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\(\frac{x+2}{x+1}=\frac{x}{x+1}+\frac{2}{x+1}\)
\(\frac{2x-3}{x-1}=\frac{2x}{x-1}+\frac{-3}{x-1}\)
\(\frac{x^2-3x+5}{x+1}=\frac{x^2}{x+1}+\frac{-3x+5}{x+1}\)
\(B=\left(\dfrac{1}{x-2}-\dfrac{2x}{4}-\dfrac{x^2}{1}+\dfrac{1}{2+x}\right)\).\(\dfrac{2}{x-1}\)
đề có phải như này không , đăng đề bài phân số gõ latex , nếu k dễ nhầm lẫn lắm
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\(x^3+8x^2+17x+10\)
\(=x^3+2x^2+x^2+5x^2+10x+5x+2x+10\)
\(=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(5x^2+5x\right)+\left(10x+10\right)\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+5x\left(x+1\right)+10\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+5x+10\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
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)
\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)
Nhớ ghi dấu ngoặc tránh giải sai.
\(a.\) \(\frac{x+4}{2x+6}+\frac{3}{x^2-9}\)
Ta có:
\(2x+6=2\left(x+3\right)\)
\(x^2-9=\left(x-3\right)\left(x+3\right)\)
nên \(MTC:\) \(2\left(x-3\right)\left(x+3\right)\)
Do đó: \(\frac{x+4}{2x+6}+\frac{3}{x^2-9}=\frac{x+4}{2\left(x+3\right)}+\frac{3}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+4\right)\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}+\frac{2.3}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2+x-12+6}{2\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2-2x+3x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x\left(x-2\right)+3\left(x-2\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-2\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{x-2}{2\left(x-3\right)}\)
tick mình đi mình giải cho nha