phan tich thanh nhan tu 3x-7canx+4
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\(x^2-3x+4\)
\(=x^2+x-4x+4\)
\(=\left(x^2+x\right)-\left(4x+4\right)\)
\(=x\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x-4\right)\left(x+1\right)\).
x^3 + 3x -4 = x^3 - 1 + 3x - 3 = (x-1)(x^2-x+1) + 3(x-1) = (x-1)(x^2-x+1+3) = (x-1)(x^2-x+4)
nhớ cho mik nha
\(1-3x-x^3+3x^2\)\(=\left(1-x^3\right)+\left(3x^2-3x\right)\)
\(=\left(1-x\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-x^2-x-1\right)=\left(x-1\right)\left(2x-x^2-1\right)\)
Câu a :
\(\dfrac{3x^2-12x+12}{x^4-8x}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
Câu b :
\(\dfrac{7x^2+14x+7}{3x^2+3x}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
\(x^8+3x^4+1=\left(x^8+\frac{2.3x^4}{2}+\frac{9}{4}\right)-\frac{5}{4}\)
\(=\left(x^4+\frac{3}{2}\right)^2-\frac{5}{4}=\left(x^4+\frac{3-\sqrt{5}}{2}\right)\left(x^4+\frac{3+\sqrt{5}}{2}\right)\)
\(3x-7\sqrt{x}+4=3x-4\sqrt{x}-3\sqrt{x}+4=\sqrt{x}\left(3\sqrt{x}-4\right)-\left(3\sqrt{x}-4\right)=\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)\)
\(3x-7\sqrt{x}+4=3x-3\sqrt{x}-4\sqrt{x}+4\)
\(=3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)\)
\(=\left(3\sqrt{x}-4\right)\cdot\left(\sqrt{x}-1\right)\)