Question

Найти корень уравнения:
$sqrt{ frac{10}{4x-26} }= frac{1}{7}$
Найти корень:
$sqrt{frac{4x+32}{7} }=6$

Answer 1

( √ 10 / ( 4X - 26) ) ^2 = ( 1/7 ) ^2 
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10 / ( 4X - 26) > 0 ---> 4X - 26 ≠ 0 ; X ≠ 6,5 
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10 / ( 4X - 26) = 1 /49 
10 * 49 = 4X - 26 
4X - 26 = 490 
4X = 516 
X = 129 
===================================
( √ ( 4X + 32) / 7 ) ^ 2 = 6 
-----------------
(4X + 32 ) / 7 > 0 
4X > - 32 
X > - 8 
--------------
( 4X + 32 ) / 7 = 36 
4X + 32 = 252 
4X = 220 
X = 55

Answer 2

$1) sqrt{ frac{10}{4x-26} } = frac{1}{7} \ frac{10}{4x-26} geq 0 \ 4x-26 neq 0 \ 4x neq 26 \ x neq 6.5$

x∈ $(- infty ;6.5)U(6.5;+infty )$

$1) sqrt{ frac{10}{4x-26} } = frac{1}{7} \ (sqrt{ frac{10}{4x-26} })^{2} = (frac{1}{7})^{2} \ frac{10}{4x-26} = frac{1}{49} /*49(4x-26) \ frac{10*49(4x-26) }{4x-26} = frac{49(4x-26)}{49} \ 10*49=4x-26 \ 490=4x-26 \ 4x-26=490 \ 4x=490+26 \ 4x=516\x=129$

Ответ: x=129

$2) sqrt{ frac{4x+32}{7} } =6 \ frac{4x+32}{7} geq 0/*7 \ 4x+32 geq 0 \ 4x geq -32 \ x geq -8$

x∈ $[-8;+ infty )$

$sqrt{ frac{4x+32}{7} } =6 \ (sqrt{ frac{4x+32}{7} } )^{2} =6^{2} \ frac{4x+32}{7} =36/*7 \ 4x+32=252\4x=252-32\4x=220\x=55$

Ответ: x=55