C-11-Q089mediumsingle_mcq

If $ax = by = cz$ , then $\dfrac{x^2}{yz} + \dfrac{y^2}{zx} + \dfrac{z^2}{xy}$ equals—

a $\dfrac{bc}{a^2} + \dfrac{ca}{b^2} + \dfrac{ab}{c^2}$
b $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}$
c $abc$
d $a + b + c$

ব্যাখ্যা

Letting $ax=by=cz=k$ gives $x=\frac{k}{a},\,y=\frac{k}{b},\,z=\frac{k}{c}$ , so $\frac{x^2}{yz}=\frac{bc}{a^2}$ . Applying this to each term produces $\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}$ .

এই অধ্যায়ের আরও প্রশ্ন

C-11-Q090If are continued proportional, then
C-11-Q091If are continued proportional, then equals—
C-11-Q092Given , the value of is—
C-11-Q093If , then
C-11-Q094The solutions of Example 5, where , include—
C-11-Q095The condition in Example 5 ensures—
C-11-Q096By componendo-dividendo on , the next simplified form is—
C-11-Q097If , then equals—
C-11-Q098If , with positive, equals—
C-11-Q099If , then equals—

Algebraic Ratio and Proportion — সব প্রশ্ন দেখুন (180টি) →

If $ax = by = cz$ , then $\dfrac{x^2}{yz} + \dfrac{y^2}{zx} + \dfrac{z^2}{xy}$ equals—

\dfrac{bc}{a^2} + \dfrac{ca}{b^2} + \dfrac{ab}{c^2}

\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}

abc

a + b + c

ব্যাখ্যা

Letting $ax=by=cz=k$ gives $x=\frac{k}{a},\,y=\frac{k}{b},\,z=\frac{k}{c}$ , so $\frac{x^2}{yz}=\frac{bc}{a^2}$ . Applying this to each term produces $\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}$ .

If ax=by=czax = by = czax=by=cz, then x2yz+y2zx+z2xy\dfrac{x^2}{yz} + \dfrac{y^2}{zx} + \dfrac{z^2}{xy}yzx2​+zxy2​+xyz2​ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

If ax=by=czax = by = czax=by=cz, then x2yz+y2zx+z2xy\dfrac{x^2}{yz} + \dfrac{y^2}{zx} + \dfrac{z^2}{xy}yzx2​+zxy2​+xyz2​ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

If $ax = by = cz$ , then $\dfrac{x^2}{yz} + \dfrac{y^2}{zx} + \dfrac{z^2}{xy}$ equals—

If $ax = by = cz$ , then $\dfrac{x^2}{yz} + \dfrac{y^2}{zx} + \dfrac{z^2}{xy}$ equals—