4abc-1-3abc- -2-3 abc - 0

The correct option is B 5abcLet the term to be filled in the blank be x. Then, 4abc+1/3abc+ x+2/3abc=0 x=0 4abc 1/3abc 2/3abc x=0 4abc (1/3+2/3)abc x= 4abc ab ...

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Byju's Answer

Standard IX

Mathematics

Commutative Property

4abc+1/3abc+ ...

Question

$4 a b c + \frac{1}{3} a b c +$ _______ + $\frac{2}{3} a b c$ = 0

5 a b c

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- 5 a b c

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3 a b c

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- 3 a b c

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Solution

The correct option is B $- 5 a b c$
Let the term to be filled in the blank be $x$ .

Then,
$4 a b c + \frac{1}{3} a b c + x + \frac{2}{3} a b c = 0$

$⟹ x = 0 - 4 a b c - \frac{1}{3} a b c - \frac{2}{3} a b c$

$⟹ x = 0 - 4 a b c - (\frac{1}{3} + \frac{2}{3}) a b c$

$⟹ x = - 4 a b c - a b c$

$⟹ x = - 5 a b c$

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Similar questions

$4 a b c + \frac{1}{3} a b c +$ ___ $+$ $\frac{2}{3} a b c$ = 0

What value of $x$ makes the equation $4 a b c + \frac{1}{3} a b c + x$ + $\frac{2}{3} a b c$ =0 true ?

Show that:

$\sum a^{3} cos (B - C) = 3 a b c$

(or)

Prove that:

$a^{3} cos (B - C) + b^{3} cos (C - A) + c^{3} cos (A - B) = 3 a b c$

15 (a^{3} b^{2} c^{2} - a^{2} b^{3} c^{2} + a^{2} b^{2} c^{3}) \div 3 a b c

(6 a b - 3 a b c + 9 a b c d) \div (- \frac{1}{3} a b)

is equal to

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4abc+13abc+ _______ +23abc = 0

The correct option is B −5abcLet the term to be filled in the blank be x. Then, 4abc+13abc+x+23abc=0 ⟹x=0−4abc−13abc−23abc ⟹x=0−4abc−(13+23)abc ⟹x=−4abc−abc ⟹x=−5abc

$4 a b c + \frac{1}{3} a b c +$ _______ + $\frac{2}{3} a b c$ = 0