(2х +1)/12x^2y + 2 -3y/18xy^2 = (2x + 1)/12xy^2 + 2 -1/6xy = =(2x + 1 + 24xy^2 - 2y)/12xy^2
1) x∈(-∞,∞)
2) IxI-7≠0 x≠7 x≠ -7 x∈(-∞,-7)∪(-7;7)∪(7;∞)
3) <span>x∈(-∞,∞)
</span>4) x≠0
5) x²-16 ≠0 x= -4 x= 4 <span>x∈(-∞,-4)∪(-4;4)∪(4;∞)</span>
1)
(3a-2b)²=9a²-12ab+4y²
2)
(3a-2)(2+3a)=6a+9a²-4-6a=9a²-4
3)
(a-b)³
4)
(x+1)(x²-x+1)=x³-x²+x+x²-x+1=x³+1
5)
99²=9801
6)
(x+1)²-(x-1)²=x²+2x+1-(x²-2x+1)=x²+2x+1-x²+2x-1=4x
7)
(x+2)(x-2)-(x+2)²=x²-2x+2x-4-(x²+4x+4)=x²-4-x²-4x-4=-4x-8
Task/26488293
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154.
(x+3)⁴ - 3(x+3)² +2 =0 ;
замена : t =(x+3)²
t² -3t +2 =0 ⇔[ t = 1 ; t =2 .⇔[ (x+3)² =1 ; (x+3)² =2.⇔ [ x+3 =±1 ;x+3 =±√2 .
x₁= - 3 -√2 ; x₂= - 4 ; x₃ = -2 ; x₄ = - 3 +√2 .
165.
5/(x +1)<span>² - 4 / (x+1) =1;
</span>замена : t =1/ (x+1)
5t<span>² - 4t -1 = 0 ;
t</span>₁= (2 -3)/5 = -1/5 ⇒ 1/ (x₁+1) = - 1/ 5 ⇔ x₁ = - 6 ;<span>
t</span>₂=(2+3) /5 =1. ⇒ 1/ (x₂+1) =1 ⇔ x₂ = 0 .
170.
x + 3√x- 4 =0 ;
замена : t =√x ≥0
t² +3t - 4 =0 ;
t₁= (-3 -5)/2 = - 4 < 0 посторонний ;<span>
t</span>₂=(-3+5) /2 = 1. ⇒ √x =1 ⇔ x = 1 .
180. хорошо, что хорошо не видно...
3(x-2) - 16/(x-3) = 1 ;
3(x-3) - 16 /(x -3 ) + 2 =0 ;
замена : t =x - 3 ≠ 0 ⇔ x ≠ 3 .
3t² +2t - 16 =0 ;
t₁= (-1 - 7)/3 <span> = - </span>8/3 ⇒ x₁ - 3 = - 8/3 ⇔ x<span>₁ =1/3 ;</span><span>
t</span>₂=(-1+7) /3 = 2 . ⇒ x₂ - 3 = 2 <span>⇔ </span>x₂ =5.
185.
1/( x-3) + 1/x =1/2; <span>ОДЗ : x</span>≠0<span> ; x </span>≠3 .
2(x +x -3) =x(x-3) ;
x² -7x +6 =0 ;
x₁=1 ;
x₂=6.
193.
(x ² -1)/ x =x² -1/x ; <span>ОДЗ : x</span>≠0 .
x -1/x = x² -1/x ;
x =x <span>² ;
</span>x(x-1)= 0
x₁=0 → посторонний корень ;
x₂=1.
195.
(x² -2x -5) /(x-3)(x-1) +1/(x-3) =1 ; <span>ОДЗ : x</span>≠1 ; x ≠ 3 .
<span>x² -2x -5 +x -1= (x-3)(x-1) ;
</span><span>x² -2x -5 +x -1= </span>x²<span>- 4x + 3 ;
</span>3x = 9 ;
x =3 ∈ ОДЗ → посторонний корень ;
x ∈ ∅
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