Question

ПОМОГИТЕ ПОЖАЛУЙСТА УМОЛЯЮ

Answer 1

$\displaystyle\bf\\1)\\\\\frac{b^{2}-6b+9 }{b-7} \cdot\frac{b-7}{b-3} =\frac{(b-3)^{2} }{b-7} \cdot\frac{b-7}{b-3} =b-3\\\\2)\\\\\sqrt{36\cdot0,81} =\sqrt{6^{2}\cdot0,9^{2} } =6\cdot 0,9=5,4\\\\\\\Big(\frac{1}{5} \sqrt{10} \ \Big)^{2} =\Big(\frac{1}{5} \Big)^{2} \cdot\Big(\sqrt{10} \Big)^{2} =\frac{1}{25} \cdot 10=\frac{2}{5} =0,4\\\\3)\\\\x^{2} +4x-21=0\\\\D=4^{2} -4\cdot(-21)=16+84=100=10^{2} \\\\\\x_{1}=\frac{-4-10}{2} =\frac{-14}{2} =-7$

$\displaystyle\bf\\x_{2} =\frac{-4+10}{2} =\frac{6}{2} =3\\\\\\Otvet \ : \ -7 \ ; \ 3$

Answer 2

Объяснение:

1.

$\dfrac{b^{2}-6b+9 }{b-7} \: \cdot \: \dfrac{b-7}{b-3}= \dfrac{b^{2}-2 \: \cdot \:3b+3^{2} }{b-7} \: \cdot \: \dfrac{b-7}{b-3}= \dfrac{(b-3)^{2} }{b-7} \: \cdot \: \dfrac{b-7}{b-3}= b-3$

2.

а) $\sqrt{36 \cdot 0,81} =\sqrt{6^{2} \cdot 0,9^{2} } =6 \cdot 0,9 = 5,4$

б) $\bigg ( \dfrac{1}{5} \cdot \sqrt{10} \bigg )^{2} =\bigg ( \dfrac{1}{5} \bigg )^{2} \cdot (\sqrt{10} )^{2} =\dfrac{1}{25} \cdot 10= 0,04 \cdot 10 = 0,4$

3.

$x^{2} +4x-21=0\\\\\boxed{D=b^{2}-4ac}\\\\D=4^{2} -4\cdot 1 \cdot (-21)=16+84=100\\\\\boxed{x_{1,2}=\dfrac{-b \: \pm \: \sqrt{D} }{2a} }\\\\\\x_{1}=\dfrac{-4-\sqrt{100} }{2 \cdot 1} =\dfrac{-4-10}{2} =\dfrac{-14}{2} =-7;\\\\x_{2}=\dfrac{-4+\sqrt{100} }{2 \cdot 1} =\dfrac{-4+10}{2} =\dfrac{6}{2} =3.$

Ответ: x₁ = -7; x₂ = 3.