Oblicz (prosze o pełne obliczenia daje 100 punktów!)
a) (1 +√2)² + (1 -√2)²
b) (√3 - 1)² - (2- √3)²
c) (2√3 - 3/2)² - (2√3 + 3/2)²
d) (1/3 + 3√2)² - (2 - 3√2)²
e) (4 - √5) (4+√5) - (√5 - 2) ( 2 + √5)
f) (√6 - √5) (√6 + √5) + (√6 - √5)²
g) (2√5 - √10)² - (2√5 + 1) (1 - 2√5)
h) (√6 - 2√3)² - (5√2 - 1) (1 + 5√2)
a) (1 +√2)² + (1 -√2)²
b) (√3 - 1)² - (2- √3)²
c) (2√3 - 3/2)² - (2√3 + 3/2)²
d) (1/3 + 3√2)² - (2 - 3√2)²
e) (4 - √5) (4+√5) - (√5 - 2) ( 2 + √5)
f) (√6 - √5) (√6 + √5) + (√6 - √5)²
g) (2√5 - √10)² - (2√5 + 1) (1 - 2√5)
h) (√6 - 2√3)² - (5√2 - 1) (1 + 5√2)
Verified answer
Odpowiedź:
a) 1 + 2 + 1 - 2 = 2
b) 3 - 1 - 4 + 3 = 1
c) 4*3 - 9/4 - 4*3 - 9/4 = 12 - 9/4 - 12 - 9/4 = -18/4 = -9/2
d) 1/9 + 9*2 - 4 + 9*2 = 1/9 + 18 - 5 +18 = - 4 8/9 + 36 = 31 1/9
e) 16 + 4[tex]\sqrt{5}[/tex] - [tex]4\sqrt{5}[/tex] - 5 - [tex]2\sqrt{5}[/tex] - 5 + 4 +[tex]2\sqrt{5}[/tex] = 16 - 5 - 5 +4 = 10
f) 6 + [tex]\sqrt{30}[/tex] - [tex]\sqrt{30}[/tex] - 5 + 6 - 5 = 6 - 5 + 6 - 5 = 2
g) 4*5 - 10 - [tex]2\sqrt{5}[/tex] + 4*5 - 1 + [tex]2\sqrt{5}[/tex] = 20 - 10 + 20 - 1 = 29
h) 6 - 4*3 - [tex]5\sqrt{2}[/tex] - 25*2 + 1 + [tex]5\sqrt{2}[/tex]= 6 - 12 - 50 + 1 = -55
[tex]a)\\\\(1+\sqrt{2})^2+(1-\sqrt{2})^2=1^2+2\cdot1\sqrt{2}+(\sqrt{2})^2+1^2-2\cdot1\sqrt{2}+(\sqrt{2})^2=\\\\=1+2\sqrt{2}+2+1-2\sqrt{2}+2=1+2+1+2=6\\\\\\b)\\\\(\sqrt{3}-1)^2-(2-\sqrt{3})^2=(\sqrt{3})^2-2\sqrt{3}\cdot1+1^2-(2^2-2\cdot2\cdot\sqrt{3}+(\sqrt{3})^2=\\\\=3-2\sqrt{3}+1-(4-4\sqrt{3}+3)=4-2\sqrt{3}-(7-4\sqrt{3})=4-2\sqrt{3}-7+4\sqrt{3}=\\\\=-3+2\sqrt{3}[/tex]
[tex]c)\\\\(2\sqrt{3}-\frac{3}{2})^2-(2\sqrt{3}+\frac{3}{2})^2=\\\\=(2\sqrt{3})^2-\not2\cdot2\sqrt{3}\cdot\frac{3}{\not2}+(\frac{3}{2})^2-((2\sqrt{3})^2+\not2\cdot2\sqrt{3}\cdot\frac{3}{\not2}+(\frac{3}{2})^2)=\\\\=4\cdot3-2\sqrt{3}\cdot3+\frac{9}{4}-(4\cdot3+2\sqrt{3}\cdot3+\frac{9}{4})=12-6\sqrt{3}+\frac{9}{4}-(12+6\sqrt{3}+\frac{9}{4})=\\\\=12-6\sqrt{3}+\frac{9}{4}-12-6\sqrt{3}-\frac{9}{4}=-6\sqrt{3}-6\sqrt{3}=-12\sqrt{3}[/tex]
[tex]d)\\\\(\frac{1}{3}+3\sqrt{2})^2-(2-3\sqrt{2})^2=\\\\=((\frac{1}{3})^2+2\cdot\frac{1}{\not3}\cdot\not3\sqrt{2}+(3\sqrt{2})^2)-(2^2-2\cdot2\cdot3\sqrt{2}+(3\sqrt{2})^2)=\\\\=\frac{1}{9}+2\sqrt{2}+9\cdot2-(4-12\sqrt{2}+9\cdot2)=\frac{1}{9}+2\sqrt{2}+18-(4-12\sqrt{2}+18)=\\\\=\frac{1}{9}+2\sqrt{2}+18-4+12\sqrt{2}-18=\frac{1}{9}+14\sqrt{2}-4=\frac{1}{9}-\frac{36}{9}+14\sqrt{2}=\\\\=-\frac{35}{9}+14\sqrt{2}[/tex]
[tex]e)\\\\(4-\sqrt{5})(4+\sqrt{5})-(\sqrt{5}-2)(2+\sqrt{5})=4^2-(\sqrt{5})^2-(\sqrt{5}-2)(\sqrt{5}+2)=\\\\=16-5-((\sqrt{5})^2-2^2)=11-(5-4)=11-1=10\\\\\\f)\\\\(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})+(\sqrt{6}-\sqrt{5})^2=(\sqrt{6})^2-(\sqrt{5})^2+(\sqrt{6})^2-2\cdot\sqrt{6}\cdot\sqrt{5}+(\sqrt{5})^2=\\\\=6-5+6-2\sqrt{6\cdot5}+5=7-2\sqrt{30}+5=12-2\sqrt{30}[/tex]
[tex]g)\\\\(2\sqrt{5}-\sqrt{10})^2-(2\sqrt{5}+1)(1-2\sqrt{5})=\\\\=(2\sqrt{5})^2-2\cdot2\sqrt{5}\cdot\sqrt{10}+(\sqrt{10})^2-(1+2\sqrt{5})(1-2\sqrt{5})=\\\\=4\cdot5-4\sqrt{5\cdot10}+10-(1^2-(2\sqrt{5})^2)=20-4\sqrt{50}+10-(1-4\cdot5)=\\\\=30-4\sqrt{25\cdot2}-(1-20)=30-4\cdot\sqrt{25}\cdot\sqrt{2}-(-19)=30-4\cdot5\sqrt{2}+19=\\\\=49-20\sqrt{2}[/tex]
[tex]h)\\\\(\sqrt{6}-2\sqrt{3})^2-(5\sqrt{2}-1)(1+5\sqrt{2})=\\\\=(\sqrt{6})^2-2\cdot\sqrt{6}\cdot2\sqrt{3}+(2\sqrt{3})^2-(5\sqrt{2}-1)(5\sqrt{2}+1)=\\\\=6-4\sqrt{6\cdot3}+4\cdot3-((5\sqrt{2})^2-1^2)=6-4\sqrt{18}+12-(25\cdot2-1)=\\\\=18-4\sqrt{9\cdot2}-(50-1)=18-4\cdot\sqrt{9}\cdot\sqrt{2}-49=18-4\cdot3\sqrt{2}-49=\\\\=18-12\sqrt{2}-49= -31-12\sqrt{2}[/tex]
Zastosowane wzory
[tex](a+b)^2=a^2-2ab+b^2\\\\(a-b)^2=a^2-2ab+b^2\\\\(a-b)(a+b)=a^2-b^2[/tex]