Răspuns :
a)
(6x³y²+5x²y)²=
(x²y(6xy+5))²=
(x²)²y²(6xy+5)²=
x⁴y²(6xy+5)²
b)
(1,2x²y+0,5x³y²)²=
(x²y(1,2+0,5xy))²=
(x²)²y²(1,2+0,5xy)²
x⁴y²(1,2+0,5xy)²
c)
(3x-2y)²=
(3x)²-2•3x•2y+(2y)²=
3²x²-2•3x•2y+(2y)²=
9x²-2•3x•2y+2²y²=
9x²-12xy+4y²
[tex]d) \\ \\ ( {5a}^{b} - 2ab) ^{2} = \\ \\ ( {5a}^{b}) ^{2} - 2 \times {5a}^{b} \times 2ab + {(2ab)}^{2} = \\ \\ {5}^{2}( {a}^{b}) ^{2} - 2 \times {5a}^{b} \times 2ab + {(2ab)}^{2} = \\ \\ 25( {a}^{b}) ^{2} - 2 \times {5a}^{b} \times 2ab + {(2ab)}^{2} = \\ \\ {25a}^{b \times 2} - 2 \times {5a}^{b} \times 2ab + {(2ab)}^{2} = \\ \\ {25a}^{2b} - 2 \times {5a}^{b} \times 2ab + {(2ab)}^{2} = \\ \\ {25a}^{2b} - 2 \times {5a}^{b} \times 2ab + {2}^{2} {a}^{2} {b}^{2} = \\ \\ {25a}^{2b} - 2 \times {5a}^{b} \times 2ab + {4a}^{2} {b}^{2} = \\ \\ {25a}^{2b} - (2 \times 5 \times 2) {a}^{b}ab + {4a}^{2} {b}^{2} = \\ \\ {25a}^{2b} - 20 {a}^{b}ab + {4a}^{2} {b}^{2} = \\ \\ {25a}^{2b} - {20a}^{b + 1}b + {4a}^{2} {b}^{2} [/tex]
(6x³y²+5x²y)²=
(x²y(6xy+5))²=
(x²)²y²(6xy+5)²=
x⁴y²(6xy+5)²
b)
(1,2x²y+0,5x³y²)²=
(x²y(1,2+0,5xy))²=
(x²)²y²(1,2+0,5xy)²
x⁴y²(1,2+0,5xy)²
c)
(3x-2y)²=
(3x)²-2•3x•2y+(2y)²=
3²x²-2•3x•2y+(2y)²=
9x²-2•3x•2y+2²y²=
9x²-12xy+4y²
[tex]d) \\ \\ ( {5a}^{b} - 2ab) ^{2} = \\ \\ ( {5a}^{b}) ^{2} - 2 \times {5a}^{b} \times 2ab + {(2ab)}^{2} = \\ \\ {5}^{2}( {a}^{b}) ^{2} - 2 \times {5a}^{b} \times 2ab + {(2ab)}^{2} = \\ \\ 25( {a}^{b}) ^{2} - 2 \times {5a}^{b} \times 2ab + {(2ab)}^{2} = \\ \\ {25a}^{b \times 2} - 2 \times {5a}^{b} \times 2ab + {(2ab)}^{2} = \\ \\ {25a}^{2b} - 2 \times {5a}^{b} \times 2ab + {(2ab)}^{2} = \\ \\ {25a}^{2b} - 2 \times {5a}^{b} \times 2ab + {2}^{2} {a}^{2} {b}^{2} = \\ \\ {25a}^{2b} - 2 \times {5a}^{b} \times 2ab + {4a}^{2} {b}^{2} = \\ \\ {25a}^{2b} - (2 \times 5 \times 2) {a}^{b}ab + {4a}^{2} {b}^{2} = \\ \\ {25a}^{2b} - 20 {a}^{b}ab + {4a}^{2} {b}^{2} = \\ \\ {25a}^{2b} - {20a}^{b + 1}b + {4a}^{2} {b}^{2} [/tex]