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A=\(\frac{x^3-7x+6}{x^3+5x^2-2x-24}\)=\(\frac{x^3-2x^2+2x^2-4x-3x+6}{x^3-2x^2+7x^2-14x+12x-24}\)=\(\frac{x^2\left(x-2\right)+2x\left(x-2\right)-3\left(x-2\right)}{x^2\left(x-2\right)+7x\left(x-2\right)+12\left(x-2\right)}\)=\(\frac{\left(x-2\right)\left(x^2+2x-3\right)}{\left(x-2\right)\left(x^2+7x+12^{^{^{^{^{^{^{^{^{ }}}}}}}}}\right)}\)=\(\frac{\left(x-2\right)\left(x^2-x+3x-3\right)}{\left(x-2\right)\left(x^2+3x+4x+12\right)}\)=\(\frac{\left(x-2\right)\left(x-1\right)\left(x+3\right)}{\left(x-2\right)\left(x+4\right)\left(x+3\right)}\)=\(\frac{x-1}{x+4}\)
\(\frac{x^2+2x+1}{5x^3+5x^2}=\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}=\frac{x+1}{5x^2};\)
b, \(\frac{2x^2+2x}{x+1}=\frac{2x\left(x+1\right)}{x+1}=2x\)
\(a)\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(3x-1)(x-3)^2}(ĐK:x\ne3,x\ne\frac{1}{3})\)
\(=\frac{2x+5}{3x-1}\)
Còn bài b bạn tự làm nhé
Điều kiện: \(x\ne\left\{-1;-2;-5\right\}\)
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x^2-4\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+5\right)}=\frac{x-2}{x+5}\)
Điều kiện: \(x\ne\left\{3;\frac{1}{3}\right\}\)
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)
\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
\(=\frac{2x^2-x-15}{3x^2-10x+3}=\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)
\(=\frac{\left(2x+5\right)\left(x-3\right)}{\left(3x-1\right)\left(x-3\right)}=\frac{2x+5}{3x-1}\)
\(\frac{x^2-5x+6}{x^2-2x}=\frac{x^2-2x-3x+6}{x.\left(x-2\right)}=\frac{x.\left(x-2\right)-3.\left(x-2\right)}{x.\left(x-2\right)}\)
\(=\frac{\left(x-3\right).\left(x-2\right)}{x.\left(x-2\right)}=\frac{x-3}{x}\)
Câu 1:
\(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{\left(x-7\right)\left(x-3\right)}{\left(x-7\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
\(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}=\dfrac{2x^2-6x+5x-15}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{\left(2x+5\right)\left(x-3\right)}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
Do đó: \(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)
bài này dễ lắm, mk làm 1 câu là bn làm câu sau dc hà
bn thấy tử số có 2x chung, vạy tử là; 2x2 +2x = 2x(x+1)
mẫu số là hằng đẳng thức (x+1)2 = x2 +2x+1
vậy ta có: tử/mẫu = 2x(x+1)/(x+1)2 = 2x/x+1
\(A=\dfrac{x^3-7x+6}{x^3+5x^2-2x-24}\)
\(=\dfrac{x^3-2x^2+2x^2-4x-3x+6}{x^3+4x^2+x^2+4x-6x-24}\)
\(=\dfrac{x^2\left(x-2\right)+2x\left(x-2\right)-3\left(x-2\right)}{x^2\left(x+4\right)+x\left(x+4\right)-6\left(x+4\right)}\)
\(=\dfrac{\left(x^2+2x-3\right)\left(x-2\right)}{\left(x^2+x-6\right)\left(x+4\right)}\)
\(=\dfrac{\left(x^2+3x-x-3\right)\left(x-2\right)}{\left(x^2+3x-2x-6\right)\left(x+4\right)}\)
\(=\dfrac{\left[x\left(x+3\right)-\left(x+3\right)\right]\left(x-2\right)}{\left[x\left(x+3\right)-2\left(x+3\right)\right]\left(x+4\right)}\)
\(=\dfrac{\left(x-1\right)\left(x-2\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)\left(x+4\right)}\)
\(=\dfrac{x-1}{x+4}\)