Question

Please solve this pleaseeee ​

Answer 1

$\large\underline{\text{Important concepts}}$

Given that $\dfrac{a}{b}=\dfrac{c}{d},\ bd\neq0$, supposing denominator is nonzero

$\red{\bigstar}$Componendo

$\implies\dfrac{a+b}{b}=\dfrac{c+d}{d}$

$\implies\dfrac{a+kb}{a+kb}=\dfrac{c+kd}{c+kd}$

$\red{\bigstar}$Dividendo

$\implies\dfrac{a-b}{b}=\dfrac{c-d}{d}$

$\implies\dfrac{a+kb}{a-kb}=\dfrac{c+kd}{c-kd}$

$\red{\bigstar}$Invertendo

$\implies\dfrac{b}{a}=\dfrac{d}{c}$

$\red{\bigstar}$Alternando

$\implies\dfrac{b}{c}=\dfrac{d}{a}$

$\red{\bigstar}$Componendo et Dividendo

From Dividendo, where $k=1$

$\implies\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}$

$\large\underline{\text{Solution, Question 22}}$

Consider

$\implies x=\dfrac{x}{1}$

It is given that

$\implies x=\dfrac{\sqrt[3]{m+1}+\sqrt[3]{m-1}}{\sqrt[3]{m+1}-\sqrt[3]{m-1}}$

$\red{\bigstar}$Componendo et Dividendo

$\implies\dfrac{x+1}{x-1}=\sqrt[3]{\dfrac{m+1}{m-1}}$

Cube both sides

$\implies\left(\dfrac{x+1}{x-1}\right)^{3}=\dfrac{m+1}{m-1}$

$\implies\dfrac{x^{3}+3x^{2}+3x+1}{x^{3}-3x^{2}+3x-1}=\dfrac{m+1}{m-1}$

$\red{\bigstar}$Componendo et Dividendo

$\implies\dfrac{x^{3}+3x}{3x^{2}+1}=m$

Cross multiplication

$\implies x^{3}+3x=m(3x^{2}+1)$

$\implies x^{3}-3mx^{2}+3x=m$

$\large\underline{\text{Solution, Question 23, A}}$

Consider

$\implies x=\dfrac{x}{1}$

Given that

$\implies\dfrac{x}{1}=\dfrac{\sqrt{2a+1}+\sqrt{2a-1}}{\sqrt{2a+1}-\sqrt{2a-1}}$

$\red{\bigstar}$Componendo et Dividendo

$\implies\dfrac{x+1}{x-1}=\dfrac{2\sqrt{2a+1}}{2\sqrt{2a-1}}$

$\implies\dfrac{x+1}{x-1}=\sqrt{\dfrac{2a+1}{2a-1}}$

Square both sides

$\implies\dfrac{x^{2}+2x+1}{x^{2}-2x+1}=\dfrac{2a+1}{2a-1}$

$\red{\bigstar}$Componendo et Dividendo

$\implies\dfrac{x^{2}+1}{2x}=2a$

Cross multiplication

$\implies x^{2}+1=4ax$

$\large\underline{\text{Solution, Question 23, B}}$

Consider

$\implies x=\dfrac{x}{1}$

It is given that

$\implies\dfrac{x}{1}=\dfrac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$

$\red{\bigstar}$Componendo et Dividendo

$\implies\dfrac{x+1}{x-1}=\sqrt{\dfrac{a+3b}{a-3b}}$

Square both sides

$\implies\dfrac{x^{2}+2x+1}{x^{2}-2x+1}=\dfrac{a+3b}{a-3b}$

$\red{\bigstar}$Componendo et Dividendo

$\implies\dfrac{x^{2}+1}{2x}=\dfrac{a}{3b}$

Cross multiplication

$\implies3b(x^{2}+1)=2ax$

$\implies3bx^{2}-2ax+3b=0$

$\large\underline{\text{Solution, Question 24}}$

Consider

$\implies b=\dfrac{b}{1}$

It is given that

$\implies\dfrac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=\dfrac{b}{1}$

$\red{\bigstar}$Componendo et Dividendo

$\implies\sqrt{\dfrac{a+x}{a-x}}=\dfrac{b+1}{b-1}$

Square both sides

$\implies\dfrac{a+x}{a-x}=\dfrac{b^{2}+2b+1}{b^{2}-2b+1}$

$\red{\bigstar}$Componendo et Dividendo

$\implies\dfrac{a}{x}=\dfrac{b^{2}+1}{2b}$

Hence

$\implies x=\dfrac{2ab}{b^{2}+1}$

$\large\underline{\text{Solution, Question 25, A}}$

It is given that

$\implies\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\dfrac{4x-1}{2}$

$\red{\bigstar}$Componendo et Dividendo

$\implies\sqrt{\dfrac{x+1}{x-1}}=\dfrac{4x+1}{4x-3}$

Squaring both sides

$\implies\dfrac{x+1}{x-1}=\dfrac{4x+1}{4x-3}$

Cross multiplication

$\implies(x+1)(4x-3)=(x-1)(4x+1)$

$\implies 4x^{2}+x-3=4x^{2}-3x-1$

$\implies4x=2$

$\implies x=\dfrac{1}{2}$

$\large\underline{\text{Solution, Question 25, B}}$

It is given that

$\implies\dfrac{\sqrt{x+4}+\sqrt{x-10}}{\sqrt{x+4}-\sqrt{x-10}}=\dfrac{5}{2}$

$\red{\bigstar}$Componendo et Dividendo

$\implies\sqrt{\dfrac{x+4}{x-10}}=\dfrac{7}{3}$

Squaring both sides

$\implies\dfrac{x+4}{x-10}=\dfrac{49}{9}$

Cross multiplication

$\implies 9(x+4)=49(x-10)$

$\implies -40x=-526$

$\implies x=\dfrac{263}{20}$

$\large\underline{\text{Solution, Question 25, C}}$

It is given that

$\implies\dfrac{x^{3}+3x}{3x^{2}+1}=\dfrac{341}{91}$

$\red{\bigstar}$Componendo et Dividendo

$\implies\dfrac{x^{3}+3x^{2}+3x+1}{-x^{3}+3x^{2}-3x+1}=\dfrac{432}{250}$

$\implies-\left(\dfrac{x+1}{x-1}\right)^{3}=\dfrac{216}{125}$

$\implies\left(\dfrac{x+1}{x-1}\right)^{3}=\left(-\dfrac{6}{5}\right)^{3}$

Putting cube root

$\implies\dfrac{x+1}{x-1}=\dfrac{6}{-5}$

$\red{\bigstar}$Componendo et Dividendo

$\implies x=\dfrac{1}{11}$

Since the answer exceeded 5000 characters, I am putting the rest on the attachment.

Answer 2

Step-by-step explanation:

topic :

• componedo and dividendo

to find :

• what is use in thse number
• componedo and dividendo

solution :

• please check the attached file where you find your answer