2x+5=x-1
xy=-31
(x-2)(y+1)=23
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\(a,2x+5=x-1+\left(x+3\right)-\left(x-3\right)\)
\(2x+5=x-1+x+3-x+3\)
\(2x-x-x+x=-1+3+3-5\)
\(3x=0\Leftrightarrow x=0\)
a, \(xy+2x+2y+y^2\)
\(=\left(xy+2x\right)+\left(y^2+2y\right)\)
\(=x.\left(y+2\right)+y.\left(y+2\right)\)
\(=\left(y+2\right).\left(x+y\right)\)
b, Sửa đề:
\(-6x^2-9xy+15y^2\)
\(=-\left(6x^2+9xy-15y^2\right)\)
\(=-\left(6x^2-6xy+15xy-15y^2\right)\)
\(=-\left[\left(6x^2-6xy\right)+\left(15xy-15y^2\right)\right]\)
\(=-\left[6x.\left(x-y\right)+15y.\left(x-y\right)\right]\)
\(=-\left[\left(x-y\right).\left(6x+15y\right)\right]\)
c, \(2x\left(x-3\right)+y\left(x-3\right)+\left(3-x\right)\)
\(=2x\left(x-3\right)+y\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right).\left(2x+y-1\right)\)
Chúc bạn học tốt!!!
Bài 1:
a) \(4x\left(3x-1\right)-2\left(3x+1\right)-\left(x+3\right)\)
\(=12x^2-4x-6x-2-x-3\)
\(=12x^2-11x-5\)
b) \(=\left(-2x^2-1xy+2y^2\right)\left(-1x^2y\right)\)
\(=\left[\left(-1x^2y\right)\left(-2x^2\right)\right]-\left[\left(-1x^2y\right).1xy\right]+\left[\left(-1x^2y\right).2y^2\right]\)
\(=\left(2x^4y\right)-\left(-1x^3y^2\right)+\left(-2x^2y^3\right)\)
\(=2x^4y+1x^3y^2-2x^2y^3\)
c) \(4x\left(3x^2-x\right)-\left(2x+3\right)^2\left(6x^2-3x+1\right)\)
\(=\left(4x.3x^2\right)-\left(4x.x\right)-\left[\left(2x\right)^2+2.2x.3+3^2\right]\left(6x^2-3x+1\right)\)
\(=12x^3-4x^2-\left(4x^2+12x+9\right)\left(6x^2-3x+1\right)\)
\(=12x^3-4x^2-\left[4x^2\left(6x^2-3x+1\right)+12x\left(6x^2-3x+1\right)+9\left(6x^2-3x+1\right)\right]\)
\(=12x^3-4x^2-\left[\left(24x^4-12x^3+4x^2\right)+\left(72x^3-36x^2+12x\right)+\left(36x^2-27x+9\right)\right]\)
\(=12x^3-4x^2-24x^4+12x^3-4x^2-72x^3+36x^2-12x-36x^2+27x-9\)
\(=-48x^3-8x^2-24x^4+15x-9\)
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
=-10x-50
4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16\)
=-6x+25
5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)
\(=y^2-6y+9-y^2+6y-9\)
=0
6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
=12x+18
a) 2x + 5 = x - 1
2x - x = -1 - 5
x = -6
~Học tốt
#Huyền Anh :)
a) 2x + 5 = x - 1
2x - x = -1 - 5
\(\Rightarrow\) x = 6