Question

7. Efectuați calculele:
a) (x + 2)² + (x - 1)²;
c)(x-√2)² + (√2x+1)²;
(x-3)²-x(x-5);
g) (x + 2)(x+3)-(x + 1)²;
(2x+3)2-4(x+3)(x-3);
b) (x + 3)² + (2x + 1)²;
d) (2x + 1)²-4x(x + 1);
f) (x - 5)²-x(x + 17);
h) (x + 2)²-(x+3)(x-3);
j) (3x - 1)²-(3x - 5)(3x + 5).

Answer 1

$\displaystyle a)~(x+2)^2+(x-1)^2=x^2+2 \cdot x \cdot 2+2^2+x^2-2 \cdot x \cdot 1+1^2=\\\\ =x^2+4x+4+x^2-2x+1=2x^2+2x+5\\\\ b)~\Big(x-\sqrt{2} \Big)^2+\Big(\sqrt{2} x+1\Big)^2=\\\\=x^2-2 \cdot x \cdot \sqrt{2} +\Big(\sqrt{2} \Big)^2+\Big(\sqrt{2} x\Big)^2+2 \cdot \sqrt{2} x\cdot 1+1^2=\\\\=x^2-2\sqrt{2}x +2+2x^2+2\sqrt{2} x+1=3x^2+3$

$\displaystyle c)~(x-3)^2-x(x-5)=x^2-2 \cdot x \cdot 3+3^2-x^2+5x=\\\\=x^2-6x+9-x^2+5x=-x+9\\\\ d)~(x+2)(x+3)-(x+1)^2=x^2+3x+2x+6-\Big(x^2+2\cdot x \cdot 1+1^2\Big)=\\\\=x^2+5x+6-\Big(x^2+2x+1\Big)=x^2+5x+6-x^2-2x-1=3x+5$

$\displaystyle e)~(2x+3)\cdot 2-4(x+3)(x-3)=4x+6-4\Big(x^2-9\Big)=\\\\=4x+6-4x^2+36=-4x^2+4x+42\\\\ f)~(x+3)^2+(2x+1)^2=x^2+2 \cdot x \cdot 3+3^2+(2x)^2 +2 \cdot 2x \cdot 1+1^2=\\\\=x^2+6x+9+4x^2+4x+1=5x^2+10x+10$

$\displaystyle g)~(2x+1)^2-4x(x+1)=(2x)^2+2 \cdot 2x \cdot 1+1^2-4x^2-4x=\\\\=4x^2+4x+1-4x^2-4x=1\\\\ h)~(x-5)^2-x(x+17)=x^2-2 \cdot x \cdot 5+5^2-x^2-17x=\\\\=x^2-10x+25-x^2-17x=-27x+25$

$\displaystyle i)~(x+2)^2-(x+3)(x-3)=x^2+2 \cdot x \cdot 2+2^2-\Big(x^2-9\Big)=\\\\=x^2+4x+4-x^2+9=4x+13\\\\ j)~(3x-1)^2-(3x-5)(3x+5)=(3x)^2-2 \cdot 3x \cdot 1+1^2-\Big(9x^2-25\Big)=\\\\=9x^2-6x+1-9x^2+25=-6x+26$