tính, theo hằng đẳng thức:
a) ((a+b)-c)^2
b) (5a-3b)(5a+3b)
c) (3x+1)(3x-1)
d) (5x^2-2)(5x^2+2)
e) (x^2+2/5y)(x^2-2/5y)
f) (x+y+z)(x+y-z)
tính, theo hằng đẳng thức:
a) ((a+b)-c)^2
b) (5a-3b)(5a+3b)
c) (3x+1)(3x-1)
d) (5x^2-2)(5x^2+2)
e) (x^2+2/5y)(x^2-2/5y)
f) (x+y+z)(x+y-z)
$\begin{array}{l} a){\left( {a + b – c} \right)^2} = {\left( {a + b} \right)^2} + {c^2} – 2\left( {a + b} \right)c\\ = {a^2} + {b^2} + {c^2} + 2ab – 2bc – 2ca\\ b)\left( {5a – 3b} \right)\left( {5a + 3b} \right) = 25{a^2} – 9{b^2}\\ c)\left( {3x + 1} \right)\left( {3x – 1} \right) = 9{x^2} – 1\\ d)(5{x^2} – 2)(5{x^2} + 2) = 25{x^2} – 4\\ e)\left( {{x^2} + \dfrac{2}{5}y} \right)\left( {{x^2} – \dfrac{2}{5}y} \right) = {x^4} – \dfrac{4}{{25}}{y^2}\\ f)(x + y + z)\left( {x + y – z} \right) = {\left( {x + y} \right)^2} – {z^2} = {x^2} + {y^2} – {z^2} + 2xy \end{array}$
a, $[(a+b)-c]^{2}$=$(a+b)^{2}$-$2(a+b)c$-$c^{2}$ =$(a+b)^{2}$-$2(a+b)c$-$c^{2}$ =$a^{2}$ +$2ab$ +$b^{2}$ -$2ac$-$2bc $ -$c^2$
b, (5a-3b)$(5a+3b)$ =$25a^2$-$9b^2$
c, $(3x+1)(3x-1)$ = $9x^2$-$1$
d)$(5x^2-2)(5x^2+2)$ = $25x^4$-4
e) $(x^2+\frac{2y}{5})(x^2-\frac{2y}{5}))$ = $x^4$ -$\frac{4y^2}{25}$
f) $(x+y+z)(x+y-z)$ =$[(x+y)+z][(x+y)-z]$=$(x+y)^2)$-$z^2$ =$x^2+2xy+y^2$-$z^2$