ĐKXĐ: x-1>=3

=>x>=4

Ta có: \(C_{x+1}^{x-2}+2\cdot C_{x-1}^3=7\left(x-1\right)\)

=>\(\frac{\left(x+1\right)!}{\left(x+1-x+2\right)!\cdot\left(x-2\right)!}+2\cdot\frac{\left(x-1\right)!}{\left(x-1-3\right)!\cdot3!}=7\left(x-1\right)\)

=>\(\frac{\left(x+1\right)!}{\left(x-2\right)!\cdot6}+2\cdot\frac{\left(x-1\right)!}{\left(x-4\right)!\cdot6}=7\left(x-1\right)\)

=>\(\frac{\left(x+1\right)\cdot x\cdot\left(x-1\right)}{6}+2\cdot\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)}{6}=7\left(x-1\right)\)

=>\(\frac{x\left(x+1\right)\left(x-1\right)}{6}+\frac{2\left(x-1\right)\left(x-2\right)\left(x-3\right)}{6}=\frac{42\left(x-1\right)}{6}\)

=>x(x+1)(x-1)+2(x-1)(x-2)(x-3)=42(x-1)

=>(x-1)\(\left\lbrack x\left(x+1\right)+2\left(x-2\right)\left(x-3\right)-42\right\rbrack=0\)

=>x(x+1)+2(x-2)(x-3)-42=0

=>\(x^2+x+2\left(x^2-5x+6\right)-42=0\)

=>\(x^2+x-42+2x^2-10x+12=0\)

=>\(3x^2-9x-30=0\)

=>\(x^2-3x-10=0\)

=>(x-5)(x+2)=0

=>x=5(nhận) hoặc x=-2(loại)

a) f'(x) = - 3sinx + 4cosx + 5. Do đó

f'(x) = 0 <=> - 3sinx + 4cosx + 5 = 0 <=> 3sinx - 4cosx = 5

<=> sinx - cosx = 1. (1)

Đặt cos φ = , (φ ∈) => sin φ = , ta có:

(1) <=> sinx.cos φ - cosx.sin φ = 1 <=> sin(x - φ) = 1

<=> x - φ = + k2π <=> x = φ + + k2π, k ∈ Z.

b) f'(x) = - cos(π + x) - sin = cosx + sin.

f'(x) = 0 <=> cosx + sin = 0 <=> sin = - cosx <=> sin = sin

<=> = + k2π hoặc = π - x + + k2π

<=> x = π - k4π hoặc x = π + k, (k ∈ Z).

mai đăng lại bài này nhé t làm cho h đi ngủ

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