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1) \(x^2-6x+3\)
\(=x^2-6x+9-6\)
\(=\left(x-3\right)^2-6\)
\(=\left(x-3+\sqrt{6}\right)\left(x-3-\sqrt{6}\right)\)
2) \(2m^2+10m+8\)
\(=2m^2+2m+8m+8\)
\(=2m\left(m+1\right)+8\left(m+1\right)\)
\(=\left(2m+8\right)\left(m+1\right)\)
\(=2\left(m+4\right)\left(m+1\right)\)
3) \(9x^2+6x-8\)
\(=\left(9x^2+6x+1\right)-9\)
\(=\left(3x+1\right)^2-9\)
\(=\left(3x+4\right)\left(3x-2\right)\)
4) \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x^2-2x+7x-14\right)\)
\(=x\left[x\left(x-2\right)+7\left(x-2\right)\right]\)
\(=x\left(x+7\right)\left(x-2\right)\)
1) Ta có : 2x2 + 3x - 5
= 2x2 - 2x + 5x - 5
= 2x(x - 1) + 5(x - 1)
= (x - 1) (2x + 5)
3) x2 + x - 6
= x2 + 2x - 3x - 6
= x(x + 2) - (3x + 6)
= x(x + 2) - 3(x + 2)
= (x - 3)(x + 2)
a,\(4x\left(2x+3\right)-x\left(8x-1\right)=5\left(x+2\right)\)
\(< =>8x^2+12x-8x^2+x=5x+10\)
\(< =>13x=5x+10< =>8x=10\)
\(< =>x=\frac{10}{8}=\frac{5}{4}\)
b, \(\left(3x-5\right)\left(3x+5\right)-x\left(9x-1\right)=4\)
\(< =>9x^2-25-9x^2+x=4\)
\(< =>x=4+29=33\)
c,\(3-4x\left(25-2x\right)=8x^2+x-300\)
\(< =>3-100x+8x^2=8x^2+x-300\)
\(< =>x+100x=3+300\)
\(< =>101x=303< =>x=\frac{303}{101}=3\)
d,\(2\left(1-\frac{3x}{5}\right)-\frac{2+3x}{10}=7-\frac{3\left(2x+1\right)}{4}\)
\(< =>2-\frac{6x}{5}-\frac{2+3x}{10}=7-\frac{6x+3}{4}\)
\(< =>-\frac{24x}{20}-\frac{4+6x}{20}+\frac{30x+15}{20}=5\)
\(< =>\frac{30x-6x-24x+15-4}{20}=5\)
\(< =>\frac{11}{5}=5< =>11=25\)(vo li)