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Suppose x = √2 + √3, y = √5 + √1, and z = 2√2 + 1. ; then which of the following is true?

Ax <y < z

Bz < y < x

Cy < x < z

Dx < z < y

Answer:

A. x <y < z

Read Explanation:

Approximate the values:

x=2+3≈1.414+1.732=3.146x = \sqrt{2} + \sqrt{3} \approx 1.414 + 1.732 = 3.146
y=5+1=2.236+1=3.236y = \sqrt{5} + \sqrt{1} = 2.236 + 1 = 3.236
z=22+1≈2(1.414)+1=3.828z = 2\sqrt{2} + 1 \approx 2(1.414) + 1 = 3.828

Therefore,

\boxed{z > y > x}


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