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a)\(\left(x+1\right)\left(x+3\right)-x\left(x-2\right)=7\)
\(x\left(x+3\right)+x+3-x^2+2x=7\)
\(x^2+3x+x+3-x^2+2x=7\)
\(6x+3=7\)
\(6x=4\)
\(x=\frac{4}{6}=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
b) \(2x\left(3x+5\right)-x\left(6x-1\right)=33\)
\(6x^2+10x-6x^2-x=33\)
\(9x=33\)
\(x=\frac{33}{9}\)
Vậy \(x=\frac{33}{9}\)
a) (x-1)*(x+2)-(x-3)*(-x+4)=19
\(\Leftrightarrow x^2+2x-x-2-\left(-x^2+4x+3-12\right)=19\)
\(\Leftrightarrow x^2+2x-x-2+x^2-4x-3+12=19\)
\(\Leftrightarrow2x^2-3x+7-19=0\)
\(\Leftrightarrow2x^2-3x-12=0\)
Đề sai??
b) (2x -1)*(3x+5)-(6x-1)*(6x+1)=(-17)
\(\Leftrightarrow6x^2+10x-3x-5-\left(36x^2+6x-6x-1\right)=-17\)
\(\Leftrightarrow6x^2+10x-3x-5-36x^2-6x+6x+1=-17\)
\(\Leftrightarrow-30x^2+7x-4+17=0\)
\(\Leftrightarrow-30x^2+7x+13=0\)
???
Bài 6:
c: \(9x^2+6x+1=\left(3x+1\right)^2\)
d: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)
e: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
a)xy(x2+2y)=xy.x2+xy.2y
=x3y+2xy2
b)-4(6x2-xy)=-4.6x2+4.xy
=-24x2+4xy
c)4x[x2+6x-1/2]
=4x.x2+4x.6x-4x.1/2
=4x3+24x2-2x
a) \(\left(x+1\right)\left(x+2\right)\left(x-3\right)=\left(x^2+2x+x+2\right)\left(x-3\right)\)
\(=\left(x^2+3x+2\right)\left(x-3\right)\)
\(=x\left(x^2+3x+2\right)-3\left(x^2+3x+2\right)\)
\(=x^3+3x^2+2x-3x^2-9x-6\)
\(=x^3-7x-6\)
b) \(\left(2x-1\right)\left(x+2\right)\left(x+3\right)=\left(2x-1\right)\left(x^2+3x+2x+6\right)\)
\(=\left(2x-1\right)\left(x^2+5x+6\right)\)
\(=2x\left(x^2+5x+6\right)-\left(x^2+5x+6\right)\)
\(=2x^3+10x^2+12x-x^2-5x-6\)
\(=2x^3+9x^2+7x-6\)