Lê Chí Cường Làm đúng mà

a) \(\dfrac{5-2\sqrt{5}}{\sqrt{5}}+\dfrac{20}{5+\sqrt{5}}\)

\(=\dfrac{\left(5-2\sqrt{5}\right)\sqrt{5}}{\sqrt{5}}+\dfrac{20}{5+\sqrt{5}}\)

\(=\dfrac{\left(5-2\sqrt{5}\right)\sqrt{5}}{5}+\dfrac{20\left(5-\sqrt{5}\right)}{20}\)

\(=\dfrac{5\sqrt{5}-10}{5}+5-\sqrt{5}\)

\(=\dfrac{5\left(\sqrt{5}-2\right)}{5}+5-\sqrt{5}\)

\(=\sqrt{5}-2+5-\sqrt{5}\)

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

\(=0+\left(-2+5\right)\)

\(=3\)

b) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)

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

\(=\dfrac{3\sqrt{3}+6}{3}+\dfrac{2\sqrt{2}+2}{2}-\sqrt{3}-\sqrt{2}\)

\(=\dfrac{3\left(\sqrt{3}+2\right)}{3}+\dfrac{2\left(\sqrt{2}+1\right)}{2}-\sqrt{3}-\sqrt{2}\)

\(=\sqrt{3}+2+\sqrt{2}+1-\sqrt{3}-\sqrt{2}\)

\(=\left(\sqrt{3}-\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{2}\right)+\left(2+1\right)\)

\(=3\)

a) \(x^3 + 8y^3\)

\(=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

b) \(a^6-b^6\)

\(=\left(a^3-b^3\right)\left(a^3+b^3\right)\)

c) \(8y^3-125\)

\(=\left(2y-5\right)\left(4y^2+10y+25\right)\)

d) \(8x^3+27\)

\(=\left(2x+3\right)\left(4x^2-6x+9\right)\)

\(x^6-y^6\)

\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)

m = 6 tạ => P = 6000 N

Để thấy đường

\(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\)

\(\Leftrightarrow2x^3\cdot\left(x-1\right)-7x^2\cdot\left(x-1\right)+7x\cdot\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^3-7x^2+7x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x^3-1\right)-7x\cdot\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x-1\right)\cdot\left(x^2+x+1\right)-7x\cdot\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[2\left(x^2+x+1\right)-7x\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2+2x+2-7x\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2-5x+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2-x-4x+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[x\cdot\left(2x-1\right)-2\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(x-2\right)\cdot\left(2x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-2\right)\cdot\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\x-2=0\\2x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{1}{2};x_2=1;x_3=2\)

Đẹp lắm!