If x, y, z are three positive numbers, then the minimum value of -Maths 9th

If x, y, z are three positive numbers, then the minimum value of -Maths 9th
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Description : If x+y =10 and x=z then show that z+y =10 -Maths 9th

Last Answer : It is given that x+y =10 Also x= z Therefore, x+y =10 Z+y =10 ( x = z)

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Last Answer : It is given that x+y =10 Also x= z Therefore, x+y =10 Z+y =10 ( x = z)

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if xylogxy/x+y, yzlogyz/y+z and zxlogzx/z+x are mutually equal, then show that x^x= y^y=z^z -Maths 9th

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Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

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if xylogxy/x+y, yzlogyz/y+z and zxlogzx/z+x are mutually equal, then show that x^x= y^y=z^z -Maths 9th

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It is known that if x+y =10, then x+y+z = 10+z.Which axiom of Euclids does this statement illustrate? -Maths 9th

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If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Description : If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Last Answer : Let l+m−2nlogx​=m+n−2llogy​=n+l−2mlogz​=k(say) So, we get logx=k(l+m−2n) ....... (i) logy=k(m+n−2l) ....... (ii) logz=k(n+l−2m) ....... (iii) ∴logx+logy+logz=k(l+m−2n)+k(m+n−2l)+k(n+l−2m) ⇒logx+logy+logz=kl+km−2kn+km+kn−2kl+kn+kl−2km ⇒log(xyz)=0 ⇒logxyz=log1 ⇒xyz=1

If x = logabc, y = logbca, z = logc ab, then -Maths 9th

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Last Answer : (a) x + y + z + 2 = xyz.x = logabc ⇒ ax = bc ⇒ ax + 1 = abc ⇒ a = (abc)1/(x +1) Similarly, b = (abc)1/(y + 1), c = (abc)1/(z + 1)∴ abc = \((abc)^{rac{1}{x+1}+rac{1}{y+1}+rac{1}{z+1}}\)⇒ \({rac{1}{x+1}+ ... + xz + x + z + 1 + xy + y + x + 1 = xyz + xy + yz + zx + x + y + z + 1⇒ x + y + z + 2 = xyz.

If y = (1/(a^(1-loga x))), z = (1/(a^(1-loga y))) and x = a^k, then k = -Maths 9th

Description : If y = (1/(a^(1-loga x))), z = (1/(a^(1-loga y))) and x = a^k, then k = -Maths 9th

Last Answer : (b) \(rac{1}{{1-log_az}}\) y = \(rac{1}{a^{1-log_ax}}\) = \(a^{-(1-log_ax)}\)⇒ logay = \(rac{1}{{1-log_ax}}\) and loga z = \(rac{1}{{1-log_ay}}\)∴ logaz = \(rac{1}{1-\bigg(rac{1}{1-log_ax}\bigg)}\) = \( ... x = \(rac{1}{{1-log_az}}\)⇒ x = \(a^{rac{1}{1-log_az}}\) = ak ⇒ k = \(rac{1}{{1-log_az}}\).

If (log x)/(a^2+ab+b^2) = (log y)/(b^2+bc+c^2) = (log z)/(c^2+ca+a^2), then x^(a-b). y^(b-c). z^(c-a) = -Maths 9th

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Last Answer : (c) 1Let each ratio = k and base = e ⇒ loge x = k(a2 + ab + b2) ⇒ (a - b) loge x = k (a - b) (a2 + ab + b2) ⇒ loge xa - b = k(a3 - b3) ⇒ xa - b = \(e^{k(a^3-b^3)}\) Similarly, yb-c = \(e^{k(b^3-c^3)}\), zc-a = \ ... (e^{k(b^3-c^3)}\) . \(e^{k(c^3-a^3)}\)= \(e^{k[a^3-b^3+b^3-c^3+c^3-a^3]}\) = e0 = 1.

If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Description : If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Last Answer : (b) 1Let \(rac{ ext{log}\,x}{l+m-2n}\) = \(rac{ ext{log}\,y}{m+n-2l}\) = \(rac{ ext{log}\,z}{n+l-2m}\) = k. Thenlog x = k(l + m – 2n), log y = k(m + n – 2l); log z = k(n + l – 2m) ⇒ log x + log y + log z = k(l + m – 2n) + k(m + n – 2l) + k(n + l – 2m)⇒ log(xyz) = 0 ⇒ log(xyz) = log 1 ⇒ xyz = 1.

If x = log2a a, y = log3a2a, z = log4a3a, then xyz – 2yz equals -Maths 9th

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Last Answer : (d) -1\(x\) = log2a a = \(rac{ ext{log}\,a}{ ext{log}\,2a}\), y = log3a 2a = \(rac{ ext{log}\,2a}{ ext{log}\,3a}\)z = log4a 3a = \(rac{ ext{log}\,3a}{ ext{log}\,4a}\)∴ xyz - 2yz = \(rac{ ext{log}\ ... \(rac{ ext{log}\,(4a)^{-1}}{ ext{log}\,(4a)}\) = \(rac{-1. ext{log}\,4a}{ ext{log}\,4a}\) = -1.

If x + y + z = 0, then what is [(y-z-x)/2]^3 + [(z-x-y)/2]^3 + [(x-y-z)/2]^3 equal to ? -Maths 9th

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If(1/(x+z))+(1/(z+x)) = (2/(x+y)), then what is ( x^2 + y^2) equal to ? -Maths 9th

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If x + y + z = 0, then x (y – z)^3 + y (z – x)^3 + z (x – y)^3 equals -Maths 9th

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If a = (xy)/(x+y), b = (xz)/(x+z), and c = (yz)/(y+z), where a, b and c are non-zero, then what is x equal to ? -Maths 9th

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If x + y + z = 2s, then what is (s – x)^3 + (s – y)^3 + 3(s – x) (s – y)z equal to : -Maths 9th

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