Question

Квадрат двучлена

Воспользуйся формулами квадрата двучлена.

Answer 1

${( {a}^{2} {b}^{3} + a )}^{2} = {( {a}^{2} {b}^{3} )}^{2} + 2 \times {a}^{2} {b}^{3} \times a + {a}^{2} = {a}^{4} {b}^{6} + 2 {a}^{3} {b}^{3} + {a}^{2}$

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${(0.2 {z}^{2} - {x}^{3}y )}^{2} = {(0.2 {z}^{2} )}^{2} - 2 \times 0.2 {z}^{2} \times {x}^{3} y + {( {x}^{3} y)}^{2} = 0.04 {z}^{4} - 0.4 {x}^{3} y {z}^{2} + {x}^{6} {y}^{2}$

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${(15a {c}^{3} + 0.6c)}^{2} = {(15a {c}^{3} )}^{2} + 2 \times 15a {c}^{3} \times 0.6c + {(0.6c)}^{2} = 225 {a}^{2} {c}^{6} + 18a {c}^{4} + 0.36 {c}^{2}$

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${( \frac{1}{2} {x}^{3}y - 4 {y}^{2} )}^{2} = {( \frac{1}{2} {x}^{3} y)}^{2} - 2 \times \frac{1}{2} {x}^{3} y \times 4 {y}^{2} + {(4 {y}^{2} )}^{2} = \frac{1}{4} {x}^{6} {y}^{2} - 4 {x}^{3} {y}^{3} + 16 {y}^{4}$

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${(2 \frac{1}{3}u + 6 {u}^{4} )}^{2} = {( \frac{7}{3}u + 6 {u}^{4} )}^{2} = {( \frac{7}{3}u )}^{2} + 2 \times \frac{7}{3} u \times 6 {u}^{4} + {(6 {u}^{4} )}^{2} = \frac{49}{9} {u}^{2} + 28 {u}^{5} + 36 {u}^{8}$