Đặt `1 - x = a ; 3 - x = b`

`=> a^3 + b^3 = (a + b)^3`

`<=> a^3 + b^3 = a^3 + 3a^2 b + 3ab^2 + b^3`

`<=> a^2 b + ab^2 = 0`

`<=> ab(a + b) = 0`

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`C = 1^2 - 2^2 + 3^2 - 4^2 + ... + 99^2 - 100^2`

`= (1 - 2).(1 + 2) + (3 - 4).(3 + 4) + ... + (99 - 100).(99 + 100)`

`= (- 1) . (1 + 2) + (- 1) . (3 + 4) + ... + ...

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`c)` ĐKXĐ:

`{(5 - 3absx >= 0),(x^2 + 4x + 3):} <=> {(- 5/3 <= x <= 5/3),(x \ne - 1):}`(vì `x` luôn `\ne - 3)`

`d)` ĐKXĐ:

`x^2 - ...

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`x^2 - 2x = 2sqrt[2x - 1] (x >= 2)`

`<=> x^4 - 4x^3 + 4x^2 = 8x - 4`

`<=> x^4 - 4x^3 + 4x^2 - 8x + 4 = 0`

`<=> (x^2 + 2)(x^2 - 4x + 2) ...

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Với `x > 0`

`A = sqrtx + 1/(sqrtx) + 3 >= 2sqrt[sqrtx . 1/(sqrtx)] + 3 = 2 + 3 = 5`(Cauchy)

Dấu "=" xảy ra `<=> sqrtx = 1/(sqrtx) <=> x = 1`

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`E = 1 + 5 + 5^2 + ... + 5^100`

`5E = 5 + 5^2 + 5^3 + ... + 5^101`

`5E - E = (5 + 5^2 + 5^3 + ... + 5^101) - (1 + 5 + 5^2 + ... + 5^100)`

`4E = 5^101 - 1`< ...

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`{(1/x - 1/y = 1),(3/x + 4/y = 5):} (x ; y \ne 0) <=> {(4/x - 4/y = 4),(3/x + 4/y = 5):}`

`<=> {(7/x = 9),(1/x - 1/y = 1):} <=> {(x = 7/9),(y = 7/2):} ...

`abs(2x - y + 5) + abs(y + 3) = 0`

Ta có:

`abs(2x - y + 5) ; abs(y + 3) >= 0 AA x ; y`

`=> abs(2x - y + 5) + abs(y + 3) >= 0 AA x ; y`

Dấu "=" xảy ra `&l ...

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(b tự viết thêm gạch trên đầu cho số nhé)

`(a \in NN`*`; b ; c \in NN ; (a ; b ; c) <= 9)`

`VT = 4b2 + a1c = 4 . 100 + b . 10 + 2 + a . 100 ...

`(a + b)^2 = 2(a^2 + b^2)`

`<=> a^2 + 2ab + b^2 = 2a^2 + 2b^2`

`<=> a^2 - 2ab + b^2 = 0`

`<=> (a - b)^2 = 0`

`<=> a - b = 0`

`<=> a ...