Question

buna seara,ma puteti ajuta la aceste cerinte?va rog frumos​

Answer 1

Răspuns:

2. 3(x-2)²-5=43

3(x-2)²=43+5

3(x-2)²=48/:3

(x-2)²=16

(x-2)²=√16

x-2=±4

x-2=4 => x=4+2=6

x-2= -4 => x=-4+2= -2

Raspuns: c) {-2,6}

3. 3(2x+3)²-16=59

3(2x+3)²=59+16

3(2x+3)²=75/:3

(2x+3)²=25

2x+3=√25

2x+3=±5

2x+3=5

2x=5-3

2x=2/:2

x=1

2x+3= -5

2x= -5-3

2x= -8/:2

x= -4

Raspuns: c) {-4,1}

4. (2x-√2)²=18

18=9·2 si √18=3√2

Atunci:

(2x-√2)²=(3√2)²

2x-√2=±3√2

2x-√2=3√2

2x=3√2+√2

2x=4√2/:2

x=2√2

2x-√2= -3√2

2x= -3√2+√2

2x= -2√2/:2

x= -√2

Raspuns: c) {-√2,2√2}

Explicație pas cu pas:

Answer 2

$\displaystyle 2.~~3(x-2)^2-5=43 \Rightarrow 3\Big(x^2-2\cdot x \cdot 2+2^2\Big)-5=43 \Rightarrow \\\\ \Rightarrow 3\Big(x^2-4x+4\Big)-5=43 \Rightarrow 3x^2-12x+12-5-43=0\Rightarrow \\\\ \Rightarrow 3x^2-12x-36=0\Big|:3\Rightarrow x^2-4x-12=0\\\\\Delta=(-4)^2-4\cdot 1 \cdot (-12)=16+48=64 > 0\\\\ x_1=\frac{-(-4)-\sqrt{64} }{2 \cdot 1} =\frac{4-8}{2} =\frac{-4}{2} =-2\\\\ x_2=\frac{-(-4)+\sqrt{64} }{2 \cdot 1} =\frac{4+8}{2} =\frac{12}{2} =6\\\\ x\in \{-2;6\}~~~~~\boxed{c}$

$\displaystyle 3.~~3(2x+3)^2-16=59 \Rightarrow 3\Big((2x)^2+2\cdot 2x\cdot 3+3^2\Big)-16=59 \Rightarrow \\\\ \Rightarrow 3\Big(4x^2+12x+9\Big)-16=59 \Rightarrow 12x^2+36x+27-16-59=0\Rightarrow \\\\ \Rightarrow 12x^2+36x-48=0 \Big|:12 \Rightarrow x^2+3x-4=0\\\\ \Delta=3^2-4\cdot1 \cdot (-4)=9+16=25 > 0\\\\ x_1=\frac{-3-\sqrt{25} }{2 \cdot 1} =\frac{-3-5}{2} =\frac{-8}{2} =-4\\\\ x_2=\frac{-3+\sqrt{25} }{2 \cdot 1} =\frac{-3+5}{2} =\frac{2}{2} =1\\\\ x\in\{-4;1\}~~~~~\boxed{c}$

$\displaystyle 4.~~\Big(2x-\sqrt{2} \Big)^2=18 \Rightarrow (2x)^2-2 \cdot 2x \cdot \sqrt{2} +\Big(\sqrt{2} \Big)^2=18 \Rightarrow \\\\ \Rightarrow 4x^2-4x\sqrt{2} +2=18 \Rightarrow 4x^2-4x\sqrt{2} +2-18 =0 \Rightarrow \\\\ \Rightarrow 4x^2-4x\sqrt{2} -16=0\Big|:4\Rightarrow x^2-x\sqrt{2} -4=0 \\\\ \Delta=\Big(\sqrt{2}\Big)^2-4 \cdot 1 \cdot (-4)=2+16=18 > 0\\\\ x_1=\frac{-\Big(-\sqrt{2} \Big)-\sqrt{18} }{2 \cdot1} =\frac{\sqrt{2}-3\sqrt{2} }{2} =\frac{-2\sqrt{2} }{2} =-\sqrt{2}$

$\displaystyle x_2=\frac{-\Big(-\sqrt{2} \Big)+\sqrt{18} }{2 \cdot1} =\frac{\sqrt{2}+3\sqrt{2} }{2} =\frac{4\sqrt{2} }{2} =2\sqrt{2} \\\\ x\in\Big\{-\sqrt{2};2\sqrt{2} \Big\}~~~~~\boxed{c}$