Simplify :
713−21313+31477\sqrt{\dfrac{1}{3}} - 2\dfrac{1}{3}\sqrt{\dfrac{1}{3}} + 3\sqrt{147}731−23131+3147 .
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Given:
713−21313+3147=713−7313+34413=713−7313+3×2113=713−7313+6313=13(7−73+63)=13(70−73)=13(210−73)=13(2033)=203333×3=203397\sqrt{\dfrac{1}{3}} - 2\dfrac{1}{3}\sqrt{\dfrac{1}{3}} + 3\sqrt{147}\\[1em] = 7\sqrt{\dfrac{1}{3}} - \dfrac{7}{3}\sqrt{\dfrac{1}{3}} + 3\sqrt{\dfrac{441}{3}}\\[1em] = 7\sqrt{\dfrac{1}{3}} - \dfrac{7}{3}\sqrt{\dfrac{1}{3}} + 3 \times 21\sqrt{\dfrac{1}{3}}\\[1em] = 7\sqrt{\dfrac{1}{3}} - \dfrac{7}{3}\sqrt{\dfrac{1}{3}} + 63\sqrt{\dfrac{1}{3}}\\[1em] = \sqrt{\dfrac{1}{3}}\Big(7 - \dfrac{7}{3} + 63\Big)\\[1em] = \sqrt{\dfrac{1}{3}}\Big(70 - \dfrac{7}{3}\Big)\\[1em] = \sqrt{\dfrac{1}{3}}\Big(\dfrac{210 - 7}{3}\Big)\\[1em] = \sqrt{\dfrac{1}{3}}\Big(\dfrac{203}{3}\Big)\\[1em] = \dfrac{203\sqrt{3}}{3\sqrt{3} \times \sqrt{3}}\\[1em] = \dfrac{203\sqrt{3}}{9}731−23131+3147=731−3731+33441=731−3731+3×2131=731−3731+6331=31(7−37+63)=31(70−37)=31(3210−7)=31(3203)=33×32033=92033
Hence, the value of 713−21313+3147=203397\sqrt{\dfrac{1}{3}} - 2\dfrac{1}{3}\sqrt{\dfrac{1}{3}} + 3\sqrt{147} = \dfrac{203\sqrt{3}}{9}731−23131+3147=92033.
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