Advertisements
Advertisements
प्रश्न
Fill in the blanks of the following.
`a/3 = b/4 = c/7 = (a-2b+3c)/("______") = ("______")/ (6 - 8 +14)`
उत्तर
⇒ `a/3 = b/4 = c/7 = [a-2b+3c]/bb16 = bb(2a -2b+2c)/{6-8+14}`
Explanation:
`a/3 = b/4 = c/7 = [-2 xx b]/[-2 xx 4] = [3 xx c]/[3 xx 7]`
⇒ `a/3 = b/4 = c/7 = [-2b]/-8 = [3c]/21 ={a +(-2b)+3c}/{3+(-8) + 21}` ...(Theorem of Equal ratios)
⇒ `a/3 = b/4 = c/7 = (a-2b +3c)/16`
Also,
`a/3 = b/4 = c/7 = (2 xx a)/(2 xx 3) = (-2 xx b)/(-2 xx 4) = [2 xx c]/[2 xx 7]`
⇒ `a/3 = b/4 = c/7 = (2a)/6 = (-2b)/-8 = (2c)/14 = [2a+(-2b)+2c]/[6+(-8)+14]`
⇒ `a/3 = b/4 = c/7 = [a-2b+3c]/16 = (2a -2b+2c)/{6-8+14}`
APPEARS IN
संबंधित प्रश्न
Fill in the blanks of the following.
`x/7 = y/3 = (3x + 5y)/("_____") = (7x -9y)/("_____")`
If a(y + z) = b(z + x) = c(x + y) and out of a, b, c no two of them are equal then show that,
`(y - z)/[a ( b - c )] = ( z - x)/[ b ( c - a)] = ( x - y)/[c ( a - b )]`
If `x/[3x - y -z] = y/[3y - z -x] = z/[3z -x -y]` and x + y + z ≠ 0 then show that the value of each ratio is equal to 1.
If `[ax + by]/( x + y) = ( bx + az )/(x + z) = (ay + bz)/[y + z]` and `x + y + z ≠ 0 ` then show that `[a + b]/2`.
If `(y + z)/a = (z + x )/b = (x + y)/c` then show that `x/[b + c - a ] = y/[c + a - b] = z/(a + b - c)`
Solve.
`[ 16x^2 - 20x +9]/[ 8x^2 + 12x + 21] = ( 4x - 5 )/( 2x + 3)`
Solve.
`( 5y^2 + 40y - 12)/( 5y + 10y^2 - 4) = ( y + 8)/( 1 + 2y)`
Solve:
`[12x^2 + 18x + 42]/[ 18 x^2 + 12x +58 ] = [ 2x + 3]/[ 3x + 2]`
If `[ 2x - 3y ]/[ 3z + y] = [ z - y ]/[ z - x ] = [ x + 3z ]/[ 2y - 3x]` then prove that every ratio = `x/y`.
If `[by + cz ]/[b^2 + c^2] = [cz + ax]/[c^2 + a^2] = [ax + by]/[a^2 + b^2]` then prove that `x/a= y/b = z/c`
