Question

va rog va dau coroana ​cu rezolvări va rog la ecuații

Answer 1

Asa se rezolva ecuatiile

Answer 2

$\displaystyle 1)~\frac{5x-10}{9} =\frac{20}{x-2} \Rightarrow (5x-10)(x-2)=20 \cdot 9 \Rightarrow \\\\ \Rightarrow 5x^2-10x-10x+20=180 \Rightarrow 5x^2-20x+20-180=0 \Rightarrow \\\\ \Rightarrow 5x^2-20x-160=0\big|:5 \Rightarrow x^2-4x-32=0\\\\ \Delta=(-4)^2-4 \cdot 1 \cdot (-32)=16+128=144 > 0$

$\displaystyle x_1=\frac{-(-4)-\sqrt{144} }{2 \cdot 1}=\frac{4-12}{2} =\frac{-8}{2} =-4\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{c)~\{-4;8\}}\\ x_2= \frac{-(-4)+\sqrt{144} }{2 \cdot 1}=\frac{4+12}{2} =\frac{16}{2} =8$

$\displaystyle 2)~3(x-2)^2-5=43 \Rightarrow 3\Big(x^2-2 \cdot x \cdot 2+2^2\Big)-5=43 \Rightarrow \\\\ \Rightarrow 3\Big(x^2-4x+4\Big)-5=43 \Rightarrow 3x^2-12x+12-5-43=0 \Rightarrow \\\\ \Rightarrow 3x^2-12x-36=0 \big|:3 \Rightarrow x^2-4x-12=0\\\\ \Delta=(-4)^2-4 \cdot 1 \cdot (-12)=16+48=64 > 0$

$\displaystyle x_1=\frac{-(-4)-\sqrt{64} }{2\cdot 1} =\frac{4-8}{2} =\frac{-4}{2} =-2\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{c)~\{-2;6\}}\\x_2=\frac{-(-4)+\sqrt{64} }{2\cdot 1} =\frac{4+8}{2} =\frac{12}{2} =6$

$3)~3(2x+3)^2-16=59 \Rightarrow 3\Big((2x)^2 +2 \cdot 2x \cdot 3+3^2\Big)-16=59 \Rightarrow \\\\ \Rightarrow 3\Big(4x^2+12x+9\Big)-16=59 \Rightarrow 12x^2+36x+27-16-59=0 \Rightarrow \\\\ \Rightarrow 12x^2+36x-48=0\big|:12 \Rightarrow x^2+3x-4=0\\\\ \Delta=3^2-4 \cdot 1 \cdot (-4)=9+16=25 > 0$

$\displaystyle x_1=\frac{-3-\sqrt{25} }{2\cdot1} =\frac{-3-5}{2} =\frac{-8}{2} =-4\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{c)~\{-4;1\}}\\ x_2=\frac{-3+\sqrt{25} }{2\cdot 1} =\frac{-3+5}{2} =\frac{2}{2} =1$

$\displaystyle 4)~\Big(2x-\sqrt{2} \Big)^2=18 \Rightarrow (2x)^2-2 \cdot 2x \cdot \sqrt{2} +\Big(\sqrt{2} \Big)^2-18=0 \Rightarrow \\\\ \Rightarrow 4x^2-4\sqrt{2} x+2-18=0 \Rightarrow 4x^2-4\sqrt{2} x-16=0 \big|:4 \Rightarrow \\\\ \Rightarrow x^2-\sqrt{2} x-4=0\\\\ \Delta=\Big(\sqrt{2} \Big)^2-4\cdot 1 \cdot (-4)=2+16=18 > 0$

$\displaystyle x_1=\frac{-\Big(-\sqrt{2}\Big)-\sqrt{18} }{2\cdot 1} =\frac{\sqrt{2} -3\sqrt{2} }{2} =\frac{\sqrt{2} \Big(1-3\Big)}{2} =\frac{\sqrt{2} \cdot (-2)}{2} =-\sqrt{2} \\\\ x_2=\frac{-\Big(-\sqrt{2}\Big)+\sqrt{18} }{2\cdot 1} =\frac{\sqrt{2} +3\sqrt{2} }{2} =\frac{\sqrt{2}\Big(1+3\Big) }{2} =\frac{\sqrt{2} \cdot 4}{2} =2\sqrt{2}\\\\ \boxed{c)~\Big\{-\sqrt{2} ;2\sqrt{2} \Big\}}$

$\displaystyle 5)~3\Big(2x+3\sqrt{3} \Big)^2-79=146 \Rightarrow \\\\ \Rightarrow 3\Big((2x)^2+2\cdot 2x \cdot 3\sqrt{3} +\Big(3\sqrt{3} \Big)^2\Big)-79=146 \Rightarrow\\\\ \Rightarrow 3\Big(4x^2+12x\sqrt{3}+27\Big)-79=146 \Rightarrow 12x^2+36x\sqrt{3} +81-79-146=0 \Rightarrow \\\\ \Rightarrow 12x^2+36x\sqrt{3} -144=0 \big|:12 \Rightarrow x^2+3x\sqrt{3} -12=0\\\\ \Delta=\Big(3\sqrt{3} \Big)^2-4 \cdot 1 \cdot (-12)=27+48=75 > 0$

$\displaystyle x_1=\frac{-3\sqrt{3}-\sqrt{75} }{2 \cdot 1} =\frac{-3\sqrt{3} -5\sqrt{3} }{2} =\frac{\sqrt{3}\Big(-3-5\Big) }{2} =\frac{\sqrt{3} \cdot (-8)}{2} =-4\sqrt{3} \\\\ x_2=\frac{-3\sqrt{3}+\sqrt{75} }{2 \cdot 1} =\frac{-3\sqrt{3} +5\sqrt{3} }{2} =\frac{\sqrt{3}\Big(-3+5\Big) }{2} =\frac{\sqrt{3} \cdot 2}{2} =\sqrt{3}$

$6)~12(x+1)^2=\Big(6+2\sqrt{3}\Big)^2 \Rightarow \\\\\Rightarrow 12\Big(x^2+2\cdot x \cdot 1+1^2\Big)=6^2+2 \cdot 6 \cdot 2\sqrt{3}+\Big(2\sqrt{3} \Big)^2 \Rightarrow \\\\ \Rightarrow 12\Big(x^2+2x+1\Big)=36+24\sqrt{3} +12 \\\\ \Rightarrow 12x^2+24x+12-36-12-24\sqrt{3} =0 \Rightarrow \\\\ \Rightarrow 12x^2+24x-36-24\sqrt{3}=0\big|:12 \Rightarrow x^2+2x-3-2\sqrt{3} =0\\\\ \Delta=2^2-4 \cdot 1 \cdot \Big(-3-2\sqrt{3} \Big)=4+12+8\sqrt{3} =16+8\sqrt{3} > 0$

$\displaystyle x_1=\frac{-2-\sqrt{16+8\sqrt{3} } }{2 \cdot 1} =\frac{-2-\sqrt{\Big(2+2\sqrt{3}\Big)^2 } }{2} =\frac{-2-\Big|2+2\sqrt{3}\Big| }{2} =\\\\ =\frac{-2-2-2\sqrt{3} }{2} =\frac{2\Big(-1-1-\sqrt{3}\Big) }{2} =-2-\sqrt{3}$

$\displaystyle x_2=\frac{-2+\sqrt{16+8\sqrt{3} } }{2 \cdot 1} =\frac{-2+\sqrt{\Big(2+2\sqrt{3}\Big)^2 } }{2} =\frac{-2+\Big|2+2\sqrt{3}\Big| }{2} =\\\\ =\frac{-2+2+2\sqrt{3} }{2} =\frac{2\Big(-1+1+\sqrt{3}\Big) }{2} =\sqrt{3}$