(b-c\right)^2+2a\left(b-c\right)\\a^2=\left(a-c\right)^2+2b\left(a-c\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\dfrac{\left(a-c\right)^2+2b\left(a-c\right)+\left(a-c\right)^2}{\left(b-c\right)^2+2a\left(b-c\right)+\left(b-c\right)^2}\)
\(=\dfrac{\left(a-c\right)\left(a+b-c\right)}{\left(b-c\right)\left(b+a-c\right)}=\dfrac{a-c}{b-c}\) (đpcm)
Ta có : \(3\sqrt{xyz}=\sqrt{x}^2+\sqrt{y}^3+\sqrt{z}^3\ge3\sqrt[3]{\sqrt{x}^3\sqrt{y}^3\sqrt{z}^3}=3\sqrt{x}\sqrt{y}\sqrt{z}=3\sqrt{xyz}.\)
Dấu = xảy ra
=> x =y =z
=> A = (1+1)(1+1)(1+1) =8
mk thấy nó sai sai . Tại sao 3\(\sqrt[3]{\sqrt{x}^3\sqrt{y}^3\sqrt{z}^3}\) = 3\(\sqrt{x}\sqrt{y}\sqrt{z}\)
\(\sqrt{x\left(1-y\right)\left(1-z\right)}=\sqrt{x\left(yz-y-z+1\right)}=\sqrt{x\left(yz-y-z+x+y+z+2\sqrt{xyz}\right)}\)
\(=\sqrt{x\left(yz+x+2\sqrt{xyz}\right)}=\sqrt{x^2+2x\sqrt{xyz}+xyz}=\sqrt{\left(x+\sqrt{xyz}\right)^2}\)
\(=x+\sqrt{xyz}\)
Tương tự: \(\sqrt{y\left(1-x\right)\left(1-z\right)}=y+\sqrt{xyz}\) ; \(\sqrt{z\left(1-x\right)\left(1-y\right)}=z+\sqrt{xyz}\)
\(\Rightarrow VT=x+y+z+3\sqrt{xyz}=1-2\sqrt{xyz}+3\sqrt{xyz}=1+\sqrt{xyz}\) (đpcm)
