Question

vă rog frumos repede tot 50 P​

Answer 1

$\displaystyle 1.\\a)~x^2=1 \Rightarrow x=\pm \sqrt{1} \Rightarrow x=\pm 1\\\\ b)~2x^2+1=-7 \Rightarrow 2x^2=-7-1 \Rightarrow 2x^2=-8 \Rightarrow x^2=-\frac{8}{2} \Rightarrow x^2=-4 \Rightarrow \\\\ \Rightarrow x=\pm\sqrt{-4} \Rightarrow x=\pm 2i$

$c)~x^2+2x^2=4x^2 \Rightarrow 3x^2-4x^2=0\Rightarrow -x^2=0 \Rightarrow x=0$

$\displaystyle 2.~~34+15(x+3)^2=(-13)^2 \Rightarrow 34+15\Big(x^2+2 \cdot x \cdot 3+3^2\Big)=169 \Rightarrow \\\\ \Rightarrow 34+15\Big(x^2+6x+9\Big)=169 \Rightarrow 34+15x^2+90x+135=169 \Rightarrow \\\\ \Rightarrow 15x^2+90x+135+34-169=0 \Rightarrow 15x^2+90x=0\Big|:15 \Rightarrow x^2+6x=0\\\\ \Delta=6^2-4 \cdot 1 \cdot 0=36-0=36 > 0\\\\ x_1=\frac{-6-\sqrt{36} }{2 \cdot 1}=\frac{-6-6}{2} =\frac{-12}{2} =-6\\\\ x_2=\frac{-6+\sqrt{36}}{2\cdot1}=\frac{-6+6}{2} =\frac{0}{2} =0$

$3.~~(3+x)^2+(3-x)^2=36 \Rightarrow (3+(-3))^2+(3-(-3))^2=36 \Rightarrow \\\\ \Rightarrow 0^2+6^2=36 \Rightarrow 36=36 \Rightarrow a=-3 ~este~solutie~a~ecuatiei.$

$\displaystyle 4.~~3\Big(x^2+4x+7\Big)=4(3x+8)+1 \Rightarrow 3x^2+12x+21=12x+32+1\Rightarrow \\\\ \Rightarrow 3x^2+12x-12x=32+1-21 \Rightarrow 3x^2=12 \Rightarrow x^2=\frac{12}{3} \Rightarrow x^2=4 \Rightarrow \\\\ \Rightarrow x=\pm \sqrt{4} \Rightarrow x=\pm 2$

$\displaystyle 5.~~(3x+4)^2=(2x+1)^2 \Rightarrow (3x)^2+2 \cdot 3x \cdot 4+4^2=(2x)^2+2\cdot 2x \cdot 1+1^2\Rightarrow \\\\ \Rightarrow 9x^2+24x+16=4x^2+4x+1 \Rightarrow 9x^2-4x^2+24x-4x+16-1=0 \Rightarrow \\\\ \Rightarrow 5x^2+20x+15=0\Big|:5 \Rightarrow x^2+4x+3=0\\\\ \Delta=4^2-4 \cdot 1 \cdot 3=16-12=4 > 0\\\\ x_1=\frac{-4-\sqrt{4} }{2 \cdot 1} =\frac{-4-2}{2} =\frac{-6}{2} =-3\\\\ x_2=\frac{-4+\sqrt{4} }{2 \cdot 1}=\frac{-4+2}{2} =\frac{-2}{2} =-1$

$6.~~A=484~cm^2\\\\ A=l^2 \Rightarrow l^2=484 \Rightarrow l=\sqrt{484}\Rightarrow l=22~cm$