zonic-the-hedgehog
zonic-the-hedgehog
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√(ab) ≤ (a + b)/2 <-- I think like this <=> 2√ab ≤ a + b <=> 0 ≤ a - 2√ab + b <=> 0 ≤ ( √a - √b )² => That is true
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(5ax²-2ax²- 7ax²) . (1/2x + 1/4x) = -3ax
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A) Vector AB = ( 6 - 2 ; -5 + 4 ) = (4 ; -1 ) Vector normal AB = (1 ; 4 ) d : (x - 2) + 4(y + 4) = 0 <=> d: x + 4y + 14 = 0 B) Vector AB = ( -2 - 5 ; 1 + 5 ) = ( -7 ; 6 ) Vector nomal AB = ( 6 ; 7 ) d: 6(x - 5) + 7(y + 5) = 0 <=&g…
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a) |2 - 3x| - 1--> Where is "=" ( the equal ) ? c) |x+3|-2x-1 --> Where is "=" ( the equal ) ? d) |x+3|-x+3 --> Where is "=" ( the equal ) ? b) |x + 3| = |2x - 1| <=> x + 3 = 2x - 1 Or x + 3 = 1…
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Lim ( 2x + 3 )^(x + 1) ( ▬▬▬ ) x -> ∞ ( 2x + 1 ) Lim ( 2x + 1 + 2 )^(x + 1) ( ▬▬▬▬▬ ) x -> ∞ ( 2x + 1 ) Lim ( 2 )^(x + 1) ( 1 + ▬▬▬ ) x -> ∞ ( 2x + 1) = e
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2^x = -2 x = ( 2iπn + iπ + log(2))/log(2) ( n of Z )
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2^x = -2 x = ( 2iπn + iπ + log(2))/log(2) ( n of Z )
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Lim √(16 - x) - 4 ▬▬▬▬▬▬▬ x -> 0 x Lim 16 - x - 16 ▬▬▬▬▬▬▬▬▬ x -> 0 x[ √(16 - x) + 4 ] Lim - x ▬▬▬▬▬▬▬▬▬ x -> 0 x[ √(16 - x) + 4 ] Lim - 1 ▬▬▬▬▬▬▬ x -> 0 …
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Lim √(16 - x) - 4 ▬▬▬▬▬▬▬ x -> 0 x Lim 16 - x - 16 ▬▬▬▬▬▬▬▬▬ x -> 0 x[ √(16 - x) + 4 ] Lim - x ▬▬▬▬▬▬▬▬▬ x -> 0 x[ √(16 - x) + 4 ] Lim - 1 ▬▬▬▬▬▬▬ x -> 0 …
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4t² - 3t - 1 = 0 <=> t = 1 <=> t = -1/4 ( Don't choose we know 2^x have positive number ) => 2^x = 1 <=> x = 0
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4t² - 3t - 1 = 0 <=> t = 1 <=> t = -1/4 ( Don't choose we know 2^x have positive number ) => 2^x = 1 <=> x = 0
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10 = 200/x + 10 <=> 200/x = 0 So that x ∈ ∅
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10 = 200/x + 10 <=> 200/x = 0 So that x ∈ ∅
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(2x+1)(x-5) < 2(2(x+2)(x-4) <=> 2x² - 10x + x - 5 < 4( x² - 4x + 2x - 8 ) <=> 2x² - 9x - 5 < 4( x² - 2x - 8 ) <=> 2x² - 9x - 5 < 4x² - 8x - 32 <=> 2x² + x - 27 > 0 -1 - √217 -1 + √217 …
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(2x+1)(x-5) < 2(2(x+2)(x-4) <=> 2x² - 10x + x - 5 < 4( x² - 4x + 2x - 8 ) <=> 2x² - 9x - 5 < 4( x² - 2x - 8 ) <=> 2x² - 9x - 5 < 4x² - 8x - 32 <=> 2x² + x - 27 > 0 -1 - √217 -1 + √217 …
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(2x+1)(x-5) < 2(2(x+2)(x-4) <=> 2x² - 10x + x - 5 < 4( x² - 4x + 2x - 8 ) <=> 2x² - 9x - 5 < 4( x² - 2x - 8 ) <=> 2x² - 9x - 5 < 4x² - 8x - 32 <=> 2x² + x - 27 > 0 -1 - √217 -1 + √217 …
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16x^2 +12x+9 = x² + 3x/4 + 9/16 = x² + 3x/4 + 9/64 + 27/64 = (x + 3/8)² - (3i√3/8)² = ( x + 3/8 - 3i√3/8 )( x + 3/8 + 3i√3 / 8 )
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n Ancol pứ = 1 . 72% = 0,72 mol mà Axit trên là axit nhị chức cho nên pứ trên là dư axit chọn theo số mol Ancol Đặt x ; y là số mol của este : HOOC - CH2 - CH2 - COOCH3 (My) ; CH3OCO - CH2 - CH2 - COOCH3 (Mz) { x + 2y = 0,72 { 132x / 146y…
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n Ancol pứ = 1 . 72% = 0,72 mol mà Axit trên là axit nhị chức cho nên pứ trên là dư axit chọn theo số mol Ancol Đặt x ; y là số mol của este : HOOC - CH2 - CH2 - COOCH3 (My) ; CH3OCO - CH2 - CH2 - COOCH3 (Mz) { x + 2y = 0,72 { 132x / 146y…
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f(x) = g(x) <=>x + 1 = 5x - 1 <=> 5x - x = 1 + 1 <=> 4x = 2 <=> x = 1/2
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f(x) = g(x) <=>x + 1 = 5x - 1 <=> 5x - x = 1 + 1 <=> 4x = 2 <=> x = 1/2
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n Ancol pứ = 1 . 72% = 0,72 mol mà Axit trên là axit nhị chức cho nên pứ trên là dư axit chọn theo số mol Ancol Đặt x ; y là số mol của este : HOOC - CH2 - CH2 - COOCH3 (My) ; CH3OCO - CH2 - CH2 - COOCH3 (Mz) { x + 2y = 0,72 { 132x / 146y…
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16x^2 +12x+9 = x² + 3x/4 + 9/16 = x² + 3x/4 + 9/64 + 27/64 = (x + 3/8)² - (3i√3/8)² = ( x + 3/8 - 3i√3/8 )( x + 3/8 + 3i√3 / 8 )
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(2x+1)(x-5) < 2(2(x+2)(x-4) <=> 2x² - 10x + x - 5 < 4( x² - 4x + 2x - 8 ) <=> 2x² - 9x - 5 < 4( x² - 2x - 8 ) <=> 2x² - 9x - 5 < 4x² - 8x - 32 <=> 2x² + x - 27 > 0 -1 - √217 -1 + √217 …
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10 = 200/x + 10 <=> 200/x = 0 So that x ∈ ∅
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10 = 200/x + 10 <=> 200/x = 0 So that x ∈ ∅
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4t² - 3t - 1 = 0 <=> t = 1 <=> t = -1/4 ( Don't choose we know 2^x have positive number ) => 2^x = 1 <=> x = 0
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2^x = -2 x = ( 2iπn + iπ + log(2))/log(2) ( n of Z )
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a) |2 - 3x| - 1--> Where is "=" ( the equal ) ? c) |x+3|-2x-1 --> Where is "=" ( the equal ) ? d) |x+3|-x+3 --> Where is "=" ( the equal ) ? b) |x + 3| = |2x - 1| <=> x + 3 = 2x - 1 Or x + 3 = 1…
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(5ax²-2ax²- 7ax²) . (1/2x + 1/4x) = -3ax