Question

5.3. Преобразуйте выражение:
1
1)
5)
» (.
у
+
4
3
2
2
1
6) 12 – т
3
т +1
2) [b+
3
n
п
3)
7) | 5т -
;
а
2
2
4)
;
| с.
8) 9р
т​

Answer 1

Объяснение:

Воспользуемся формулами сокращенного умножения

$(a+b)^{2} =a^{2} +2ab+b^{2} ;\\(a-b)^{2} =a^{2} -2ab+b^{2} .$

$1)\left(x-\dfrac{1}{2}\right )^{2} =x^{2} -2\cdot x\cdot\dfrac{1}{2} +\left(\dfrac{1}{2}\right)^{2} =x^{2} -x+\dfrac{1}{4};$

$2)\left(b+\dfrac{1}{3}\right )^{2} =b^{2} +2\cdot b\cdot\dfrac{1}{3} +\left(\dfrac{1}{3}\right)^{2} =b^{2} +\dfrac{2}{3} b+\dfrac{1}{9};$

$3)\left(a-\dfrac{1}{5}\right )^{2} =a^{2} -2\cdot a\cdot\dfrac{1}{5} +\left(\dfrac{1}{5}\right)^{2} =a^{2} -\dfrac{2}{5}a+\dfrac{1}{25};$

$4)\left(\dfrac{a}{2} +\dfrac{b}{3}\right )^{2} =\left(\dfrac{a}{2} \right)^{2} +2\cdot \dfrac{a}{2} \cdot\dfrac{b}{3} +\left(\dfrac{b}{3}\right)^{2} =\dfrac{a^{2} }{4} +\dfrac{ab}{3} +\dfrac{b^{2} }{9};$

$5)\left(\dfrac{x}{4} +\dfrac{y}{3}\right )^{2} =\left(\dfrac{x}{4} \right)^{2} +2\cdot \dfrac{x}{4} \cdot\dfrac{y}{3} +\left(\dfrac{y}{3}\right)^{2} =\dfrac{x^{2} }{16} +\dfrac{xy}{6} +\dfrac{y^{2} }{9};$

$6)\left(2\dfrac{1}{3} m+1\dfrac{1}{2}n\right )^{2} =\left(\dfrac{7}{3}m \right)^{2} +2\cdot \dfrac{7}{3}m \cdot\dfrac{3}{2}n +\left(\dfrac{3}{2}n\right)^{2} =\dfrac{49 }{9} m^{2} +7mn +\dfrac{9 }{4}n^{2} ;$

$7)\left(5m-\dfrac{n}{2}\right )^{2} =(5m)^{2} -2\cdot 5m \cdot\dfrac{n}{2} +\left(\dfrac{n}{2}\right)^{2} =25m^{2} -5mn +\dfrac{n^{2} }{4};$

$8)\left(9p-\dfrac{q}{3}\right )^{2} =(9p)^{2} -2\cdot 9p \cdot\dfrac{q}{3} +\left(\dfrac{q}{3}\right)^{2} =81p^{2} -6pq +\dfrac{q^{2} }{9}.$