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câu c nè
\(\frac{x^2-3x+5}{x+1}=\frac{\left(x^2+2x+1\right)-5x+4}{x+1}=\frac{\left(x+1\right)^2-5\left(x+1\right)+9}{x+1}\)
Ta có \(\frac{x+2}{x+1}=\frac{\left(x+1\right)+1}{x+1}=1+\frac{1}{x+1}\)
a) x^4 + 2^3-x -2
=x^4 - x^3 + 3x^3 - 3x^2 + 3x^2 - 3x + 2x-2
=x^3.(x-1) + 3x^2.(x-1) + 3x.(x-1)+2.(x-1)
=(x-1).( x^3+ 3x^2 + 3x+2)
=(X+1).(X^3 + 2X^2 + X^2 +2X +X+2)
=(X+1).(X+2).(X^2 +X + 1)
\(\frac{4x^2-6x+5}{2x-1}=2x-2+\frac{3}{2x-1}\)
Để biểu thức có giá trị nguyên thì \(\left(2x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
Với 2x - 1 = 1 => 2x = 2 => x = 1
2x - 1 = -1 => 2x = 0 => x = 0
2x - 1 = 3 => 2x = 4 => x = 2
2x - 1 = -3 => 2x = -2 => x = -1
Vậy x = {1;0;2;-1}
\(x^3+8x^2+17x+10\)
\(=x^3+2x^2+x^2+5x^2+10x+5x+2x+10\)
\(=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(5x^2+5x\right)+\left(10x+10\right)\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+5x\left(x+1\right)+10\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+5x+10\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
b)\(\frac{10}{x + 2} ; \frac{5}{2 x - 4} ; \frac{1}{6 - 3 x}\)
Giải:
a)
\(x^{3} - 1 = \left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right)\)
Mẫu chung: \(\left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right)\)
\(\frac{4 x^{2} - 3 x + 5}{\left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right)} ; \frac{\left(\right. 1 - 2 x \left.\right) \left(\right. x - 1 \left.\right)}{\left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right)} ; \frac{- 2}{\left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right)}\)
b)
\(2 x - 4 = 2 \left(\right. x - 2 \left.\right) , 6 - 3 x = 3 \left(\right. 2 - x \left.\right) = - 3 \left(\right. x - 2 \left.\right)\)
Mẫu chung:
\(6 \left(\right. x + 2 \left.\right) \left(\right. x - 2 \left.\right)\) \(\frac{60 \left(\right. x - 2 \left.\right)}{6 \left(\right. x + 2 \left.\right) \left(\right. x - 2 \left.\right)} ; \frac{15 \left(\right. x + 2 \left.\right)}{6 \left(\right. x + 2 \left.\right) \left(\right. x - 2 \left.\right)} ; - \frac{2 \left(\right. x + 2 \left.\right)}{6 \left(\right. x + 2 \left.\right) \left(\right. x - 2 \left.\right)}\)
\(\frac{x+2}{x+1}=\frac{x}{x+1}+\frac{2}{x+1}\)
\(\frac{2x-3}{x-1}=\frac{2x}{x-1}+\frac{-3}{x-1}\)
\(\frac{x^2-3x+5}{x+1}=\frac{x^2}{x+1}+\frac{-3x+5}{x+1}\)