Đáp án:
Giải thích các bước giải:
`c) x³ - x² - 21x + 45 = 0`
`-> x³ -3x² + 2x² - 6x - 15x + 45 = 0`
`-> (x³ -3x²) + (2x² - 6x) - (15x - 45) = 0`
`-> x²(x-3) + 2x(x-3) - 15(x-3) =0`
`-> (x-3)(x² + 2x - 15) = 0`
`-> (x-3)(x² - 3x + 5x - 15) = 0`
`-> (x-3)[(x²-3x) + (5x-15)] = 0`
`-> (x-3)[ x(x-3) + 5(x-3)] = 0`
`-> (x-3)(x-3)(x+5) = 0`
`->` \(\left[ \begin{array}{l}x-3=0\\x+5=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=3\\x=-5\end{array} \right.\)
`d) x³ + 3x² + 4x + 2 = 0`
`-> x³ + x² + 2x² + 2x + 2x + 2 = 0`
`-> (x³+x²)+(2x²+2x)+(2x+2)=0`
`-> x²(x+1) + 2x(x+1) + 2(x+1) =0`
`-> (x+1)(x²+2x + 2)=0 `
`-> (x+1)(x² + 2x + 1 + 1) = 0`
`-> (x+1)[(x+1)²+1] = 0`
`-> x+1 = 0`
`-> x = -1`
`e) x^4 + x² + 6x - 8 = 0`
`-> x^4 + 2x² + 1 - x² + 6x - 9 = 0`
`-> (x²+1)² - ( x - 3)² = 0`
`-> (x²+1-x+3)(x²+1+x-3)=0`
`-> (x²-x+4)(x²+x-2)=0`
`->(x²-x+4)(x²+2x-x-2)=0`
`->(x²-x+4)[x(x+2)-(x+2)]=0`
`->(x²-x+4)(x+2)(x-1)=0 ( x²-x+4=(x-1/2)² + 15/4 > 0)`
`->` \(\left[ \begin{array}{l}x+2=0\\x-1=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.\)
`g) (x²+1)² = 4(2x-1)`
`-> x^4 + 2x² + 1 = 8x - 4`
`-> x^4 + 2x² + 1 - 8x + 4 = 0`
`-> x^4 + 2x² - 8x + 5 = 0`
`-> x^4 - x² + 3x² - 3x - 5x + 5 = 0`
`-> (x^4 - x²) + (3x² - 3x) - (5x - 5) = 0`
`-> x²(x²-1) + 3x(x-1) -5(x-1)=0`
`->x²(x-1)(x+1) + 3x(x-1) - 5(x-1) = 0`
`-> (x-1)[x²(x+1) + 3x - 5] = 0`
`-> (x-1)(x³ + x² + 3x - 5) = 0`
`-> (x-1)(x³ - x² + 2x² - 2x + 5x - 5) = 0`
`-> (x-1)[(x³-x²)+(2x²-2x)+(5x-5)] = 0`
`-> (x-1)[x²(x-1)+2x(x-1)+5(x-1)] = 0`
`-> (x-1)(x-1)(x²+2x+5) = 0 [x²+2x+5=(x+1)² + 4 > 0]`
`-> (x-1)²=0`
`-> x-1=0`
`-> x = 1`
`h) (x-1)³ + (2x+3)³ = 27x³ + 8`
`-> x³ - 3x² + 3x - 1 + 8x³ + 36x² - 54x + 27 - 27x³ - 8 = 0`
`-> 18x³ - 33x² - 57x - 18 = 0`
`-> (3x+2) (6x² -15x-9)=0`
`-> 3(3x+2)(2x+1)(x-3)=0`
`->` \(\left[ \begin{array}{l}3x+2=0\\2x+1=0\\x-3=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=-2/3\\x=-1/2\\x=3\end{array} \right.\)
`i) 6x^4 - x³ - 7x² + x + 1 = 0`
`-> 6x^4 - 6x³ + 5x³ - 5x² - 2x² + 2x - x + 1 = 0`
`-> (6x^4 - 6x³) + (5x³ - 5x²) - (2x² - 2x) - (x - 1) = 0`
`-> 6x³ (x-1) + 5x² (x-1) - 2x ( x-1) - (x-1) = 0`
`-> (x-1)(6x³ + 5x² - 2x - 1) = 0`
`-> (x-1)(6x³ -3x² + 8x² - 4x + 2x - 1) = 0`
`-> (x-1)[(6x³ -3x²) + (8x² - 4x) + (2x - 1)] = 0`
`-> (x-1)[3x²(2x - 1) + 4x(2x-1) + (2x-1) ] = 0`
`-> (x-1)(2x-1)(3x²+4x + 1) = 0`
`-> (x-1)(2x-1)(3x² + x + 3x + 1) =0`
`-> (x-1)(2x-1)[(3x²+x)+(3x+1)] = 0`
`-> (x-1)(2x-1)[x(3x+1)+(3x+1)] = 0`
`-> (x-1)(2x-1)(3x+1)(x+1) = 0`
`->` \(\left[ \begin{array}{l}x-1=0\\2x-1=0\\3x+1=0\\x+1=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=1\\x=1/2\\x=-1/3\\x=-1\end{array} \right.\)
