Поможіть ДАМ 20 БАЛІВ

Question

Алгебра

Ishnswyn

6 год назад

Поможіть ДАМ 20 БАЛІВ

ОТВЕТЫ

Harrington Calvin

Nov 13, 2020

Объяснение:

$a)\ \\\sqrt{x-9+2\sqrt{x+6} }=3\\ (\sqrt{x-9+2\sqrt{x+6} })^2=3^2\\x-9+2\sqrt{x+6}=9\\2\sqrt{x+6} =18-x\\ (2\sqrt{x+6})^2 =(18-x)^2\\4*(x+6)=324-36x+x^2\\4x+24=324-36x+x^2\\x^2-40x+300=0\\D=400\ \ \ \ \sqrt{D}=20\\x_1=10\ \ \ \ x_2=30.$

Проверка:

$x_1=10.\\\sqrt{10-9+2*\sqrt{10+6} }=\sqrt{1+2*\sqrt{16} } =\sqrt{1+2*4}=\sqrt{1+8}=\sqrt{9} =3.\in\\x_2=30.\\\sqrt{30-9+2\sqrt{30+6} }=\sqrt{21+2*\sqrt{36} }=\sqrt{21+2*6}=\sqrt{21+12}=\sqrt{33}\notin. \Rightarrow\\$

: x=10.

$b)\ \sqrt{7+x}-\sqrt{3-x}=2$

ОДЗ:

$\left \{ {{7+x\geq 0} \atop {3-x\geq 0}} \right. \ \ \ \ \left \{ {{x\geq -7} \atop {x\leq 3}} \right. \ \ \ \ \Rightarrow\ \ \ \ x\in[-7;3].\\(\sqrt{7+x}-2)^2=(\sqrt{3-x} )^2 \\7+x-2*2*\sqrt{7+x}+4=3-x\\ 2x+8=4*\sqrt{7+x}\ |:2\\x+4=2*\sqrt{7+x}\\(x+4)^2=(2*\sqrt{7+x})^2\\x^2+8x+16=4*(7+x)\\x^2+8x+16=28+4x\\x^2+4x-12=0\\D=64\ \ \ \ \sqrt{D} =8\\x_1=2\ \ \ \ x_2=-6.$

Проверка:

$x=2.\\\sqrt{7+2}-\sqrt{3-x}=\sqrt{9}-\sqrt{1}=3-1=2\in.\\x=-6.\\\sqrt{7-6} -\sqrt{3-(-6)}=\sqrt{1}-\sqrt{9} =1-3=-2\notin.\ \ \ \ \Rightarrow\\$

: x=2.

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