Question

please solve all the questions right who will answer all questions correctly l will mark as biralist ​

Answer 1

Answer:

i think so this is algebra

Answer 2

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

$\bold{1) a^2 + 4b^2 + c^2 + 4ab + 4bc + 2ac} \\\implies (a)^2 + (2b)^2 + (c)^2 + 2(a)(2b) + 2(2b)(c) + 2(a)(c) \\\implies \bold{(a + 2b + c)^2}$

$\bold{2) 4a^2 + 9b^2 + c^2 - 12ab + 6bc - 4ac} \\\implies (-2a)^2 + (3b)^2 + (c)^2 + 2(-2a)(3b) + 2(3b)(c) + 2(-2a)(c) \\\implies \bold{(-2a + 3b + c)^2}$

$\bold{3) 9x^2 + y^2 + z^2 - 6xy + 2yz - 6zx} \\\implies (-3x)^2 + (y)^2 + (z)^2 + 2(-3x)(y) + 2(y)(z) + 2(-3x)(z) \\\implies \bold{(-3x + y + z)^2}$

$\bold{4) m^2 + 4n^2 + 25p^2 + 4mn - 20np - 10pm} \\\implies (m)^2 + (2n)^2 + (-5p)^2 + 2(m)(2n) + 2(2n)(-5p) + 2(-5p)(m) \\\implies \bold{(m + 2n - 5p)^2}$

$\bold{5) x^2 + 4y^2 + 4 + 4x - 4xy - 8y} \\\implies (x)^2 + (-2y)^2 + (2)^2 + 2(x)(2) - 2(-2y)(x) + 2(-2y)(2) \\\implies \bold{(x - 2y + 2)^2}$

$\bold{6) a^4 + b^4 + c^4 + 2a^2b^2 + 2b^2c^2 + 2c^2a^2} \\\implies (a^2)^2 + (b^2)^2 + (c^2)^2 + 2(a^2)(b^2) + 2(b^2)(c^2) + 2(c^2)(a^2) \\\implies \bold{(a^2 + b^2 + c^2)^2}$

$\bold{7) a^2b^2 + b^2c^2 + c^2a^2 + 2a^2bc + 2ab^2c + 2abc^2} \\\implies (ab)^2 + (bc)^2 + (ca)^2 + 2(ab)(ac) + 2(ab)(bc) + 2(ac)(bc) \\\implies \bold{(ab + bc + ca)^2}$

$\bold{8) \dfrac{x^2}{y^2} + \dfrac{y^2}{z^2} + \dfrac{z^2}{x^2} + 2\dfrac{x}{z} + 2 \dfrac{y}{x} + 2 \dfrac{z}{y}} \\\implies \left(\dfrac{x}{y}\right)^2 + \left(\dfrac{y}{z}\right)^2 + \left(\dfrac{z}{x}\right)^2 + 2 \left(\dfrac{x}{y}\right) \left(\dfrac{y}{z}\right) + 2 \left(\dfrac{z}{x}\right) \left(\dfrac{y}{z}\right) + 2 \left(\dfrac{x}{y}\right) \left(\dfrac{z}{x}\right) \\\implies \bold{ \left(\left(\dfrac{x}{y}\right) + \left(\dfrac{y}{z}\right) + \left(\dfrac{z}{x}\right)\right)^2}$

$\bold{9) \dfrac{a^2}{b^2c^2} + \dfrac{b^2}{c^2a^2} + \dfrac{c^2}{a^2b^2} + \dfrac{2}{a^2} + \dfrac{2}{b^2} + \dfrac{2}{c^2}} \\\implies \left(\dfrac{a}{bc}\right)^2 + \left(\dfrac{b}{ac}\right)^2 + \left(\dfrac{c}{ab}\right)^2 + 2 \left(\dfrac{b}{ac}\right) \left(\dfrac{c}{ab}\right) + 2 \left(\dfrac{a}{bc}\right) \left(\dfrac{c}{ab}\right) + 2 \left(\dfrac{a}{bc}\right) \left(\dfrac{b}{ac}\right) \\\implies \bold{ \left(\left(\dfrac{a}{bc}\right) + \left(\dfrac{b}{ac}\right) + \left(\dfrac{c}{ab}\right)\right)^2}$

$\bold{10) x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx} \\\implies (x)^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(x)(4z) \\\implies \bold{(x + 2y + 4z)^2}$

$\bold{11) 4x^2 + y^2 + z^2 - 4xy - 2yz + 6zx} \\\implies (2x)^2 + (-y)^2 + (z)^2 + 2(2x)(y) + 2(-y)(z) + 2(2x)(z) \\\implies \bold{(2x - y + z)^2}$

$\bold{12) 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8zx } \\\implies (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(-2x)(2z) \\\implies \bold{(-2x + 3y + 2z)^2}$

Formula Used:

• $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2$

Hope it helps!