Math, asked by shreyash1505, 7 months ago

PLS.. SOLVE THESE......
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Answered by Anonymous
17

4^x -3^(x-1/2) =3^(x+1/2) -2^(2x-1)

=> 4^x +2^(2x-1) = 3^(x-1/2) +3^(x+1/2)

2^2x +2^2x * 2^-1 = (3^x-1/2 )* 3^x + 3^(1/2)* 3^x

=> 2^2x (1+1/2) = 3^x ( 3^(-1/2) + 3^(1/2))

=> 4^x * (3/2)= 3^ x * ( 4 / sqrt(3))

=> 4^x / 3^x = 8/ (3 sqrt(3)) = 8 sqrt(3) / 9

=> (4/3)^x = 8 sqrt(3) /9

Answered by ravanji786
1

Answer:

HELLO DEAR?!!!

HERE IS UR ANSWER::-(sin o + cos o ) ( sec o+ cosec o)

(sin o + cos o ) ( sec o+ cosec o)=(sin o+ cos o) (1/ cos o + 1/ sin o)

(sin o + cos o ) ( sec o+ cosec o)=(sin o+ cos o) (1/ cos o + 1/ sin o)= (sin o + cos o)(sin o+ cos o/sin o. cos o)

(sin o + cos o ) ( sec o+ cosec o)=(sin o+ cos o) (1/ cos o + 1/ sin o)= (sin o + cos o)(sin o+ cos o/sin o. cos o)=(sin o+ cos o)^2/ sin o . cos o

(sin o + cos o ) ( sec o+ cosec o)=(sin o+ cos o) (1/ cos o + 1/ sin o)= (sin o + cos o)(sin o+ cos o/sin o. cos o)=(sin o+ cos o)^2/ sin o . cos o=1+2sin o cos o/ sin o.cos o

(sin o + cos o ) ( sec o+ cosec o)=(sin o+ cos o) (1/ cos o + 1/ sin o)= (sin o + cos o)(sin o+ cos o/sin o. cos o)=(sin o+ cos o)^2/ sin o . cos o=1+2sin o cos o/ sin o.cos o=1/sin o. cos o+2 sin o cos o/ sin o cos o

(sin o + cos o ) ( sec o+ cosec o)=(sin o+ cos o) (1/ cos o + 1/ sin o)= (sin o + cos o)(sin o+ cos o/sin o. cos o)=(sin o+ cos o)^2/ sin o . cos o=1+2sin o cos o/ sin o.cos o=1/sin o. cos o+2 sin o cos o/ sin o cos o=1/sin o . 1/cos o+2

(sin o + cos o ) ( sec o+ cosec o)=(sin o+ cos o) (1/ cos o + 1/ sin o)= (sin o + cos o)(sin o+ cos o/sin o. cos o)=(sin o+ cos o)^2/ sin o . cos o=1+2sin o cos o/ sin o.cos o=1/sin o. cos o+2 sin o cos o/ sin o cos o=1/sin o . 1/cos o+2=2+cosec o . sin o

6\√5+√2×√5-√2\√5-√2

6\√5+√2×√5-√2\√5-√26√5-6√2\5-2

6\√5+√2×√5-√2\√5-√26√5-6√2\5-26√5-6√2\3

6\√5+√2×√5-√2\√5-√26√5-6√2\5-26√5-6√2\33[2√5-2√2]\1

6\√5+√2×√5-√2\√5-√26√5-6√2\5-26√5-6√2\33[2√5-2√2]\12√5-2√2.

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