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Question Number 204733 by hardmath last updated on 26/Feb/24

$$\mathrm{Find}:\frac{{59}^2+{48}^2+{41}^2-{30}^2}{{68}^2+{52}^2+{32}^2-{48}^2}=?$$

Answered by Rasheed.Sindhi last updated on 27/Feb/24

$$\frac{{({59}^2+{41}^2)}^{★1}+{48}^2-{(100-70)}^2}{{({68}^2+{32}^2)}^{★2}+{52}^2-{(100-52)}^2}$$$${}^{★1}{59}^2+{41}^2={(59+41)}^2-2\left(59\right)\left(41\right)$$$$={100}^2-4838$$$${}^{★2}{68}^2+{32}^2={(68+32)}^2-2\left(68\right)\left(32\right)$$$${100}^2-4352$$$$\frac{\cancel{{100}^2}-4838+{48}^2-\cancel{{100}^2}-{70}^2+2\left(100\right)\left(70\right)}{\cancel{{100}^2}-4352+\cancel{{52}^2}-\cancel{{100}^2}-\cancel{{52}^2}+2\left(100\right)\left(52\right)}$$$$\frac{-4838+{48}^2-{70}^2+14000}{-4352+10400}$$$$\frac{9162+(48-70)(48+70)}{-4352+10400}$$$$\frac{9162+(-22)\left(118\right)}{6048}$$$$=\frac{9162-2596}{6048}=\frac{6566}{6048}=\frac{469}{432}$$

Answered by Rasheed.Sindhi last updated on 27/Feb/24

$$\frac{({59}^2+{41}^2)+{48}^2-{30}^2}{({68}^2+{32}^2)+{52}^2-{48}^2}$$$$\frac{{(59+41)}^2-2\left(59\right)\left(41\right)+{48}^2-{30}^2}{{(68+32)}^2-2\left(68\right)\left(32\right)+{52}^2-{48}^2}$$$$\frac{{(59+41)}^2-{30}^2-4838+{48}^2}{{(68+32)}^2-{48}^2-4352+{52}^2}$$$$\frac{(59+41-30)(59+41+30)-4838+{48}^2}{(68+32-48)(68+32+48)-4352+{52}^2}$$$$\frac{\left(70\right)\left(130\right)-4838+{48}^2}{\left(52\right)\left(148\right)-4352+{52}^2}$$$$\frac{9100-4838+{48}^2}{7696-4352+{52}^2}$$$$\frac{4262+{48}^2}{3344+{52}^2}-1+1$$$$\frac{4262-3344+{48}^2-{52}^2}{3344+{52}^2}+1$$$$\frac{918+(48-52)(48+52)}{3344+2704}+1$$$$\frac{918+(-4)\left(100\right)}{6048}+1$$$$\frac{918+(-4)\left(100\right)}{6048}+1$$$$\frac{518}{6048}+1=1\frac{37}{432}$$

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