Question

Find rational numbers a and b such that ​

Answer 1

Answer:

The value are a=3 b =-2

Step-by-step explanation:

Given expression

\frac{\sqrt2-1}{\sqrt2+1}=a+b\sqrt2

2

+1

2

−1

=a+b

2

we have to find the rational number a and b

\frac{\sqrt2-1}{\sqrt2+1}=a+b\sqrt2

2

+1

2

−1

=a+b

2

Rationalizing left side

\frac{\sqrt2-1}{\sqrt2+1}\times \frac{\sqrt2-1}{\sqrt2-1}=a+b\sqrt2

2

+1

2

−1

×

2

−1

2

−1

=a+b

2

\frac{(\sqrt2-1)^2}{(\sqrt2+1)(\sqrt2-1)}=a+b\sqrt2

(

2

+1)(

2

−1)

(

2

−1)

2

=a+b

2

\frac{2+1-2\sqrt2}{2-1}=a+b\sqrt2

2−1

2+1−2

2

=a+b

2

3-2\sqrt2=a+b\sqrt23−2

2

=a+b

2

Comparing above we get

a=3 and b=-2

Answer 2

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