Question

PRECISO DE AJUDA URGENTE!!!!!!!!!!!!!!!!

Answer 1

a. L = 4-7+5 = 2

b. L =$\frac{9-6-3}{5+9}= \frac{0}{14} = 0$

c. L= $( \frac{12-4-5}{-4+6+4} )^{3} = ( \frac{3}{6} )^{3} = ( \frac{1}{2} )^{3} = \frac{1}{8}$

d. L = $\sqrt{ \frac{2-3-3}{-5-4} } = \sqrt{ \frac{-4}{-9} } = \sqrt{ \frac{4}{9} } = \frac{2}{3}$

e. L = $\sqrt[3]{ \frac{-24-20+2+3}{-8+3} } = \sqrt[3]{ \frac{-39}{-5} } = \sqrt[3]{ \frac{39}{5} } =1,98$

f. L =$\frac{ \sqrt{8+6+2} }{6-8} = \frac{ \sqrt{16} }{-2} =-2$



a. L = $\frac{(x-1)(x+1)}{(x-1)}= x+1 = 1+1 = 2$

b. L = $\frac{(2-x)(2+x)}{(2+x)} =2-x=2+2 = 4$

c. L = $\frac{(2x-1)(x+3)}{(2x-1)(x-2)} = \frac{x+3}{x-2} = \frac{1/2+3}{1/2-2} = \frac{7/2}{-3/2} =- \frac{7}{3}$

d. L =$\frac{3 x^{2} }{2x} = \frac{3x}{2} = \frac{3}{2}$

e. L =$\frac{3 x^{2} }{-2x} =- \frac{3x}{2} =- \frac{-6}{2} = 3$

f. L = $\frac{3 x^{2} -6x+6}{3 x^{2} -8x+8} = \frac{3-6+6}{3-8+8} = \frac{3}{3} =1$