Tính: (4x-5)^2; (3x+1/3y)^2; (-x+0,3)^2

Giải SBT Toán 8 Bài 3: Hằng đẳng thức đáng nhớ

Bài 1 trang 13 SBT Toán 8 Tập 1: Tính:

a) ${\left(4x-5\right)}^2$;

b) ${\left(3x+\frac{1}{3}y\right)}^2$;

c) ${\left(-x+0,3\right)}^2$;

d) ${\left(-x-10y\right)}^2$;

e) ${\left(a^3-3a\right)}^2$;

g) ${\left(a^4+\frac{1}{2}{a}^2\right)}^2$.

Lời giải:

a) ${\left(4x-5\right)}^2={\left(4x\right)}^2-2.4x.5+5^2=16x^2-40x+25$;                      

b) ${\left(3x+\frac{1}{3}y\right)}^2={\left(3x\right)}^2+2.3x.\frac{1}{3}y+{\left(\frac{1}{3}y\right)}^2=9x^2+2xy+\frac{1}{9}{y}^2$;                 

c) ${\left(-x+0,3\right)}^2={\left(-x\right)}^2+2.\left(-x\right).0,3+0,3^2=x^2-0,6x+0,09$;

d) (–x – 10y)² = (‒x)² + 2.(‒x).(‒10y) + (‒10y)² = x² + 20xy + 100y².             

e) ${\left(a^3-3a\right)}^2={\left(a^3\right)}^2-2.a^3.3a+{\left(3a\right)}^2=a^6-6a^4+9a^2$;                 

g) ${\left(a^4+\frac{1}{2}{a}^2\right)}^2={\left(a^4\right)}^2+2.a^4.\frac{1}{2}{a}^2+{\left(\frac{1}{2}{a}^2\right)}^2=a^8+a^6+\frac{1}{4}{a}^4$.