C-11-Q084mediumsingle_mcq

The proof of $\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}$ given $\dfrac{a}{b} = \dfrac{c}{d}$ uses the substitution—

a $a = bk,\ c = dk$
b $a + b = c + d$
c $abcd = 1$
d $a = c$

ব্যাখ্যা

Since $\frac{a}{b}=\frac{c}{d}=k$ , both equal a common ratio $k$ , allowing $a=bk$ and $c=dk$ . Substituting these expressions reduces both sides of the identity to the same function of $k$ , completing the proof.

এই অধ্যায়ের আরও প্রশ্ন

C-11-Q085If , then—
C-11-Q086The proof of given uses the substitution and the expansion—
C-11-Q087If , then—
C-11-Q088If and are not mutually equal, the value of each ratio is—
C-11-Q089If , then equals—
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—

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

The proof of $\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}$ given $\dfrac{a}{b} = \dfrac{c}{d}$ uses the substitution—

a = bk,\ c = dk

a + b = c + d

abcd = 1

a = c

The proof of a2+b2a2−b2=ac+bdac−bd\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}a2−b2a2+b2​=ac−bdac+bd​ given ab=cd\dfrac{a}{b} = \dfrac{c}{d}ba​=dc​ uses the substitution—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The proof of a2+b2a2−b2=ac+bdac−bd\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}a2−b2a2+b2​=ac−bdac+bd​ given ab=cd\dfrac{a}{b} = \dfrac{c}{d}ba​=dc​ uses the substitution—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The proof of $\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}$ given $\dfrac{a}{b} = \dfrac{c}{d}$ uses the substitution—

The proof of $\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}$ given $\dfrac{a}{b} = \dfrac{c}{d}$ uses the substitution—