- Please review the new safety procedures and  direct any questions to Mr.Bae at extention 2528.

direct

242424+252+2524

=242676+2524

=245200

242424+252+2524

=242676+2524

=245200

242424 + 252 + 2524

= 242676 + 2524

= 245200

Mẫu số bằng \(\sqrt{50}+\sqrt{250}=5\sqrt{2}+5\sqrt{10}=5\sqrt{2}\left(1+\sqrt{5}\right).\)

Kí hiệu tử số là \(A\) thì ta có 

\(A^2=\left(\sqrt{8+\sqrt{40+8\sqrt{5}}}+\sqrt{8-\sqrt{40+8\sqrt{5}}}\right)^2\)

       \(=8+\sqrt{40+8\sqrt{5}}+2\sqrt{8+\sqrt{40+8\sqrt{5}}}\cdot\sqrt{8-\sqrt{40+8\sqrt{5}}}+8-\sqrt{40+8\sqrt{5}}\)

       \(=16+2\sqrt{\left(8+\sqrt{40+8\sqrt{5}}\right)\left(8-\sqrt{40+8\sqrt{5}}\right)}\)

      \(=16+2\sqrt{8^2-\left(40+8\sqrt{5}\right)}=16+2\sqrt{24-8\sqrt{5}}\)

      \(=16+2\sqrt{4-2\cdot2\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2}=16+2\sqrt{\left(2-2\sqrt{5}\right)^2}\)

      \(=16+2\left|2-2\sqrt{5}\right|=16-4+4\sqrt{5}=12+4\sqrt{5}=4\left(3+\sqrt{5}\right).\)

Vậy  \(A=4\left(3+\sqrt{5}\right)=2\left(6+2\sqrt{5}\right)=2\left(\sqrt{5}+1\right)^2.\)

Thành thử  \(B=\frac{2\left(\sqrt{5}+1\right)^2}{5\sqrt{2}\left(1+\sqrt{5}\right)}=\frac{\sqrt{2}\left(\sqrt{5}+1\right)}{5}=\frac{\sqrt{10}+\sqrt{2}}{5}.\)

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\(B=\dfrac{\left|x^3-3x^2+3x-1\right|}{x^3-3x^2+x}\)

\(=\dfrac{\left|\left(x-1\right)^3\right|}{x\left(x^2-2x+1\right)}\)

\(=\dfrac{\left|\left(x-1\right)^3\right|}{x\left(x-1\right)^2}=\pm\dfrac{x-1}{x}\)

\(B=\dfrac{\left|x^3-3x^2+3x-1\right|}{x^2-2x^3+x}\)

\(B=\dfrac{\left|x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\right|}{x\left(x^2-2x+1\right)}\)

\(B=\dfrac{\left|\left(x-1\right)^3\right|}{x\left(x-1\right)^2}\)

Khi \(x< 1,x\ne0\) 

\(B=\dfrac{1-x}{x}\) 

Khi \(x>1\) 

\(B=\dfrac{x-1}{x}\)