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\(1.x^3+2x+x^2=x\left(x^2+x+2\right)\)
\(2.2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
\(3.-3x^3-5x^2+8x=-3x^3+3x^2-8x^2+8x\)
\(=-3x^2\left(x-1\right)-8x\left(x-1\right)=\left(3x^2+8x\right)\left(1-x\right)\)
\(=x\left(3x+8\right)\left(1-x\right)\)
\(4.x^2+4x-5=x^2-x+5x-5=\left(x-1\right)\left(x+5\right)\)
\(5.6x^2-3x-3=6x^2-6x+3x-3=3\left(x-1\right)\left(2x+1\right)\)
\(6.3x^2-2x-5=3x^2+3x-5x-5=\left(x+1\right)\left(3x-5\right)\)
\(8.x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)\(=\left(x+2y\right)\left(x-y-2\right)\)
\(9.x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
\(10.x^2-y^2+6x+9=\left(x+3-y\right)\left(x+3+y\right)\)
g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)
\(\Leftrightarrow-5\left(4x+3\right)=0\)
\(\Leftrightarrow4x+3=0\)
\(\Leftrightarrow4x=-3\)
\(\Leftrightarrow x=\frac{-3}{4}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)
h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)
\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)
\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)
\(\Leftrightarrow-9x+2x-3-10x=30\)
\(\Leftrightarrow-17x-3=30\)
\(\Leftrightarrow-17x=33\)
\(\Leftrightarrow x=\frac{-33}{17}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)
Bài 1:
a: \(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)-\left(4x^2+12x+9\right)-5+20x\)
\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)
\(=4x^2-8x-16-5+20x-4x^2-12x-9\)
\(=-30\)
b: \(B=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-\left(x^3+4x^2-246x\right)-175\)
\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)
\(=-175\)
d: \(D=25x^2-20x+4-36x^2-12x-1+11\left(x^2-4\right)-48+32x\)
\(=-11x^2-32x+3-48+32x+11x^2-44\)
=-89
a: \(=\dfrac{2x^3-6x^2+10x-2x^2+6x-10}{x^2-3x+5}=2x-2\)
b: \(=\dfrac{x^3-2x^2+7x-2x^2+4x-14}{x^2-2x+7}=x-2\)
c: \(=2x^4+4x^3-4x^3-8x^2+6x^2+12x+44x+88\)
\(=2x^4-2x^2+56x+88\)