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EXERCISE 63. REVIEW.

When a = 2, b = - 2, and c = 4; find the value of :

1. √a3+b3 + c3 - (a - b - c).

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3. (b - 5) (b - 4) – 3 (b - 2) (b − 1) + 3 (b + 1) (6 + 2).

3 x3 + 10 x2 + 7x-2

4. Reduce to lowest terms

3x2 + 13x2 + 17 x + 6

5.

Simplify:

1

x-3

(x - 3) (x - 2) (x - 1)(x - 3) (x - 1) (x - 2)

- - 1)(x-3) + (x-1)(x-2)

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HINT. Add the first two fractions; then their sum and the third fraction and this result to the fourth fraction.

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12.

X

x2 + 6 xy + 5 y2 ^^ x2 + 4x +4x+2

z2 + xy

13.

x2 + yz

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zx + zh

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(x - y) (x - 2) (y - z) (y - x) (z - x) (z - y)

14. (1+1)+(1-1).

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18.(+++)+(+24)

x+y

2

19.(1-1)(1+2+1-243).

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(x + 1) (x + 2)

1+2x2

24.(1-1+1)(1).

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Multiply by 33, the L. C. M. of the denominators.

Then,

11x-3x + 3 = 33x - 297,

11x-3x-33 x = 297 3,

- 25 x = 300.

.. x = 12.

NOTE. Since the minus sign precedes the second fraction, in removing the denominator the sign of every term of the numerator is changed.

8

2. Solve 2+1-2-1-4-1

The L. C. D. = (2x + 1) (2x – 1).

Multiply by the L. C. D., and we have,

Reducing,

4x2 + 4x + 1 - (4x2 - 4x + 1) = 8. .. 4 x2 + 4x + 1-4x2+4x-1= 8.

x = 1.

174. To Clear an Equation of Fractions, therefore, Multiply each term by the L. C. M. of the denominators.

If a fraction is preceded by a minus sign, the sign of every term of the numerator must be changed when the denominator is removed.

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