Question

Va rog sa ma ajutati la acest exercitiu. Dau coroana si 99 puncte

Answer 1

$a) {(4x - \sqrt{5} )}^{2} = {(4x)}^{2} - 2 \times 4 x\times \sqrt{5} + {( \sqrt{5} )}^{2} = 16 {x}^{2} - 8 \sqrt{5} x + 5$

$b) {(a \sqrt{3} - 3)}^{2} = {(a \sqrt{3} )}^{2} -2 \times a \sqrt{3} \times 3 + {3}^{2} = 3 {a}^{2} - 6 \sqrt{3} a + 9=3({a}^{2}-2\sqrt{3}a+3)$

$c) {(y \sqrt{10} - 1)}^{2} = {(y \sqrt{10} )}^{2} - 2 \times y \sqrt{10} \times 1 + {1}^{2} = 10 {y}^{2} - 2 \sqrt{10} y + 1$

$d) {(6x - \sqrt{6}) }^{2} = {(6x)}^{2} - 2 \times 6x \times \sqrt{6} + {( \sqrt{6} )}^{2} = 36 {x}^{2} - 12 \sqrt{6} x + 6=6(6{x}^{2}-2\sqrt{6}x+1)$

$e) {(x \sqrt{8} - 4x)}^{2} = {(2 \sqrt{2} x - 4x)}^{2} = {(2 \sqrt{2} x)}^{2} - 2 \times 2 \sqrt{2} x \times 4x + {(4x)}^{2} = 8 {x}^{2} - 16 \sqrt{2} {x}^{2} + 16 {x}^{2} = 24 {x}^{2} - 16\sqrt{2} {x}^{2} = 8 {x}^{2} (3 - 2\sqrt{2} )$

$f) {(a \sqrt{7} - a \sqrt{6} )}^{2} = {(a \sqrt{7} )}^{2} - 2 \times a \sqrt{7} \times a \sqrt{6} + {(a \sqrt{6} )}^{2} = 7 {a}^{2} - 2 \sqrt{42} {a}^{2} + 6 {a}^{2} = 13 {a}^{2} - 2 \sqrt{42} {a}^{2}$

$g) {(y \sqrt{12} - y \sqrt{2} )}^{2} = {(2 \sqrt{3} y - y \sqrt{2}) }^{2} = {(2 \sqrt{3}y) }^{2} - 2 \times 2 \sqrt{3} y \times y \sqrt{2} + {(y \sqrt{2}) }^{2} = 12 {y}^{2} - 4 \sqrt{6} {y}^{2} + 2 {y}^{2} = 14 {y}^{2} - 4 \sqrt{6} {y}^{2} = 2 {y}^{2} (7 - 2 \sqrt{6} )$

$h) {(x \sqrt{20} - x \sqrt{2} ) }^{2} = {(2 \sqrt{5}x - x \sqrt{2} )}^{2} = {(2 \sqrt{5} x)}^{2} - 2 \times 2 \sqrt{5} x \times x \sqrt{2} + {(x \sqrt{2} )}^{2} = 20 {x}^{2} - 4 \sqrt{10} {x}^{2} + 2 {x}^{2} = 22 {x}^{2} - 4 \sqrt{10} {x}^{2} = 2 {x}^{2} (11 - 2 \sqrt{10} )$

$i) {(a \sqrt{14} - a \sqrt{7}) }^{2} = {(a \sqrt{14}) }^{2} - 2 \times a \sqrt{14} \times a \sqrt{7} + {(a \sqrt{7}) }^{2} = 14 {a}^{2} - 2 \sqrt{98} {a}^{2} + 7 {a}^{2} = 21 {a}^{2} - 2 \times 7 \sqrt{2} {a}^{2} = 21 {a}^{2} - 14 \sqrt{2} {a}^{2} = 7 {a}^{2} (3 - 2 \sqrt{2} )$

$j) {(y \sqrt{24} - y \sqrt{6} )}^{2} = {(2 \sqrt{6} y - y \sqrt{6} )}^{2} = {(2 \sqrt{6}y) }^{2} - 2 \times 2 \sqrt{6} y \times y \sqrt{6} + {(y \sqrt{6} )}^{2} = 24 {y}^{2} - 4 \sqrt{36} {y}^{2} + 6 {y}^{2} = 30 {y}^{2} - 4 \times 6 {y}^{2} = 30 {y}^{2} - 24 {y}^{2} = 6 {y}^{2}$

Answer 2

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