task/29412801 --------------------
1 . f(x) =√(3x²+2x -8)
3x²+2x - 8 ≥0 ⇔3(x+2)(x -4/3) ≥ 0 методом интервалов
"+" "- " "+"
//////////////////// [-2] -------------- [4/3] //////////////////////
ответ: D(f) = (-∞ ; -2 ] ∪ [ 4/3 ; ∞) .
2. f(x) = Log_0,4 (x²-14x +40) / (2 - x)
(x²- 14x +40) / (2-x) > 0 ⇔ (x-4)(x-10 ) / (x-2) <0 ⇔ (x-2)(x-4)(x-10) <0 </p>
" - " "+" "- " "+"
///////////////// (2) --------------(4)/////////////// (10)--------------
ответ:D(f) = (-∞ ; 2 ) ∪ ( 4 ; 10) .
3. f(x) =√(49-x²) / Log₃₁ (x +4)
{ 49-x² ≥0 ; x +4 >0 ; x+4 ≠1 .⇔ { (x+7)(x-7) ≤ 0 ; x > - 4, x ≠ -3 .⇔
{ x ∈[ -7 ;7] ; x ∈( -4 ,-3) ∪(-3 ;∞) . ⇒x ∈( - 4 ,-3) ∪(-3 ;7 ]
ответ: D(f) = ( - 4 ,-3) ∪(-3 ;7 ]
4. f(x) =√cosx +Log₄₇ (-x² +3x -2)
{ cosx ≥ 0 ; - x² +3x - 2 > 0 . ⇔ { cosx ≥ 0 ; x²- 3x+2 < 0.
x²- 3x+2 < 0 ⇔ (x - 1)(x -2) < 0 ⇒ 1 < x < 2
и
cosx ≥ 0 ⇔ 2πn- π/2 ≤ x ≤ π/2+ 2πn ,n ∈ ℤ .
{ 1 < x < 2 ; 2πn - π/2 ≤ x ≤ π/2+ 2πn ,n ∈ ℤ. ⇒ 1 < x < π/2. * * * n = 0 * * *
ответ: D(f) = ( 1 ; π/2 ) .