सरलीकरण MCQ Quiz in मराठी - Objective Question with Answer for Simplification - मोफत PDF डाउनलोड करा

दिलेले आहे:

\(\rm x = \frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}}\)

वापरलेली संकल्पना:

जर x + 1/x = a असेल, तर x² + 1/x² = a² - 2

गणना:

⇒ \(\rm x = \frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}}\)

⇒ \(\rm \frac{1}{x} = \frac{{\sqrt 5 + 2}}{{\sqrt 5 - 2}}\)

⇒ x + 1/x = \(\frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}} + \frac{{\sqrt 5 + 2}}{{\sqrt 5 - 2}}\)

⇒ x + 1/x = \(\frac{(\sqrt 5 - 2)^2 + (\sqrt 5 + 2)^2}{(\sqrt 5 +2)(\sqrt 5 - 2)}\)

⇒ x + 1/x = 18

⇒ x2 + 1/x2 = a2 - 2

⇒ x2 + 1/x² = 18² - 2

⇒ 322

∴ पर्याय 4 योग्य आहे.

गणना:

\(√{(159-√{(244-√{(375-√{196})})})}\)

\(√{(159-√{(244-√{(375- 14)})})}\)

\(√{(159-√{(244-√{(361)})})}\)

\(√{(159-√{(244-19)})}\)

\(√{(159-√{(225)})}\)

\(√{(159-15)}\)

√144

12

∴ पर्याय 4 हे योग्य उत्तर आहे.

गणना:

⇒ (24 × 36 × 24 × 36) ÷ (2(\(\sqrt {1296 \div 2)} {)^2}\)

वापरलेली संकल्पना:

a ^m xa ⁿ = a ^{m + n}

आणि a ^m = (1/a) ^-m

गणना:

\(\left( {\frac{{67}}{{63}}} \right)^2 \times \left( {\frac{{67}}{{63}}} \right)^{-1 } \times \left( {\frac{{67}}{{63}}} \right)^{-2} \times \left( {\frac{{63}}{{67}}} \right) \)

\( ⇒ \left( {\frac{{67}}{{63}}} \right)^2 \times \left( {\frac{{67}}{{63}}} \right)^{- 1} \times \left( {\frac{{67}}{{63}}} \right)^{-2} \times \left( {\frac{{67}}{{63}}} \right )^{-1} \)

\( ⇒ \left( {\frac{{67}}{{63}}} \right)^{2-1-2-1} = \left( {\frac{{67}}{{63}} } \right)^{-2}= \left( {\frac{{63}}{{67}}} \right)^{2}\)

\( ⇒ \left(\frac{63^2}{67^2} \right) = \frac{3969}{4489}\)

0.7

\(0.\bar 7 = 0.77777 \ldots\)

\(0.0\bar 7 = 0.077777 \ldots\)

\(0.\overline {07} = 0.070707 \ldots\)

आता, 0.7777…  किंवा \(0.\bar 7\) ही संख्या सर्वात मोठी आहे.

उकल:

\(12\frac{1}{2} + 12\frac{1}{3} + 12\frac{1}{6}\)

= 25/2 + 37/3 + 73/6

= (75 + 74 + 73)/6

= 222/6

= 37

Shortcut Trick

\(12\frac{1}{2} + 12\frac{1}{3} + 12\frac{1}{6}\)

= 12 + 12 + 12 + (1/2 + 1/3 + 1/6)

= 36 + 1 = 37

वापरलेले सुत्र:

(a + b)² = a² + b² + 2ab

गणना:

दिलेली पदावली आहे:

​\(\sqrt {8\; + \;2\sqrt {15} \;} \)​​​

⇒ \(\sqrt {5\; + \;3\; + \;2\times \sqrt 5 \times \sqrt 3 \;} \)

⇒  \(\sqrt {{{(\sqrt 5 )}^2}\; + \;{{\left( {\sqrt 3 } \right)}^2}\; + \;2 \times \sqrt 5 \times \sqrt 3 \;} \)

⇒  \(\sqrt {{{\left( {\;\sqrt 5 \; + \;\sqrt 3 \;} \right)}^2}\;} \)

⇒  \(\sqrt 5 + \sqrt 3 \)

\(\sqrt {{{\left( {0.65} \right)}^2} - {{\left( {0.16} \right)}^2}} \)

त्याचप्रमाणे,

a2 - b2 = (a - b) ( a + b)

\(\begin{array}{l} \Rightarrow \sqrt {\left( {0.65 + 0.16} \right)\left( {0.65 - 0.16} \right)} \\ \Rightarrow \sqrt {\left( {0.81} \right)\left( {0.49} \right)} \\ \Rightarrow \sqrt {\left( {0.9} \right)\left( {0.9} \right) \times \left( {0.7} \right)\left( {0.7} \right)} \end{array}\)

⇒ 0.9 × 0.7 = 0.63

∴ उत्तर 0.63 आहे

संकल्पना:

आपण गुणखंडन पद्धतीने x शोधू शकतो.

हिशोब:

√[(10 + √25) (12 - √49)]

⇒ √(10 + 5)(12 – 7)

⇒ √(15 × 5)

⇒ √(3 × 5 × 5)

दिलेले आहे,

2³ × 3⁴ × 1080 ÷ 15 = 6^x

⇒ 2³ × 3⁴ × 72 = 6^x

⇒ 2³ × 3⁴ × (2 × 6²) = 6^x

⇒ 2⁴ × 3⁴ × 6² = 6^x

⇒ (2 × 3)⁴ × 6² = 6^x           [∵ x^m × y^m = (xy)^m]

⇒ 6⁴ × 6² = 6^x

⇒ 6^{(4 + 2)} = 6^x

⇒ x = 6

दिलेले आहे:

√3n = 729

वापरलेले सूत्र:

(x^a)^b = x^ab

जर x^a = x^b तर a = b 

गणना:

√3n = 729

⇒ √3n = (3²)³

⇒ (3n)^1/2 = (32)3

⇒ (3n)1/2 = 3⁶

⇒ n/2 = 6 

∴  n = 12 

दिलेले समीकरण,

(81.84 + 118.16) ÷ 53 = 1.2 × 2 + ?

⇒ 200 ÷ 53 = 1.2 × 2 + ?

⇒ 200 ÷ 125 = 1.2 × 2 +?

⇒ 1.6 = 2.4 + ?

पद्धत I: (3 + 2√5)²

= (32 + (2√5)2 + 2 × 3 × 2√5)

= 9 + 20 + 12√5 = 29 + 12√5

तुलना केल्यास, 29 + 12√5 = 29 + K√5

आपल्याकडे,

K = 12

Additional Information

29 + 12√5 = 29 + K√5

⇒ K√5 = 29 - 29 + 12√5

⇒ K√5 = 12√5

∴ K = 12