Question

please please solve it fast​

Answer 1

Answer:

1/√15+√13+1/√13+√11+1/√11-3-1/6-√32.

1/√15+√13.

Rationalising denominator of √15+√13 is √15-√13.

Multiply numerator & denominator by √15-√13.

1/√15+√13×√15-√13/√15-√13.

√15-√13/(√15)²-(√13)². (Because (a+b)(a-b)=a²-b² where a=√15 & b=√13).

√15-√13/15-13. (Because √ & ² will cancel).

√15-√13/2.

1/√13+√11.

Rationalising denominator of √13+√11 is √13-√11.

Multiply numerator & denominator by √13-√11.

1/√13+√11×√13-√11/√13-√11.

√13-√11/(√13)²-(√11)². (Because (a+b)(a-b)=a²-b² where a=√13 & b=√11).

√13-√11/13-11. (Because √ & ² will cancel).

√13-√11/2.

1/√11-3.

Rationalising denominator of √11-3 is √11+3.

Multiply numerator & denominator by √11+3.

1/√11-3×√11+3/√11+3.

√11+3/(√11)²-(3)². (Because (a-b)(a+b)=a²-b² where a=√11 & b=3).

√11+3/11-9. (Because √ & ² will cancel).

√11+3/2.

1/6-√32.

Rationalising denominator of 6-√32 is 6+√32.

Multiply numerator & denominator by 6+√32.

1/6-√32×6+√32/6+√32.

6+√32/(6)²-(√32)². (Because (a-b)(a+b)=a²-b² where a=6 & b=√32).

6+√32/36-32. (Because √ & ² will cancel).

6+√32/4.

1/√15+√13+1/√13+√11+1/√11-3-1/6-√32=(√15-√13/2)+(√13-√11/2)+(√11+3/2)-(6+√32/4).

2(√15-√13)+2(√13-√11)+2(√11+3)-(6+√32)/4. (Because LCM of 2, 2, 2 & 4 is 4).

(2√15-2√13)+(2√13-2√11)+(2√11+2(3))-(6-√32)/4. (Because +×-=- & -×+=-).

2√15-2√13+2√13-2√11+2√11+6-6-√32/4.

2√15-√32/4. (Because -×+=- & +×-=- so, -2√13+2√13, -2√11+2√11 & +6-6 will cancel i.e., it will be zero(0)).

I think this is your answer.

Answer 2

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1/√15+√13+1/√13+√11+1/√11-3-1/6-√32.

1/√15+√13.

Rationalising denominator of √15+√13 is √15-√13.

Multiply numerator & denominator by √15-√13.

1/√15+√13×√15-√13/√15-√13.

√15-√13/(√15)²-(√13)². (Because (a+b)(a-b)=a²-b² where a=√15 & b=√13).

√15-√13/15-13. (Because √ & ² will cancel).

√15-√13/2.

1/√13+√11.

Rationalising denominator of √13+√11 is √13-√11.

Multiply numerator & denominator by √13-√11.

1/√13+√11×√13-√11/√13-√11.

√13-√11/(√13)²-(√11)². (Because (a+b)(a-b)=a²-b² where a=√13 & b=√11).

√13-√11/13-11. (Because √ & ² will cancel).

√13-√11/2.

1/√11-3.

Rationalising denominator of √11-3 is √11+3.

Multiply numerator & denominator by √11+3.

1/√11-3×√11+3/√11+3.

√11+3/(√11)²-(3)². (Because (a-b)(a+b)=a²-b² where a=√11 & b=3).

√11+3/11-9. (Because √ & ² will cancel).

√11+3/2.

1/6-√32.

Rationalising denominator of 6-√32 is 6+√32.

Multiply numerator & denominator by 6+√32.

1/6-√32×6+√32/6+√32.

6+√32/(6)²-(√32)². (Because (a-b)(a+b)=a²-b² where a=6 & b=√32).

6+√32/36-32. (Because √ & ² will cancel).

6+√32/4.

1/√15+√13+1/√13+√11+1/√11-3-1/6-√32=(√15-√13/2)+(√13-√11/2)+(√11+3/2)-(6+√32/4).

2(√15-√13)+2(√13-√11)+2(√11+3)-(6+√32)/4. (Because LCM of 2, 2, 2 & 4 is 4).

(2√15-2√13)+(2√13-2√11)+(2√11+2(3))-(6-√32)/4. (Because +×-=- & -×+=-).

2√15-2√13+2√13-2√11+2√11+6-6-√32/4.

2√15-√32/4. (Because -×+=- & +×-=- so, -2√13+2√13, -2√11+2√11 & +6-6 will cancel i.e., it will be zero(0)).