Advertisements
Advertisements
प्रश्न
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
Advertisements
उत्तर
Let 2x = 3y = 6-z = k
`rArr2=k^(1/x),` `3=k^(1/y),` `6=k^(1/-z)`
Now,
`6 = 2xx3=k^(1/-z)`
`rArrk^(1/x)xxk^(1/y)=k^(1/-z)`
`rArrk^(1/x+1/y)=k^(1/-z)`
`rArr1/x+1/y=1/-z`
`rArr1/x+1/y+1/z=0`
APPEARS IN
संबंधित प्रश्न
Simplify:-
`2^(2/3). 2^(1/5)`
Solve the following equations for x:
`2^(2x)-2^(x+3)+2^4=0`
Solve the following equations for x:
`3^(2x+4)+1=2.3^(x+2)`
Show that:
`(x^(a-b))^(a+b)(x^(b-c))^(b+c)(x^(c-a))^(c+a)=1`
Solve the following equation:
`3^(x+1)=27xx3^4`
Simplify:
`root(lm)(x^l/x^m)xxroot(mn)(x^m/x^n)xxroot(nl)(x^n/x^l)`
Write the value of \[\sqrt[3]{125 \times 27}\].
The seventh root of x divided by the eighth root of x is
If a, b, c are positive real numbers, then \[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}\] is equal to
When simplified \[(256) {}^{- ( 4^{- 3/2} )}\] is
