Алгебра: Найдите S квадрата со стороной 7√2-2√5

//////////////////////_____-5____________-4__________(-3-√7)/2________(-3+√7)/2___0______////////////////////////////////////////////////////////// //////////////////////// решение первой системы -5 ≤ Х ≤ -4////////// ///////////////_____-3_____(-3-√7)/2________(-3+√7)/2___0________ 2________ /////////////////////////////////решений нетТаким образом...

Найдите S квадрата со стороной 7√2-2√5

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Ответ:

Gadik228

04.05.2023 22:39

1) log₁₂3 + log₁₂4 = log₁₂(3*4) = log₁₂12 = 1
2) log₇98 - log₇2 = log₇(98/2) = log₇49 = 2
3) log₂5-log₂35 + log₂56 = log₂(5/35) + log₂56 = log₂( $\frac{5*56}{35})=$ log₂8 = 3
4) log₁/₃5 - log₁/₃5 + log₁/₃ 9 = log₁/₃9 = -2

1) lg4 + lg250= lg(4*250) = lg1000 = 3
2) log₂6 - log₂ $\frac{6}{32}$ = log₂( $\frac{6*32}{6})$ = log₂32 = 5
3) (log₁₂4 + log₁₂36)² = (log₁₂144)² = 2² = 4
4) lg13 - lg 130 = lg $\frac{13}{130}$ = lg $\frac{1}{10}$ = -1
5) (log₂13-log₂52)⁵ = (log₂ $\frac{13}{52}$ )⁵ = (log₂ $\frac{1}{4}$ )⁵ = (-2)⁵ = -32
6) (log₀.₃9 - 2log₀.₃10)⁴ = (log₀.₃9 - log₀.₃100)⁴ = (log₀.₃ $\frac{9}{100}$ )⁴ = (log₀.₃0.09)⁴ = 2⁴ = 16

1) log₃x = -1
x = 3⁻¹ = 1/3
2) log₂x = -5
x = 2⁻⁵ = 1/32
3) log₃x = 2
x = 3² = 9
4) log₄x = 3
x = 4³ = 64
5) log₄x = -3
x = 4⁻³ = 1/64
6) log₇x = 0
x = 7° = 1
7) log₁/₇x = 1
x = 1/7
8) log₁/₂x = -3
x = (1/2)⁻³ = 8

1) log₂log₂log₃81 = log₂log₂4 = log₂2 = 1
2) log₂log₃log₁/₃(1/27) = log₂log₃3 = log₂1 = 0
3) $log_{ \sqrt{3} }$ log₅125 = $log_{ \sqrt{3} }$ 3 = 2
4) log₄log₃81 = log₄4 = 1

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Ответ:

ник4774

09.09.2021 01:59

$|2x^{2} +6x+1| \leq x^{2} -3x-19 \\
 \\ 
 \left \{ {{2 x^{2} +6x+1 \leq x^{2} -3x-19} \atop {2 x^{2} +6x+1 \geq 0}} \right. \\ 
 \left \{ {{x^{2} +9x+20 \leq 0} \atop {2 x^{2} +6x+1 \geq 0}} \right. \\ 
x^{2} +9x+20 =0, x_{1}=-4,x_{2}=-5 \\ 
2 x^{2} +6x+1=0, \\ 
D=28, \sqrt{D} = \sqrt{28}=2 \sqrt{7} \\ 
 x_{1}= \frac{-3+ \sqrt{7} }{2} ,x_{2}= \frac{-3- \sqrt{7} }{2} \\$
//////////////////////
_____-5____________-4__________(-3-√7)/2________(-3+√7)/2___0______
////////////////////////////////////////////////////////// ////////////////////////
решение первой системы -5 ≤ Х ≤ -4

$\\ \left \{ {{-2 x^{2} -6x-1 \leq x^{2} -3x-19} \atop {2 x^{2} +6x+1< 0}} \right. \\ \left \{ {{-3x^{2} -3x+18 \leq 0} \atop {2 x^{2} +6x+1 < 0}} \right. \\ 
\left \{ {{x^{2} +x-6 \geq 0} \atop {2 x^{2} +6x+1 < 0}} \right. \\ 
x^{2} + x - 6 =0, x_{1}=-3,x_{2}=2 \\$
////////// ///////////////
_____-3_____(-3-√7)/2________(-3+√7)/2___0________ 2________
/////////////////////////////////
решений нет

Таким образом решение системы -5 ≤ Х ≤ -4.
Тогда сумма целых решений: (-4) + (-5) = -9
ответ: -9.

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