Đáp án: + Giải thích các bước giải:
`(x^2 - 7x + 11)^{x^2-13x+42} = 1`
Trường hợp 1 :
`a^0 = 1` với mọi `a \ne 0`
`\to x^2 - 13x + 42 = 0`
`\to (x-7)(x-6) = 0`
`\to x = 7(TM) , x = 6(TM)`
Trường hợp 2 :
`1^b = 1``\to x^2 - 7x + 11 = 1`
`\to x^2 - 7x + 10 = 0`
`\to (x-5)(x-2) = 0`
`\to x = 5 , x = 2`
Trường hợp 3 :
`(-1)^b = -1`
`\to x^2 - 7x + 11 = -1`
`\to x^2 - 7x + 12 = 0`
`\to (x-4)(x-3)=0`
`\to x = 4(TM) , x=3(TM)`
Vậy `x = 7 , x = 6 , x= 5,x=2,x=4,x=3`
Dạng: a^b = 1
TH1: a^0 = 1 (a ≠ 0)
<=> x² - 13x + 42 = 0
<=> x = 6 hoặc x = 7
Thế vào a, ta thấy thoả điều kiện
TH2: 1^b = 1
<=> x² - 7x + 11 = 1
<=> x² - 7x + 10 = 0
<=> x = 2 hoặc x = 5
TH3: (-1)^b = 1 (b là số chẵn)
<=> x² - 7x + 11 = -1
<=> x² - 7x + 12 = 0
<=> x = 3 hoặc x = 4
Thế vào b, ta thấy thoả điều kiện
Kết hợp 3TH, vậy x € {2;3;4;5;6;7}
2KOH + ZnSO4 → Zn(OH)2 + K2SO4 Thuộc phản ứng gì?
Help me nhé giải thích ct luôn tks
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