Type 1
*
यदि $$a + b + c = 0$$ तो $$\frac{1}{(a+b)(b+c)} + \frac{1}{(a+c)(b+a)} + \frac{1}{(c+a)(c+b)} = ?$$
- 1
- 0
- -1
- -2
यदि $$a + b + c = 0$$ तो $$\frac{a^2}{b^2 – c a} + \frac{b^2}{c^2 – a b} + \frac{c^2}{a^2 – b c} = ?$$
- 0
- 1
- 3
- 2
यदि $$a + b + c = 0$$ तो $$(a-b)^2 + (b-c)^2 + (c-a)^2 = ?$$
- 1
- 3
- $$\frac{1}{3}$$
- 0
यदि $$a + b + c = 0$$, तो $$\frac{1}{a^2 + b^2 – c^2} + \frac{1}{b^2 + c^2 – a^2} + \frac{1}{c^2 + a^2 – b^2} = ?$$
- $$a^2 + b^2 + c^2$$
- 1
- -1
- 0
यदि $$a + b + c = 0$$, तब $$\frac{a+b}{ab} + \frac{b+c}{bc} + \frac{c+a}{ca} = ?$$
- 1
- -1
- 0
- 2
यदि $$a + b + c = 2s$$, तब $$(s-a)^2 + (s-b)^2 + (s-c)^2 + s^2 = ?$$
- $$a^2 + b^2 + c^2$$
- 0
- 1
- 2
यदि $$xy + yz + zx = 0$$ ($$x,y,z \neq 0$$), तब $$\frac{x^2}{y z} + \frac{y^2}{z x} + \frac{z^2}{x y} = ?$$
- 3
- 1
- $$x + y + z$$
- 0
यदि $$pq + qr + rp = 0$$, तब $$\frac{q^2}{r p} + \frac{r^2}{p q} + \frac{p^2}{q r} = ?$$
- 3
- 1
- 2
- 0
यदि $$\sqrt{a^2+b^2+ab} + \sqrt{a^2+b^2-ab}=1$$ है, तो $$(1-a^2)(1-b^2)$$ का मान ज्ञात करें।
- $$\frac{1}{4}$$
- $$\frac{4}{7}$$
- $$\frac{5}{4}$$
- $$\frac{3}{4}$$
यदि $$a + b + c = abc$$, तो $$\frac{1}{(1-a^2)(1-b^2)} + \frac{1}{(1-b^2)(1-c^2)} + \frac{1}{(1-c^2)(1-a^2)} = ?$$
- 0
- 2
- 4
- 3
Given: \( x + \frac{1}{x} = \sqrt{7} \) , then Find \( x^2 + \frac{1}{x^2} \)
- 5
- 8
- 9
- 4
Given: \( x (x-4) = 1 \) , then Find \( x^2 + \frac{1}{x^2} \)
- 18
- 8
- 9
- 4
Given: \( x + \frac{1}{x} = \sqrt{2} \) , then Find \( x^2 + \frac{1}{x^2} + 3 \)
- 3
- 8
- 9
- 4
Given: \( x + \frac{1}{x} = 7 \) , then Find \( √x + \frac{1}{√x} \)
- 3
- 8
- 9
- 4