Question

Решите пожалуйста нечего не могу понять ,а от этого зависит моя оценка за четверть.
√4x+√63x–√81x

√12y–0.5√48y+2√108y

(5√7–√63+√14)×√7

(2+√3)(1–√3)

(√14+2)(2–√14)

(1–2√3) в квадрате

Заранее спасибо

Answer 1

$sqrt{4x}+sqrt{63x}-sqrt{81x}=sqrt{4}*sqrt{x}+sqrt{63x}-sqrt{81}*sqrt{x}= \ =2sqrt{x}-9sqrt{x}+sqrt{63x}=sqrt{63x}-7sqrt{x};$
$sqrt{12y}-0,5sqrt{48y}+2sqrt{108y}=sqrt{4*3y}-0,5sqrt{16*3y}+2sqrt{36*3y}= \ =2sqrt{3y}-0,5*4sqrt{3y}+2*6sqrt{3y}=2sqrt{3y}-2sqrt{3y}+12sqrt{3y}=12sqrt{3y};$
$(5sqrt{7}-sqrt{63}+sqrt{14})*sqrt{7}=5sqrt7}*sqrt{7}-sqrt{63}*sqrt{7}+sqrt{14}*sqrt{7}= \ =5(sqrt{7})^2-sqrt{63*7}+sqrt{14*7}= 5*7-sqrt{441}+sqrt{2*7*7}= \ =35-21+sqrt{7^2*2}=14+|7|sqrt{2}=14+7sqrt{2}=7*(2+sqrt{2});$
$(2+sqrt{3})(1-sqrt{3})=2-2sqrt{3}+sqrt{3}-(sqrt{3})^2=-sqrt{3}+2-3=-1-sqrt{3};$
$(sqrt{14}+2)(2-sqrt{14})=(2+sqrt{14})(2-sqrt{14})=2^2-(sqrt{14})^2= \ =4-14=-10;$
$(1-2sqrt{3})^2=1^2-2*1*2sqrt{3}+(2sqrt{3})^2=1-4sqrt{3}+2^2(sqrt{3})^2= \ =1+4*3-4sqrt{3}=1+12-4sqrt{3}=13-4sqrt{3};$