Question

AJUTOR!! PE POZA, TOATE 4 PUNCTELE!! ​

Answer 1

a.

$a = ( \sqrt{3} - 1) ^{2} = 3 - 2 \sqrt{3} + 1 = 4 - 2 \sqrt{3}$

$b = 5 - 2 \sqrt{3}$

=>b este mai mare

b.

${x}^{2} + {y}^{2} = 5$

$xy = 2$

$(x - y) ^{2} = {x}^{2} - 2xy + {y}^{2} = {x}^{2} + {y}^{2} - 2xy = 5 - 2 \times 2 = 5 - 4 = 1$

c.

$\frac{(4x - 5) ^{2} - 9 }{(4x - 3)^{2} - 25 } = \frac{16 {x}^{2} - 40x + 25 - 9}{16 {x}^{2} - 24x + 9 - 25 } = \frac{16 {x}^{2} - 40x + 16 }{16 {x}^{2} - 24x - 16 }$

m am blocat aici, vezi tu cum te grupezi pentru a mai face simplificări

d.

${2}^{x} + {2}^{x + 1} + {2}^{x + 2} = 14 \\ {2}^{x} ({2}^{1} + {2}^{2} )= 14 \\ {2}^{x} \times 6 = 14 \\ {2}^{x} = 14 \div 6 = 2.3 \\ {2}^{x} = {2}^{1.23} \\ deci \: x = 1.23$

e posibil să fi greșit ceva la calcul, da' verifici tu când o scrii

Answer 2

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

$b)~x^2+y^2=5,~xy=2\\\\ (x-y)^2=x^2-2 \cdot x \cdot y+y^2=x^2+y^2-2xy=5-2 \cdot 2=5-4=1$

$\displaystyle c)~\frac{(4x-5)^2-9}{(4x-3)^2-25}=\frac{(4x)^2-2\cdot 4x \cdot 5+5^2-9}{(4x)^2-2 \cdot 4x \cdot 3+3^2-25}=\frac{16x^2-40x+25-9}{16x^2-24x+9-25} =\\\\=\frac{16x^2-40x+16}{16x^2-24x-16} =\frac{8(2x-1)(x-2)}{8(2x+1)(x-2)}=\frac{2x-1}{2x+1} \\\\\frac{2x-1^{(8x-16}}{2x+1} = \frac{\cfrac{2x-1}{8x-16} }{\cfrac{2x+1}{8x-16} } =\frac{2x-1}{8x-16} \cdot \frac{8x-16}{2x+1} =\frac{2x-1}{2x+1}$

$\displaystyle d)~2^x+2^{x+1}+2^{x+2}=14 \Rightarrow 2^x\Big(1+2^1+2^2\Big)=14 \Rightarrow 2^x\Big(1+2+4\Big)=14 \Rightarrow \\\\ \Rightarrow 2^x \cdot 7=14 \Rightarrow 2^x=\frac{14}{7} \Rightarrow 2^x=2 \Rightarrow x=1$